Game Pricing and Double Sequence of Random Variables

Journal of Modern Mathematics Frontier Volume 2 Issue 2, June 2013 www.sjmmf.org

33

Game Pricing and Double Sequence of Random Variables Yukio Hirashita

Department of International Liberal Studies, Chukyo University Nagoya, 466-8666, Japan [email protected]

Abstract

In this paper, we study a random payoff with positive or plus infinite expectation and determine the optimal proportion of investment for maximizing the limit expectation of growth rate per attempt. With this objective, we introduce a new pricing method in which the price is different from that obtained by the Black-Scholes formula for a European option. We will price the St. Petersburg game nicely.

Keywords

Proportion of Investment; Pricing of Random Payoff; Black-Scholes Formula

Introduction

The portfolio pricing equation (Luenberger 1998) is useful for determining prices of securities only if the optimal portfolio has been already known. In this paper, we determine both the price and optimal proportion of investment for a random payoff.

The determination of the utility function is more experimental than mathematical. With an additional reason that exp(log )X X= frequently provides

[ ] exp( [log ])E X E X≠ , the log utility approach (Luenberger 1998) is not preferable.

The investor should repeatedly invest a fixed proportion of his or her own current capital without borrowing. As a rule, if the investor invests 1 dollar, then he or she receives ( )a x dollars (including the invested 1 dollar) with a cumulative distribution function ( )F x defined on an interval I . For simplicity, we omit the currency notation. Let 0M > be the investor’s capital, 0u > the price of the random payoff ( ( )a x , ( ))F x , and 0 1t≤ ≤ the proportion of investment. Then, after one attempt, he or she has capital of ( ) (1 )Mta x u M t/ + − if x occurs. It should be noted that the reserved part (1 )M t− does not include the interest, which is the custom, for example, in

foreign exchange accounts.

Let 0nM > be the capital after n attempts. In general, growth rate implies 1 1n nM M+ / − or 1log( )n nM M+ / after one attempt. However, for the purposes of succinctness in this paper 1n nM M+ / is used to define the growth rate. In this context, the growth rate per

attempt is defined as ( )1

0n

nM M/ .

Without dealing with ( )1

0n

nM M/ directly, this paper defines a double sequence of random variables N nX , with respect to the bounded step functions ( )Nf x such that limN→+∞ ( ) ( )Nf x a x= . It is shown that the finite limit lim [ ]n N n

NE X→+∞ ,

→+∞ exists if, and only if, the

random payoff is effective. In this case, the equalities

lim [ ]n N nN

E X→+∞ ,→+∞

( )uG t= exp( log( ( ) 1) ( ))I

a x t u t dF x:= / − +∫

= exp( [log( ( ) 1)])E a x t u t/ − + and lim [ ] 0n N nN

V X→+∞ ,→+∞

=

are obtained. These equalities again support the well-known assertion that although in principle an investor may choose any utility function, a repetitive situation tends to hammer the utility into one that is close to the logarithm (Kelly 1956, Luenberger 1998).

We study the optimal proportion of investment ut , for the price 0u > in order to maximize the limit expectation of growth rate per attempt. In order to determine the price of the random payoff, we require a risk-free interest rate, 0r > , for a particular period. The equation ( ) 1u uG t r= + (if r is simple) or

( ) ru uG t e= (if r is continuously compounded) is used

to determine the price of a random payoff. If ( ) 0a x ≥ for each x I∈ , then the existence and uniqueness of the price are guaranteed by the fact that ( )u uG t is continuous and strictly decreases from +∞ to 1 with respect to 0 ( ) ( )

Iu E a x dF x< < := ∫ (Theorem 4.1). In

this context, the price of the St. Petersburg game

www.sjmmf.org Journal of Modern Mathematics Frontier Volume 2 Issue 2, June 2013

34

(Bernoulli 1954) is determined to be 5 0815. if the risk-free interest rate is 4% (Example 6.4). On the other hand, the Black-Scholes formula is deduced from the equation rE u e/ = , where E is the expectation of a European option (Example 6.6).

Any random variable (X : Ω, )P R→ reduces a random payoff (x, ( ))F x , where ( )F x (P ω:= ∈Ω |

( ) )X xω ≤ .

Optimal Proportion of Investment

Assume that the payoff function ( )a x is measurable with the distribution function ( )F x defined on an interval (I ⊆ −∞ , )+∞ . Set inf ( )x I a xξ ∈:= . We also assume that ξ > −∞ and ξ is the essential infimum of

( )a x , that is, ( )

( )a x

dF xξ ε< +∫ 0> for each 0ε > . Further,

assume that ( )a x is not a constant function (a.e.), that

is, ( )

( ) 1a x

dF xξ δ< +

<∫ for some 0δ > .

We use the following notation with the Lebesgue-Stieltjes integral.

1( ) ( ) ( )( )I I

E a x dF x H dF xa x

:= , := ,∫ ∫

1 ( ) (1)( )I

H dF xa xξ ξ

:= .−∫

In this paper, we assume that 0E > . If ( )

( ) 0a x

dF xξ=

>∫ ,

we define Hξ = +∞ and 1 Hξ/ 0= .

Since ( )a x is not constant, we have Eξ < , 0Hξ > , and 1 Hξ/ < +∞ . From the relation

211 ( ) ( )

( )1( ( ) ) ( ) ( ) ( )

( )

I

I I

a x dF xa x

a x dF x dF x E Ha x ξ

ξξ

ξ ξξ

= − × −

< − × = − ,−

∫

∫ ∫

we have 1 H Eξξ + / < . In particular, if 0ξ = , 0 1 H E≤ / < . If 0ξ > , then using 1 1 ( )a xξ/ ≥ / and

1 ( ) (1 ( ))a x a x= × / , we have 1 H Eξ < / < .

For price 0u > , let [0ut ∈ , 1] be the optimal proportion of investment. The precise definition of the term "optimal" and its significance are shown beneath Lemma 5.1. Here, we present certain properties of ut in order to explain the approximate outline of the paper.

(a) If u E> , 0ut = .

Assume that u E> and (0t∈ , 1] , then the expectation

of payoffs,

( )I

Mt a x u/∫ ( ) (1 ) (1 )dF x M t M M E u t+ − = − − / ,

is less than M . More precisely, using Jensen’s inequality, we have ( ) 1 (1 ) 1 (0)u uG t E u t G< − − / < = for each (0t∈ , 1] . Therefore, 0ut = .

In the proof of Theorem 5.1, we will show that: [ ) if

0 (2)ifu

E Eu t

Eφ,+∞ , < +∞,

| = = , = +∞.

(b) If 0ξ > and 0 u ξ< ≤ , then 1ut = .

From 0 ( )u a xξ< ≤ ≤ and [0t∈ , 1) , after one attempt, we have ( ) (1 )Mta x u M t/ + − ( )Ma x u= /

(1 )( ( ) 1)M t a x u− − / − ( )Ma x u≤ / for each x I∈ . This implies that ( ) (1)u uG t G< for each [0t∈ , 1), that is,

1ut = .

Accordingly, in the proof of Theorem 5.1, we will also show that

(0 1 ] if 0 or 0 and 1 (3)

if 0 or 0 andu

H Hu t

Hξ ξ

φ ξ ξ, / , > , = < +∞,

| = = , < , = = +∞, which yields a maximum price of 1 H/ at which all the capital should be repeatedly invested.

(c) If max(0 ) uξ, < , ( )ut u u ξ≤ / − .

If ( )t u u ξ> / − , 0u u tξ− − / > . Therefore, the negative result ( )Mta x u/ (1 ) 0M t+ − < occurs with a positive

probability ( ) ( )

( ) 0a x u u t

dF xξ ξ ξ≤ < + − − /

> .∫ This contradicts

the concept of continual investment without borrowing.

In the proof of Theorem 5.1, the condition for ( )ut u u ξ= / − is shown such that:

(0 1 ] if 0 and 1 0 (4)

if 0 or 1 0u

H Huu tHu

ξ ξ

ξ

ξ ξ ξφ ξ ξξ

, + / , ≤ + / > ,| = = , > + / ≤ .−

(d) Theorem 5.1 also shows that 0ut ∉ , 1, ( )u u ξ/ − if and only if 1 H/ u E< < (if 0ξ ≥ ) or max(0 ,

1 )H u Eξξ + / < < (if 0ξ < ). In this case, ut can be uniquely determined by the property:

( ) ( ) 0 (5)( )I

u u

a x u dF xa x t ut u

−= .

− +∫

Here, we offer 5 graphics to facilitate reader’s understanding. Recall the definitions:

inf ( )x I a xξ ∈:= , ( ) ( )I

E a x dF x:= ,∫ 1( ) ( )a xI

H dF x:= ,∫ and 1

( ) ( )a xIH dF xξ ξ−:= .∫

Case 1: 0ξ > .


35

exp( log ( ) ( ))

IM a x dF x:= ∫ and

( ) exp( log( ( ) 1) ( )).u IG t a x t u t dF x:= / − +∫

Case 2: 0ξ = and H = +∞.

( ) 0( )

a xK dF x

>:= .∫

Case 3: 0ξ = and H < +∞.

Case 4: 0ξ < and 1 0Hξξ + / > .

exp( log( ( ) ) ( ))

IM a x dF xξ ξ:= − .∫

Case 5: 0ξ < and 1 0Hξξ + / ≤ .

0lim uu

t uη +→:= / and exp( log( ( ) 1) ( ))

IM a x dF xη η:= + .∫

Pre-Optimal Proportion

We denote the integral ( ( ) ) ( ( ) ) ( )I

a x a x z z dF xβ β β− / − +∫by ( )w zβ , in which z and β are complex variables.

Lemma 3.1. The function ( )w zβ is holomorphic with respect to two complex variables z t si:= + and u hiβ := + such that, (a) max(ε , ) u Lξ < < < +∞,

(b) 6 2(32( 1) )h L Rε| |< / + ,

(c) z R| |< and z s ε∉ ≤ ∩ t ε≤ or ( ) t u u ξ ε≥ / − − ,

where 0 min(1 2ε< < / , (2( )))u u ξ/ − , max(2 , (u u/

)) Rξ− < < +∞ , 1i := − , Im( )z s= and Im( ) hβ = .

Proof. We obtain certain operator exchange properties such as


36

( )( ) ( )( )I

a xw z dF xt t a x z zβ

ββ β

∂ ∂ −= ∂ ∂ − + ∫

by proving that the related integrands are bounded. Because ( )( ) ( ( )a x a x zβ− / )zβ β− + satisfies the Cauchy-Riemann equations, ( )w zβ is shown to be holomorphic due to Hartogs’s theorem.

It should be noted that the condition (a) above leads to 0β ≠ , and if ( )a x β≠ , then we have

( )1

1

( ) 1( ) a x

a xa x z z z

β

ββ β

−

−= .

− + −

The partial derivatives of ( )( ) ( ( )a x a x zβ− / )zβ β− + with respect to t , s , u , and h are 0 , 0 , 1− and i− , respectively, at x | ( ) a x β= . In the following four cases, we assume that ( )a x β≠ .

In this proof, we will frequently use the inequality 1 (1 ) 2z z/ − ≤ / if 2z| |≥ .

<Case 1> ( ) 8( 1)a x L ε| |≥ + / .

As a result of the conditions, we have ( ) ( ) ( ) ( 1)a x u a x L a x L/ > / > / + 8 ε≥ / 16> , which

leads to 1 (1 ( ) ) 2 ( ( ) )) 2( 1) ( )a x u a x u L a x/ − / ≤ / / < + /

4ε≤ / . On the other hand, the inequality ( )

1( ) 8 16a xLa x β ε+/ > ≥ / > leads to 1 (1 ( ) )a x β/ − /

2 ( ( ) ) 2( 1) ( ) 4a x L a xβ ε≤ / / < + / ≤ / , where u hβ ≤ +

1L< + . Moreover, from ( )a xξ ≤ we have 1 ( ) ( )a x u u uξ− / ≤ − / . If 1 ( ) 0a x u− / > then ( ) 1 (1 ( ) )u u a x uξ/ − ≤ / − / , which

leads to 1 (1 ( ) ))z a x u ε− / − / > due to (c). If

1 ( ) 0a x u− / < then 1 (1 ( ) ) 0a x u/ − / < , which leads to

1 (1 ( ) ))z a x u ε− / − / > due to (c). If 1 ( ) 0a x u− / = then

( ) 8( 1)L u a x L ε> =| |≥ + / , which is a contradiction. Therefore, we have

( ) ( ) ( ) ( )

1 1 1 121 1 1 1a x a x a x a x

u u

z zβ β

ε− = − + − > ,

− − − −

which establishes ( )

11

1 2

a xzβ

ε−

< .−

Moreover, using ( ) (1 ( ) ) 2a x a x β β/ − / ≤ ,

1 (1 ( ) ) 4a x β ε/ − / < / , and 1 β/ 1 ε< / , we have

( ) ( )( )

2 22( )2 11

( ) 2

1a xa x

a x

zβ

βεβ

−

< .− −

<Case 2> ( ) 8( 1)a x L ε| |< + / and ( ) 1a x Rβ ε/ − ≤ / .

Since ( )( ) 1a x zβ ε/ − ≤ , we have

( ) 1 1a x z zβ ε/ − + ≥ − . Therefore, ( )

( )

1 211

a xR

a x Rz z

εβ

β

εε

−≤ < .

−− +

Moreover, using 1 1β ε/ < / and 1 (1 ) 2ε/ − < , we have

( ) ( ) ( )( )

2 2 3( )22( )2 11

( ) ( ) 32( 1)

11a x

a xa x

a x a x L

z zzβ

ββεββ

−

+= < .

− +− −

< Case 3 > ( ) 8( 1)a x L ε| |< + / , ( ) 1a x Rβ ε/ − > / , and

( ) 1a x u/ − (2 )Rε> / .

From 1 1β ε/ < / , 1 1u ε/ < / , and condition (b)

mentioned above, we have

( ) ( ) ( ) ( )

1 1 ( )21 1 (1 )(1 )a x a x a x a x

u u

a x hiuβ β

εβ

− = < .− − − −

Therefore, as ( ) 1a x u/ ≠ and 1 (1 ( ) )z a x u ε− / − / > , we obtain

( )1

1

1 2

a xzβ

ε−

< .−

Moreover, this implies that

( ) ( )( )

2

2 72( )2 11

( ) 32( 1)

1a xa x

a x L R

zβ

βεβ

−

+< .

− −

< Case 4 > ( ) 8( 1)a x L ε| |< + / , ( ) 1a x Rβ ε/ − > / , and

( ) 1 (2 )a x u Rε/ − ≤ / . This case is void as shown below:

3

2

( )( ) ( ) ( ) ( )1 12 4

ha xa x a x a x a xR u u u Rε ε

β β β< − − − ≤ − = < ,

which leads to the contradiction 24 2 1 4R ε< < < / .

From the inequalities mentioned above, the four integrands on the right-hand side of the following equalities are bounded. Therefore, the Cauchy-Riemann equations for ( )w zβ hold:

( )( )

21

1

1( ) ( ) (6)a x

Iw z dF x

t zβ

β

−

∂ −= ,

∂−

∫


37

( )( )

21

1

( ) ( ) ( )a x

I

iw z dF x i w zs tz

β

β β

−

∂ − ∂= = ,

∂ ∂−

∫

( )2( )2

( )( ) ( )1I a x

a xw z dF xu z z

β

ββ

∂ −= ,

∂ − +∫

( )2( )2

( )( ) ( ) ( )1I a x

a x iw z dF x i w zh uz z

β β

ββ

∂ − ∂= = .

∂ ∂− +∫

Moreover, the integrand of ( )w zβ is bounded, which insures that ( )w zβ is continuous with respect to β and z , respectively. Thus, ( )w zβ is holomorphic with respect to (z, )β . Henceforth, in this Section, we assume that max(0 , ) u Eξ < < and 0 t< ( )u u ξ< / − . It should be noted that

( ) ( )( ( ) ) 0a x t ut u t ut u u u u tξ ξ ξ− + ≥ − + = − / − − > for each x I∈ .

Specially, when ( )F x is a finite step function, we can write ( jX a= , 1 2)j j np = , ,..., with 1 min j ja aξ = = , 1 0p > ,

10n

j jjE a p

== >∑ ,

1

nj jj

H p a=

= /∑ , Hξ = +∞ ,

1( ) ( ) ( )n

u j j jjw t a u p a t ut u

== − / − + ,∑ and so on.

Lemma 3.2. ( )uw t is strictly decreasing with respect to t .

Proof. According to Lemma 3.1, we have 2

( )( ) ( ) 0( )u I

a x uw t dF xt a x t ut u

∂ −= − < . ∂ − + ∫

Lemma 3.3. 0

lim ( ) 1utw t E u+→

= / − .

Proof. Since ( ( ) ) ( ( ) )a x u a x t ut u− / − + is decreasing with respect to t , using Lebesgue (monotone convergence) theorem, we obtain

0 0

( ) ( )lim ( ) lim ( ) ( )( )u I It t

a x u a x uw t dF x dF xa x t ut u u+ +→ →

− −= =

− +∫ ∫

1Eu

= − .

Lemma 3.4. ( ( ))

lim ( ) (1 ) ( 1 )ut u uw t u H H uξ ξξ

ξ ξ−→ / −= − / + / − .

Proof. Using the same principle as above, we obtain

( ( ))

( )lim ( ) ( )( )

(1 ( ) ) (1 ) ( 1 )

u It u u

u a x uw t dF xu a x

u u H u H H uu

ξ

ξ ξ ξ

ξξ

ξ ξ ξ ξ

−→ / −

− −=

−−

= − − = − / + / − .

∫

If 0ξ = and H = +∞, then 1

lim ( ) 1 1 ( ) ( )utw t u a x dF x−→

= − /∫

= −∞.

From the above lemmas, if max(0 1 )H u Eξξ, + / < < ,

then 0

lim ( ) 0utw t+→

> and ( ( ))lim ( ) 0ut u u

w tξ −→ / −

< . Thus,

the equation ( ) 0uw t = has the only solution (0ut ∈ , ( ))u u ξ/ − , and we refer to it as pre-optimal proportion.

Note that, due to Lemma 3.1 and the inverse mapping theorem, ut is analytic with respect to u .

Lemma 3.5. If 0ξ > , 1 1H Hξξ + / < / .

Proof. If Hξ = +∞, the assertion 1H ξ< / is obvious. Since ( )uw t is strictly decreasing and 1 ( )u u ξ< / − ,

( ( ))(1) limu t u u

wξ −→ / −

> ( )uw t , that is,

( ) ( ) 1 (1 ) ( 1 )

( )I

a x u dF x Hu u H H ua x u u ξ ξξ ξ−

= − > − / + / −− +∫

for each u Eξ < < . If 1 1H Hξξ/ ≤ + / , then selecting 1u H Eξξ= + / < leads to the contradiction that

0 1 (1 ) ( 1 ) 0Hu u H H uξ ξξ ξ≥ − > − / + / − = .

Lemma 3.6. If 0ξ ≥ , then ut is strictly decreasing with respect to ( 1u Hξξ∈ + / , )E .

Proof. From Lemma 3.1, ut is analytic. Using 2

2

( ) ( ( ) )( )( ) ( ( ) )

ud tdu

u u u u

a x a x ua x uu a x u u a x u ut t t t

+ −∂ −= − ,

∂ − + − +

( ) 0u udw dut / = , and ( ) 0a x ξ≥ ≥ , we obtain

2

2

2

( )( ( ) )

( ( ) )( ( ) )

( )0

( )u u

u u

a xa x u uI t tu

a x ua x u uI t t

dF xdtdu dF x

− +

−

− +

−= < .∫∫

Lemma 3.7. lim 0uu E t−→= .

Proof. < Case 1 > . Assume that E = +∞ . From lim ( ) 1u u u ξ→+∞ / − = , for any 0 1 3ε< < / , there exists N such that 1 ( ) 1u uε ξ ε− < / − < + for each u N> . This implies that

( )

( )

0 if ( )( ) 1 1 1 if ( )

( )

1 1 1 if ( )

uuuu a x

ua x u

a x ua x u a x u

a x t ut u tt

a x utt

ξε

ε

−−

−

, = ,

− = ≤ < , < ,− + −−

≤ ≤ , > , +

for 1 2tε ε≤ ≤ − , u N> . Therefore, by Lebesgue (dominated-convergence) theorem, we obtain

1 1lim ( ) ( ) 01 1u Iu

w t dF xt t→+∞

−= = < .

− −∫

In particular, lim ( ) 1 (1 ) 0u uw ε ε→+∞ = − / − < . Therefore, there exists 0M > such that ( ) 1 (2(1 ))uw ε ε< − / − for


38

each u M> . On the basis of the fact that ( )uw t is strictly decreasing with respect to t , we have 0 ut ε< < for each u M> . This implies that lim 0u ut→+∞ = .

<Case 2> Assume that E < +∞ . By Lemma 3.1, the analytic function ( )Ew t is well defined with respect to

(0t∈ , ( ))E E ξ/ − . Even when u E= < +∞, Lemmas 3.2 and 3.3 are valid. So, ( )Ew t is strictly decreasing and

0lim ( ) 0Et

w t+→= . Therefore, we have ( ) 0Ew t < .

If 0 (2(E Eε< < / ))ξ− and (E max(0 )) 2 u Eξ+ , / < < , then due to 0 ε< ( )u u ξ< / − , ( )uw ε is well defined. By Lemma 3.1 we have lim ( )uu E

w ε−→ ( ) 0Ew ε= < .

Therefore, there exists 0δ > such that ( ) 0uw ε < for each ( )u E Eδ∈ − , . This implies that 0 ut ε< < and

lim 0uu E t−→= .

Lemma 3.8. If 0ξ > , 1 1Ht / = .

Proof. If 0ξ > , then by Lemma 3.5 we have 1 1H H Eξξ + / < / < and 1 ( )u u ξ< / − . Therefore, ( )uw t and ut are analytic near ( ) (1u t H, = / , 1) . The

conclusion follows from the equality

1

1 1 1

( )(1) ( ) 0

( )H

H IH H

a xw dF x

a x/

−= = .

− +∫

Lemma 3.9. If 0ξ > , ( 1 )

lim 1uu HHt

ξξξ

ξ+→ + /= + .

Proof. Due to Lemma 3.6, ( 1 )

lim uu H tξξ +→ + / exists, and

we denote it by γ . It is clear that 1γ > . According to the inequality ( )u u ut ξ< / − , we have 1 Hξγ ξ≤ + .

Assume Hξ < +∞ and 1 Hξγ ξ< + , then, for any 0 min((1 Hξε ξ< < + ) 3 2)γ γ− / , / , there exists 0δ > such that

and (1 )uu Ht u ξγ ε ξ εξ

− < − + <−

for each ( 1 1 )u H Hξ ξξ ξ δ∈ + / , + / + . This implies that

( )

( )

0 if ( )( ) 1 1 1 if ( )

( )

1 1 1 if ( )

u uu uu a x u

uu ua x u

a x ua x u a x u

a x t ut u t t

a x ut t

ξ ε

ε

− −

−

, = ,− = < < , < ,

− + − − < < , > .

+

By Lebesgue theorem, we obtain

( ) ( )1

1 1( 1 )1 1

1 10 lim ( ) ( )a x a x

u H

I Iu Hu

dF x dF xtξ

ξ ξ

ξ γ+

+ /

→ + /− −

= = .− −∫ ∫

This is a contradiction because the term on the right is positive, which is deduced from the fact that the function

( )1

11

1 ( )a x

H

IdF x

tξ ξ+ /−

−∫

is strictly decreasing from ( 1 ) 1 0E Hξξ/ + / − > to 0 with respect to (0t∈ , 1 ) .Hξξ+ Assume Hξ = +∞ and γ < +∞ . Then, we have 0 1 ( ( ) )a x γ ξγ ξ< / − + 1 ( ( ) ) 1u ua x t tξ ξ ξ≤ / − + ≤ / for each u Eξ < < . Therefore, by Lebesgue theorem, we obtain

( )0 lim ( )( )

1 1lim (1 ( ))( )

1 1 1 1(1 ( )) (1 ( )) 0( )

Iuu u

Iuu u u

I I

a x u dF xa x u ut t

u dF xa x u ut t t

dF x dF xa x

ξ

ξ

ξ ξγ γ ξγ ξ γ ξ

+

+

→

→

−=

− +

= −− +

= − > − = ,− +

∫

∫

∫ ∫

which is a contradiction. This implies that if Hξ = +∞ ,

1 Hξγ ξ= + = +∞.

Lemma 3.10. If 0ξ < and 1 0Hξξ + / > , ( 1 )

lim uu H tξξ +→ + /

1 Hξξ= + .

Proof. Due to 1 0Hξ ξ/ > − > , we have Hξ < +∞ . It should be noted that there exists 0δ > such that ut is strictly increasing or decreasing in the interval

( 1u Hξξ∈ + / , 1 )Hξξ δ+ / + , which is demonstrated in the proof of Lemma 3.16. Therefore,

( 1 )lim uu H t

ξξ +→ + /

exists and denoted by γ . By the inequality ( )u u ut ξ< / − , we have 1 Hξγ ξ≤ + . Assume γ <

1 Hξξ+ , then, as in the proof of Lemma 3.9, we have a

contradiction.

Lemma 3.11. If 0ξ = and 1 0H/ > , (1 )lim 1uu H t+→ /

= .

Proof. It should be noted that H < +∞ . Due to Lemma 3.6, (1 )

lim uu H t+→ / exists, and is denoted by γ .

According to the relation ( ) 1u u ut ξ< / − = , we have 1γ ≤ . Assume γ < 1 , then as in the proof of Lemma

3.9, the function ( ( ) )I

a x H−∫ ( ( ) ) ( )a x t Ht H dF x/ − + is

strictly decreasing from 1 0HE − > to 0 in the interval (0t∈ , 1), which leads to a contradiction.


39

Lemma 3.12. If 0ξ < and 1 0Hξξ + / ≤ , 0

lim 0uu t+→= .

Proof. On the basis of the definition, 0 ( )u u ut ξ< < / − and max(0 , 1 ) 0H uξξ + / = < E< . Therefore,

00 lim uu t+→≤

0lim ( ) 0

uu u ξ+→

≤ / − = .

Lemma 3.13. If 0ξ ≤ , ( ) 0

( )u a xdF xt ≠

< .∫

Proof. As 0ξ ≤ , we have 0 ( ) 1u u ut ξ< < / − ≤ for each (max(0u∈ , 1 )Hξξ + / , )E , and ( ) u ua x u ut t− +

0 if ( )( ( )) ( ( ) ) ( ) 0 if ( )

( ( ) ) 0 if ( )u

u

u a x uu u a x u a x u a x ut

a x u u u a x ut

ξ ξ> , = ,

= − − > − / − ≥ , < , − + > > , > .

In this case, the equation ( ) 0u uw t = is equivalent to

( ) 0 ( ) 0

1( ) ( ) 1( ) 1a x a x

u u u

u dF x dF xa x u ut t t≠ =

+ = .− + −∫ ∫

And so,

( ) 0 ( ) 0

( ) 0

1 1( ) ( )1

1 ( ) 0( )

a x a xu u

a xu u u

dF x dF xt tu dF x

a x u ut t t

= ≠

≠

−+

− +

= > ,− +

∫ ∫

∫

which leads to ( ) 0

( )u a xdF xt ≠

< .∫

Lemma 3.14. If 0ξ = and H = +∞ , 0lim uu t+→

( ) 0( )

a xdF x

>= ∫ .

Proof. Due to Lemma 3.6, 0

lim uu t+→ exists, and we

denote it by γ . We can choose 0δ > such that 2 utγ γ/ < < for each (0u∈ , )δ . It should be noted

that the equation ( ) 0u uw t = is equivalent to

( )1( ) 0 ( ) 0

1

1 1( ) ( ) 01 a x

u

a x a xu u

dF x dF xt t= >

−

−+ = .

− + −∫ ∫

Assume that 1γ < , we have

( )1

1

1 if 0 ( )11

2 if ( )a xu

u

a x u

t a x u

γ

γ−

, < < , −< − , > ,

for each (0u∈ , )δ . In this case, by Lebesgue theorem, we obtain

( ) 0 ( ) 0

1 1( ) ( ) 01 a x a x

dF x dF xγ γ= >

−+ = ,

− + ∫ ∫

which implies that ( ) 0

( )a x

dF xγ>

= .∫

In the case in which 1γ = , due to Lemma 3.13, we have

( ) 0( )

a xdF xγ

>≤ ∫ 1≤ . Thus,

( ) 0( )

a xdF xγ

>= .∫

Lemma 3.15. The function u ut / is strictly decreasing with respect to (max(0u∈ , 1 )Hξξ + / , )E .

Proof. Using the equality

( )

2 2

2

( ( ) ) ( )( ) ( ) ( ) 0( ) ( )

uI I

u u u u

a x u a x u ua x u tdF x dF xa x u u a x u ut t t t

− + −−= =

− + − +∫ ∫

and by the proof of Lemma 3.6, we obtain

2

2 2

2

2

2

2

2

( ) ( ( ) )( ( ) ) ( ( ) )

( ( ) )2( ( ) )

1( ( ) )

( ( ) )( ( ) )

( ) ( )

( )

( )0

( )

u u u u

u u

u u

u u

a x a x uua x u u a x u uI It t t tu

a x ua x u uI t t

a x u uI t ta x u

a x u uI t t

u dF x dF xtd tdu u u dF x

dF x

dF x

−

− + − +

−

− +

− +

−

− +

+ = −

= − < .

∫ ∫∫

∫∫

We define the continuous function ut in the interval [0 , )+∞ as follows:

1 if 0 1If 0 then if 1 (7)

0 ifuu

u HH u Ettu E

ξ, ≤ ≤ / ,

> , := , / < < , , ≥ .

( ) 0( ) if 0

1 if 0 1If 0 then (8)if 1

0 if

a x

u

u

dF x u

u HtH u Etu E

ξ

> , = . , < ≤ / ,= , :=

, / < < , , ≥ .

∫

if 0 max(0 1 )

If 0 then if max(0 1 ) (9)0 ifuu

u u Hu

H u Ettu E

ξ

ξ

ξξ

ξ ξ

, ≤ ≤ , + / , −< , := , , + / < < , , ≥ .

Using Lemma 3.15 and the above extensions, we have that u ut / is strictly decreasing in the interval 0 u E< < . We denote the value of

0lim uu

ut+→/ by η.

Lemma 3.16. If 0ξ < , then the value max0 u E< < exists, which satisfies the following properties: (a) ut is strictly increasing in the interval max0 u u< < .

(b) ut is strictly decreasing in the interval maxu u E< < .

Proof. 1 ( )u u ut ξ/ = / − is strictly decreasing in the interval 0 1u Hξξ< < + / , if 1 0Hξξ + / > . As the function uy ut:= / is strictly decreasing with respect to u , ut can be considered to be a function with a variable (0y∈ , )η .

If (max(0u∈ , 1 )Hξξ + / , )E , uu tt = . From ( ) 0u uw t = we have

1 ( ) 1( ) 1I

u

dF xa x y t

= .− +∫


40

Thus,

( )2

( )( ) 0

( ) 1

ud tdy

Iu

a xdF x

a x y t

−= .

− +∫

This implies that

2

2

( )( ( ) 1)

1( ( ) 1)

( )

( )u

u

a xa x yI tu

a x yI t

dF xdtdy dF x

− +

− +

= .∫∫

Denoting 21

( ( ) 1)( )

ua x yI tdF x

− +∫ by s , we obtain

32

2 2

2 3

22

3

3 2 3

( )( ( ) )2 ( )

( ( ) 1)1( )( )2 ( ) ( )

( ( ) 1) ( ( ) 1)

( ) ( )( ( ) 1)

2 ( ) ( )2 ( )( ( ) 1) ( ( ) 1)

u

u

d tdy

Iuu

d tdy

I Iu u

Iu

I Iu u

a x a xs dF x

a x yd ttdy s a xa x dF x dF x

a x y a x yt t

a xs dF xa x y t

a x a xs dF x ds a x y a x yt t

− −

− + =

− + × − + − +

− +

−= −

− + − +

∫

∫ ∫

∫

∫ ∫2

2 3

( )

( ) 1( ) ( )( ( ) 1) ( ( ) 1)I I

u u

F x

a x dF x dF xa x y a x yt t

. + × − + − + ∫ ∫

As a quadratic function with respect to s ,

2 23 2 us d dyt− / × / has the determinant given by

2

2

2

3

2

3 3

( ) ( )( ( ) 1)

( ) ( )( ( ) 1)

.( ) 1( ) ( )

( ( ) 1) ( ( ) 1)

Iu

Iu

I Iu u

a x dF xa x y t

a x dF xa x y t

a x dF x dF xa x y a x yt t

×

− +

− + − × − + − +

∫

∫

∫ ∫

Due to Hölder inequality with respect to the two functions 3 2

( )( ( ) 1)u

a xa x y t /− +

and 3 21

( ( ) 1)ua x y t /− +, this

determinant is negative. Therefore, we have 2

2 0ud tdy

< ,

which implies that ud tdy is strictly decreasing.

First, we assume that 1 0Hξξ + / ≤ . Assign

0lim uy

d dytα +→:= / and β lim uy

d dytη−→:= / .

If 0α ≤ , then 0ud dyt / < for each 0 y η< < . This contradicts the fact that 0ut = if 0y = (Lemma 3.7), and 0ut > if 0 y η< < . Therefore, 0α > .

If 0β ≥ , 0ud dyt / > for each 0 y η< < . This contradicts the fact that 0ut = if y η= (Lemma 3.11), and 0ut > if 0 y η< < . Therefore, 0β < .

The value max0 y η< < such that max

0y yud dyt =/ | = can

then be determined. The value max0 u E< < required is determined using maxy .

Second, we assume that 1 0Hξξ + / > . If (0u∈ , 1 )Hξξ + / , then ut ( )u u ξ= / − is strictly increasing.

Moreover, we have 1( ) u Hu uξξξ = + // − |

( 1 )1 lim uu H

H tξ

ξ ξξ +→ + /

= + = (Lemma 3.10) and 1u Hyξξ= + /|

Hξ= . Redefine β as lim uy Hd dyt

ξ−→

/ . If 0β < , then as

above, we obtain the required value 1 Hξξ + / maxu< E< . If 0β ≥ , 0ud dyt / > for each 0 y Hξ< < . This

implies that ud dut / 0ud dy dy dut= / × / < since 0dy du/ < for each ( 1u Hξξ∈ + / , )E (Lemma 3.15).

Thus, ut is strictly decreasing. Therefore, the required

value is max 1u Hξξ= + / .

Pre-Growth Rate

In this Section we assume that (max(0u∈ , )ξ , )E and 0 ρ< < t < ( )u u ξ/ − unless otherwise mentioned. Define ( )uG tρ, by the value

( ) ( )exp ( ) exp log ( ) (10)( )

t

u I

a x t ut uw t dt dF xa x u uρ ρ ρ

− += , − +

∫ ∫

which can be verified using the following inequalities:

( ) ( )( ) ( )

u u t a x t ut u tu u a x u u

ξξ ρ ρ ρ ρ

/ − − − +< < ,

/ − − − +

( ) 1 1max( )( ) ( )

a x ua x s us u u u tρ ξ

−< ,

− + / − −

for each x I∈ and (min(s ρ∈ , )t , max(ρ , ))t . As ( )uw t is strictly decreasing with respect to t from the positive value 1 0E u/ − > , to the value (1 ) ( 1 )u H H uξ ξξ ξ− / + / − (Lemmas 3.2, 3.3, and 3.4),

( )t

uw t dtρ∫ is strictly decreasing with respect to ρ near

0+ . Therefore, the limit

( )0 0

( )limexp ( ) exp lim log ( ) (11)( )

t

u I

a x t ut uw t dt dF xa x u uρρ ρ ρ ρ+ +→ →

− += − +

∫ ∫

( )( )exp log ( ) 1 ( )I

a x t u t dF x= / − +∫

finitely exists or +∞ , which we denote by ( )u tG and refer to as pre-growth rate. The equality mentioned above is obtained using Lebesgue theorem because the integrand is monotone with respect to ρ in x |

( )a x u> or x | ( )a x < u .

Lemma 4.1. ( )u t E uG < / , if 0 min(1 ( ))t u u ξ< < , / − .

Proof. By Jensen’s inequality we have

( ) ( )log ( ) 1 ( ) log ( ) 1 ( )

log( 1) log( )I I

a x t u t dF x a x t u t dF x

Et u t E u

/ − + ≤ / − +

= / − + < / .∫ ∫


41

Lemma 4.2.

( )( )

log ( ) 1 ( )a x u

a x t u t dF x<

/ − + < +∞.∫

Proof. In general, ( ) 1 ( )( ( ) ) 0a x t u t u u u t uξ ξ/ − + ≥ − / − − / > . If ( )a x u< , ( ) 1a x t u t/ − + < 1 . Therefore, we obtain

( )( )

log ( ) 1 ( ) loga x u

u ua x t u t dF x tu uξ

ξ<

−/ − + ≤ − < +∞. −

∫

Lemma 4.3. The following three statements are equivalent.

(1) ( ) 1

log ( ) ( )a x

a x dF x>

< +∞.∫

(2) ( )u tG < +∞ for each u and t.

(3) 1 1( )u tG < +∞ for some 1u and 1t .

Proof. (1) ⇒ (2). The function ( )log ( ) 1a x t u t/ − + satisfies the following inequalities.

( ) 1log ( ) ( ) log ,

u a xa x dF x u

< ≤≤ < +∞∫

( )( ) max(1 ) ( ) max(1 )

( ) max(1 )

log ( ) 1 ( ) log ( ) ( )

(1 )log log 1 ( )( )

log log

a x u a x u

a x u

a x t u t dF x a x dF x

t u t dF xu a x t

t tu

> , > ,

> ,

/ − + −

−= + +

≤ + < +∞.

∫ ∫

∫

Based on Lemma 4.2, we obtain the result. (3) ⇒ (1). It should be noted that 1 (max(0u ∈ , )ξ , )E and 1 (0t ∈ , 1 1( ))u u ξ/ − . The result can be obtained in a similar manner as above.

(2) ⇒ (3). It is clear.

If one of the above three statements is satisfied, we can write G < +∞ .

Lemma 4.4. If G < +∞,

0lim ( ) 1ut

tG+→= .

Proof. Since 0

lim ( ) 1 0utw t E u+→

= / − > (Lemma 3.3),

0( )

t

uw t dt∫ is strictly increasing and bounded with

respect to t near 0+ . Therefore, we obtain that 0

limt +→

0( ) 0

t

uw t dt = .∫

Lemma 4.5. If (max(0u∈ , 1 )Hξξ + / , )E ,

0 ( )max ( ) ( )t u u u u uG t G tξ ρ ρ< < / − , ,= .

Proof. It is clear from the facts that 0 ( )uG tρ,< < +∞,

(1 ) ( 1 )E u H H uξ ξξ− / + / − ( )uw t< 1E u< / − , and

( )( ) exp ( )

( ) ( ) ( ) ( )

t

u u

t

u u u u

G t t w t dtt

G t w t dt G t w tt

ρ ρ

ρ ρρ

,

, ,

∂∂ / ∂ =

∂∂ = = . ∂

∫

∫

Lemma 4.6. If G < +∞ and (max(0u∈ , 1 )Hξξ + / , )E ,

0 ( )max t u u ξ< < / − ( ) ( )u u utG G t= .

Proof. In a similar manner as that of the proof of Lemma 4.5, we have ( ) ( )

0 0

0

( )limexp ( ) lim exp ( )

lim ( ) ( ) ( ) ( )

t tuu u

u u uu

tG w t dt w t dtt t t

G t w t t w tG

ρ ρρ ρ

ρρ

+ +

+

→ →

,→

∂ ∂ ∂= =

∂ ∂ ∂= = ,

∫ ∫

which implies the conclusion.

Lemma 4.7. Two functions ( )uG tρ, and ( )u tG ( )< +∞ are concave with respect to t .

Proof. Using Lemmas 3.2, 4.5, and Hölder inequality, we have

( )

2 2 2

2

2 2

( ) ( )( ( ) ( ) )

( ( ) ) ( ( ) ) ( )( ) 0

( ( ) ) ( ( ) ) ( )

u u u u

Iu

I

G t t G t w t w t t

a x u a x t ut u dF xG t

a x u a x t ut u dF x

ρ ρ

ρ

, ,

,

∂ / ∂ = + ∂ / ∂

− / − + = < . − − / − +

∫∫

Along the similar way of discussions, we also have

2 2( ) 0u t tG∂ / ∂ < .

Lemma 4.8. If 0ξ ≥ and t ρ> , ( )uG tρ, is strictly decreasing with respect to u .

Proof. From ( ) 0a x ≥ , we obtain

( )2

( )exp ( ) ( ) ( )

( )( ) ( ) 0( ( ) )

t tuu u u

t

u I

G tw t dt G t w t dt

u u ua xG t dF x dt

a x t ut u

ρρρ ρ

ρ ρ

,,

,

∂ ∂ ∂= = ×

∂ ∂ ∂

= − × < . − +

∫ ∫

∫ ∫

Lemma 4.9. If 0ξ ≥ and G < +∞ , ( )u tG is strictly decreasing with respect to u .

Proof. Using Lemma 4.8, we obtain the conclusion.

Lemma 4.10. If G < +∞ ,

( ( ))lim ( )ut u u

tGξ −→ / −

( )exp( log ( ) ( ))I

a x dF xξ= −∫ ( )u ξ/ − for each u.

Proof. If ( )a x u> , ( ) 1a x t u t/ − + is strictly increasing with respect to t . Therefore, using Lebesgue theorem, we have

( )( ( ))

( )

0 lim log( ( ) 1) ( )

( )log ( )

a x ut u u

a x u

a x t u t dF x

a x dF xu

ξ

ξξ

− >→ / −

>

≤ / − +

−= < +∞.

−

∫

∫

On the other hand, if ( )a x u< , then ( ) 1a x t u t/ − + is


42

strictly decreasing with respect to t . Hence, using Lebesgue theorem, we have

( )( ( ))

( )

lim log( ( ) 1) ( )

( )log ( ) 0

a x ut u u

a x u

a x t u t dF x

a x dF xu

ξ

ξξ

− <→ / −

<

/ − +

−= < ,

−

∫

∫

which implies that

( )( )

( ( ))

( ) ( )

lim ( )

( ) ( )exp log ( ) log ( )

exp log ( ) ( ) ( )

ut u u

a x u a x u

I

tG

a x a xdF x dF xu u

a x dF x u

ξ

ξ ξξ ξ

ξ ξ

−→ / −

> <

− −= + − −

= − / − .

∫ ∫

∫

As an expansion of the definition of ( )u tG , we define (( ( )) )u u uG ξ −/ − by ( )exp( log ( ) ( ))

Ia x dF xξ−∫ ( )u ξ/ −

for each (max(0u∈ , )ξ , )E .

Lemma 4.11. ( ) 1u uG t > if (max(0u∈ , 1 )Hξξ + / , )E .

Proof. If 0 ut t< < , ( ) 0uw t > . Hence, we have

0

0( ) exp( ( ) ) 1ut

uu u w t dt eG t = > = .∫

Lemma 4.12. ( )u uG t ( )< +∞ is strictly decreasing with respect to (max(0u∈ , 1 )Hξξ + / , )E .

Proof. If ( ) 2a x ξ≤ , we have

2( ) ( )( ) ( ( ) ) ( )( ( ) )

a x a xa x t ut u a x t t ut u u u u t

ξξ ξ ξ ξ

= ≤ .− + − + − + − / − −

On the other hand, if ( ) 2a x ξ> , we have

( ) ( ) ( ) 2( ) ( ( ) ) ( ( ) )

a x a x a xa x t ut u a x t t ut u a x t tξ ξ ξ

= ≤ < .− + − + − + −

Thus, by the definition of ( )u tG , we have

( )( ) log( ( ) 1) ( )

( )( ) ( )( )

uu I

u I

tG t a x t u t dF xGu ut a xt dF xGu a x t ut u

∂ ∂= / − +

∂ ∂

= − × . − +

∫

∫

The definition ( ) 0u uw t = leads to

( ) ( ( ) ) ( ) 1u uIa x a x u u dF xt t/ − + =∫ . Therefore,

( )( ) 0u u u

u uG t t

G tu u∂

= − < .∂

Lemma 4.13. If G < +∞ , lim ( ) 1u uu E G t−→= .

Proof. From Lemmas 4.11 and 4.12, lim ( ) 1u uu E G t−→≥

exists. Assume that 0ξ ≥ , then from Lemmas 3.6 and 3.7, (1 )u ut t/ − is strictly decreasing near u E−= , and lim 0uu E t−→

= . Applying Lebesgue theorem to the

equality

( )( ) 1 (1 ) exp log 1 ( ) 1

(1 )u

u u u u Iu

a x t dF xG t t t u t

− + = − + − , −

∫

we have

( ) ( )( )lim ( ) 1 exp log 0 1 ( ) 1 0u u u Iu EdF xG t t−→

− + = + − = .∫

This implies that lim ( ) 1u uu E G t−→= .

In the case in which 0ξ < , Lemma 3.16 can be used as a substitution of Lemma 3.6 near u E−= , where ut is strictly decreasing to 0 (Lemma 3.7). In order to apply Lebesgue theorem, it is sufficient to divide the above integration into two parts x | ( ) 0a x ≥ and x |

( ) 0a x < .

Lemma 4.14. If G < +∞ and (max(0u∈ , 1 )Hξξ + / ,

( )u uG t ( )expE

uuudut= / .∫

Proof. Using ( ) ( )u u u u uu uG t t G t∂ / ∂ = − / × (Lemmas 4.12 and 4.13), we can solve the differential equation.

Lemma 4.15 If 0ξ = and H = +∞ ,

0lim ( )u uu G t+→

= +∞.

Proof. Lemma 4.12 ensures the existence of


, which is finite or +∞ . If ( ) 0a x > ,

( )a x u/ is strictly decreasing with respect to u, and

0lim ( )

ua x u+→

/ = +∞ . Using Lebesgue theorem and

( ) 0( ) 0

a xdF x

>>∫ , we have

00

0 ( ) 0

1lim ( ) ( )lim2

1 ( )exp log ( )lim2 2

u u u uu

u a x

G t G

a x u dF xu

++

+

→→

→ >

≥

+ ≥ = +∞. ∫

Lemma 4.16. If 0ξ = , H < +∞, and G < +∞ ,

(1 )lim ( )u uu H G t+→ /

( )exp log ( ) ( )I

H a x dF x= .∫

Proof. By definition, G < +∞ implies that ( ) 1a x >∫

log ( ) ( )a x dF x < +∞ . From Jensen’s inequality, we have

1 1log log ( ) log ( )( ) ( )I I

H dF x dF xa x a x

+∞ > = ≥ ,∫ ∫

which implies that log ( ) ( )I

a x dF x > −∞∫ . Therefore,

log ( )a x is integrable.

It should be noted that (1 )lim 1uu H t+→ /

= (Lemma 3.11).

Using the equalities (1 )lim

u H +→ / u u Ht / = and

(1 )lim (1 ) 0u uu H

ut t+→ /− / = , we can choose

0 min(1 Hδ< < / , 1 )E H− / , such that


43

2 3 2uH u Ht/ < / < / and (1 ) 1 2u uut t− / < / for each (1u H∈ / , 1 )H δ/ + . Therefore, we have the following

properties.

(1) If ( ) 1 2a x ≥ / , then

( ) (1 )log ( ) 1 log log ( )

3max( log log ) log 2 log ( )2 2

u uu u

u

ut ta x u a xt t u tH H a x

−/ − + = + +

< , + + .

(2) If ( ) 1 2a x < / ,

( ) 3log ( ) 1 max( log log ) log ( )2 2u uH Ha x u a xt t/ − + < , + .

Using the above properties, we can apply Lebesgue theorem as follows:

( )( )( )

(1 ) (1 )lim ( ) lim exp log ( ) 1 ( )

exp log ( ) ( )

u u u uIu H u H

I

a x u dF xG t t t

H a x dF x

+ +→ / → /= / − +

= .

∫

∫

Lemma 4.17. If 0ξ = , H < +∞, and G < +∞ ,

(1 )lim (1 )uu H G−

−→ /

( )exp log ( ) ( )I

H a x dF x= .∫

Proof. In the case in which 0ξ = , based on the definition which is mentioned beneath the proof of

Lemma 4.10, we have ( )(1 ) exp log ( ) ( )u Ia x dF x uG

− = /∫ .

Thus, we obtain the conclusion.

Lemma 4.18. If 0ξ = , H < +∞, and G < +∞ ,

0lim (1 )uu G+

−→

= +∞.

Proof. Due to Lemmas 4.11, 4.12, 4.13, and 4.16, we have H exp( log ( ) ( ))

Ia x dF x∫ 1> . Therefore, by the

definition of (1 )uG− we obtain

( )0 0 0

lim (1 ) limexp log ( ) ( ) lim 1u Iu u ua x dF x u H uG+ + +

−

→ → →= / ≥ / / = +∞.∫

Lemma 4.19. If 0ξ > and Hξ = +∞ ,

lim ( )u uu G tξ +→= +∞.

Proof. Lemma 4.12 ensures the existence of lim ( )u uu G tξ +→ , which is finite or +∞ . If ( )a x ξ> , then

( ( ) ) (2( ))a x uξ ξ− / − is strictly decreasing with respect to (u ξ∈ , )E . Using Lebesgue theorem, we have

( )

lim ( ) ( )lim2( )

1 ( ) 1exp log ( )lim2 2( ) 2

u u u uu

u a x

uG t G u

a x dF xu

ξξ

ξ ξ

ξ

ξξ

++

+

→→

→ >

≥−

−≥ + = +∞. −

∫

Lemma 4.20. If 0ξ > , Hξ < +∞, and G < +∞ ,

( 1 )lim ( )u uu H G t

ξξ +→ + / ( )exp( log ( ) ( ))

IH a x dF xξ ξ= − .∫

Proof. An argument similar to that in the proof of Lemma 4.16 ensures that log( ( ) )a x ξ− is integrable.

It should be noted that ( 1 )

lim 1uu HHt

ξξξ

ξ+→ + /= +

(Lemma 3.9). From the fact that ( 1 )

lim uu Hu Ht

ξξξ +→ + /

/ =

and ( 1 )

lim ( ) 0u u uu Hu ut t t

ξξξ+→ + /

− + / = , we can choose

0 min( 1 Hξδ ξ< < + / , 1 )E Hξξ− − / , such that 2 3 2uH u Htξ ξ/ < / < / and ( ) 1 2u u uu ut t tξ − + / < / for

each ( 1u Hξξ∈ + / , 1 )Hξξ δ+ / + . Therefore, we have the following properties.

(1) If ( ) 1 2a x ξ≥ + / ,

( )

( )

log ( ) 1

log log ( )

3max( log log ) log 2 log ( )

2 2

u u

u u u

u

a x ut t

u ut t ta xu t

H Ha xξ ξ

ξξ

ξ

/ − +

− += + − +

< , + + − .

(2) If ( ) 1 2a x ξ< + / ,

( ) ( )3

log ( ) 1 max( log log ) log ( )2 2u u

H Ha x u a xt t

ξ ξ ξ/ − + < , + − .

Using the above properties, we can apply Lebesgue theorem as follows.

( )( )( )

( 1 ) ( 1 )lim ( ) lim exp log ( ) 1 ( )

exp log( ( ) ) ( )

u u u uIu H u H

I

a x u dF xG t t t

H a x dF x

ξ ξξ ξ

ξ ξ

+ +→ + / → + /= / − +

= − .

∫

∫

Lemma 4.21. If 0ξ > , 1 1( )H HG t/ / ( )exp log ( ) ( )

IH a x dF x= .∫

Proof. It should be noted that 0 1 H Eξ< < / < and 1 1 (1 )H H ξ< / / / − . From Lemma 3.8, we have 1 1Ht / = . Thus, ( )( )

( )1 1( ) exp log ( ) (1 ) 1 1 ( )

exp log ( ) ( )

H H I

I

a x H dF xG t

H a x dF x

/ / = / / − +

= .

∫∫

Lemma 4.22. If 0ξ > , (1 )lim

u H −→ / (1)uG

( )exp log ( ) ( )I

H a x dF x= .∫ Proof. From 0 uξ< < , we have 1 ( )u u ξ< / − . Thus, by Lemma 4.21 we obtain

( )1(1 )lim (1) (1) exp log ( ) ( )u H Iu H

H a x dF xG G− /→ /

= = .∫

If 0ξ > and 0 u ξ< ≤ , ( ) 1 1a x t u t/ − + ≥ for each 0t > .


44

Therefore, we can expand the definition of ( )( ) exp( log ( ) 1 ( ))u I

t a x t u t dF xG = / − +∫ , which is

greater than 1 and finite or +∞ , in the domain 0 u ξ< ≤ and 0t > .

Lemma 4.23. If 0ξ > , 0

limu +→

(1)uG = +∞.

Proof. It should be noted that 1 1( )H HG t/ /

( )exp log ( ) ( ) 1I

H a x dF x= >∫ (Lemma 4.21). From the

expansion of ( )u tG which is defined beneath the proof of Lemma 4.22, we have

( )0 0 0

lim (1) limexp log ( ) ( ) lim 1u Iu u ua x dF x u H uG+ + +→ → →

= / ≥ / / = +∞.∫

Lemma 4.24. If 0ξ < , 1 0Hξξ + / > and G < +∞ ,

( 1 )lim ( )u uu H G t

ξξ +→ + / ( )( )exp log ( ) ( )

IH a x dF xξ ξ= − .∫

Proof. It should be noted that Hξ < +∞ and

( 1 )lim 1uu H

Htξ

ξξξ+→ + /

= + (Lemma 3.10). The proof is

formally the same as that of Lemma 4.20.

Lemma 4.25. If 0ξ < , 1 0Hξξ + / > , and G < +∞ ,

( 1 )lim

u Hξξ −→ + / ( )( )(( ( )) ) exp log ( ) ( )u I

u u H a x dF xG ξξ ξ−/ − = − .∫

Proof. We obtain the conclusion using the definition which is mentioned beneath the proof of Lemma 4.10.

Lemma 4.26. If 0ξ < , 1 0Hξξ + / > , and G < +∞ ,

0lim ( ( ) )uu

u uG ξ+−

→/ − ( )( )exp log ( ) ( )

Ia x dF xξ= −∫ ( )ξ/ − .

Proof. We obtain the conclusion by applying the same process as in Lemma 4.25.

Lemma 4.27. If 0ξ < and 1 0Hξξ + / < , 1η ξ< − / .

Proof. It should be noted that the definition ( ) 0u uw t = implies that

1 ( ) 1( ) 1utI

uu

dF xa x t

= .− +∫

From Lemma 3.12, we have 0

lim 0uu t+→= . Using

Fatou’s lemma, we obtain

01 1( ) ( ) 1lim

( ) 1 ( ) 1uu tI Iuu

dF x dF xa x a x tη

+→= ≤ .+ − +∫ ∫

Observe that 0

0 lim uuutη +→

< = /

0lim ( ( )) 1

uu u uξ ξ+→

≤ / − / = − / . Assume that 1η ξ= − / ,

we have

1 ( ) 1( ) 1I

dF x Ha x ξξ

ξ= − ≤ .

− / +∫

This implies that 1 0Hξξ + / ≥ , which is a contradiction.

Lemma 4.28. If 0ξ < and 1 0Hξξ + / = , 1η ξ= − / .

Proof. Since Hξ < +∞ , 1 ( ( ) )a x ξ/ − is integrable. Thus,

from 1 ( ( ) u uIa x ut t/ / −∫ 1) ( ) 1dF x+ = and by Lebesgue

theorem, we have 1 ( ( ) 1) ( ) 1I

a x dF xη/ + =∫ . On the

other hand, it is clear that 1 ( ( ) 0 1) ( ) 1I

a x dF x/ × + =∫

and 1 ( ( )( 1 )I

a x ξ/ − /∫ 1) ( ) 1dF x Hξξ+ = − = . This

implies that the equation 1 ( ( ) 1) ( )I

a x y dF x/ +∫ ＝ 1

with respect to [0y∈ , 1 ]ξ− / has three solutions 0y = , y η= , and 1y ξ= − / . Note that

( )

2 2

2 3

1 ( )( ) ( ) 0( ) 1 ( ) 1I I

a xdF x dF xa x yy a x y

∂= > .

+∂ +∫ ∫

Therefore, the equation 1 ( ( ) 1) ( ) 1I

a x y dF x/ + =∫ has at

most two solutions. This implies that 1η ξ= − / .

Lemma 4.29. If 0ξ < , 1 0Hξξ + / ≤ , and G < +∞ ,


( )exp log( ( ) 1) ( )I

a x dF xη= + .∫

Proof. Lemma 3.12 implies that 0

lim 0uu t+→= . From

Lemma 3.15, u ut / is strictly decreasing with respect to (0u∈ , )E . Due to Lemma 3.16, ut is strictly increasing with respect to (0u∈ , max )u . Therefore, if

( ) 0a x > then ( ) 1utuua x t− +

is strictly decreasing with respect to (0u∈ , max )u . This ensures that

( ) 0 ( ) 00

( )lim log( 1) ( ) log( ( ) 1) ( )u ua x a xu

a x dF x a x dF xt tuη

+ > >→− + = + .∫ ∫

If Hξ < +∞, using Jensen’s inequality, we see that log( ( ) )a x ξ− is integrable. If ( ) 0a x ≤ and 0 min(u ξ< < − , )E , then we have

( ) ( )0 ( ) 1 12

uu

a x a x ta x tu uξ ξ

ξ ξ− −

≤ ≤ < − + < ,− −

and ( ) ( )log ( ) 1 log ( ) log( 2 )u ua x u a xt t ξ ξ/ − + < − + − .

Therefore, we can apply Lebesgue theorem to the following equality.

( ) 0 ( ) 00

( )lim log( 1) ( ) log( ( ) 1) ( )u ua x a xu

a x dF x a x dF xt tuη

+ ≤ ≤→− + = + .∫ ∫

Thus, we accomplish


45

( )0

lim ( ) exp log( ( ) 1) ( )u u Iua x dF xG t η

+→= + .∫

If Hξ = +∞, from Lemma 4.27, 1η ξ< − / . If we assign

( 1)ε ξη:= + 2 0/ > , then, there exists 0δ > such that

ut ε< for each (0u∈ , )δ . Hence, if ( ) 0a x ≤ , then we have

( )1 1 1u ua x

t tuξη ε ε> − + ≥ − + = .

Thus, we can apply Lebesgue theorem in the domain ( ) 0x a x| < and obtain the conclusion.

Summing up the above-mentioned Lemmas, we obtain the following

Theorem 4.1. If G < +∞ , ( )u uG t is continuous and strictly decreasing with respect to (0u∈ , )E . The range of ( )u uG t is (1, )+∞ (if 0ξ ≥ ) or (1,

( )( )exp log ( ) 1 ( ) )I

a x dF xη +∫ (if 0ξ < ).

Double Sequence of Random Variables

It should be noted that a series of step functions exists such that ( )Nf xξ ≤ 1( ) ( )Nf x a x+≤ ≤ and lim ( ) ( )N Nf x a x→+∞ = for each x I∈ , in which ξ

inf ( )x I a x∈= > −∞ is the essential infimum.

For example, for each positive integer N , assign 2 1NM N:= + and

1 1if ( ) (1 1)( ) 2 2 2

if ( )

j N N NN

M

j j ja a x j Mf x

a N a x N

ξ ξ ξ

ξ ξ

− − := + , + ≤ < + ≤ ≤ − ,:= := + , ≥ + .

(12)

In general, suppose ( )Nf x | jx I a∈ ⊂ | 1 j M= ,..., ,

where 1aξ = 2 Ma a< < < < +∞ . Set

1( )( )

j jj a a x a

p dF x+≤ <

:= ∫ (1 1)j M≤ ≤ − and

( )( )

MM a x a

p dF x≥

:= ∫ , then we have 1

1Mjj

p=

=∑ .

Assume 0 u< (price), 0 t< (proportion of investment), and 0 1t u tξ< / − + . Then 0 1ja t u t< / − + for each

1 j M≤ ≤ . For the random payoff ( ja , )jp |

1 2 j M= , , , , the growth rate per attempt after n attempts is

( )1

1

1 (13)n

jM m

jj

a t u t=

/ − + ,

∏

where ja occurs jm times 1 2( )Mm m m n+ + + = with

probability 11

Mm mMp p . Such event has n! 1 2( )Mm m m/ ! ! !

permutation patterns. We denote N nX , by this random

variable. Then, the expectation [ ]N nE X , is expressed as

( )1

1

1 2

111 2

1n

j M

M

M m m mj M

m m m n jM

n a t u t p pm m m+ + + = =

!/ − +

! ! ! ∑ ∏

(14)

( )( )

1

1 1

11

log 11 exp

n

n

M

n j jMj

j jj n

a t u t pa t u t p

=

=

/ − +

= / − + =

∑∑

( )( )1

1

log ( )( ) 1exp

nNI

n

dF xf x t u t / − + = .

∫

Moreover, the variance [ ]N nV X , is expressed as

( )2

1

1 2

111 2

1n

j M

M

M m m mj M

m m m n jM

n a t u t p pm m m+ + + = =

!/ − +

! ! ! ∑ ∏

2[ ] (15)N nE X ,−

( )( )2

1

log ( )( ) 1exp

nNI

n

dF xf x t u t / − + =

∫

( )( )1

1

2 log ( )( ) 1exp

nNI

n

dF xf x t u t / − + − .

∫

Lemma 5.1.

( ) ( )( )lim lim [ ] exp log ( ) 1 ( )N n IN nE X a x t u t dF x,→+∞ →+∞

= / − + .∫

Proof. If n approaches +∞ , using l’Hôpital’s theorem, we obtain

( ) ( )

( )

( )( )

1

0

1

1 log 1lim [ ] limexp

1

exp log ( ) 1 ( )

M h

j j jj

N n Mn hh

j jj

NI

a t u t a t u t pE X

a t u t p

f x t u t dF x

+

=,→+∞ →

=

/ − + / − +

=

/ − +

= / − + .

∑

∑

∫

Since 1( ) ( ) ( )N Nf x f x a x+≤ ≤ , and lim ( ) ( )N Nf x a x→+∞ = , we obtain the conclusion.

It is easily verified that if G < +∞ , lim [ ] 0n N nV X→+∞ , = for each N .

Lemma 5.2. If 0ν > , the following statements are equivalent.

(1) ( ) 1

( ) ( )a x

a x dF xν

>< +∞∫ .

(2) ( ( ) 1) ( )I

a x t u t dF xν

/ − + < +∞∫ for each 0u > , 0t > ,

and 1 0t u tξ / − + > .


46

(3) 1 1 1( ( ) 1) ( )I

a x t u t dF xν

/ − + < +∞∫ for some 1 0u > ,

1 0t > , and 1 1 1 1 0t u tξ / − + > .

Proof. If ( ) 2 1a x u t t> / × − , we have ( ) 2 ( ( ) 1)a x u t a x t u t< / × / − + and ( ) 1 2 ( )a x t u t t u a x/ − + < / × .

This implies the conclusion.

We say that a random payoff ( ( )a x , ( ))F x is effective

when ( ) 1

( ) ( )a x

a x dF xν

>< +∞∫ for some 0ν > , with the

additional conditions 0E > and ξ > −∞ .

It should noted that for each 0 1ν< < , there exists hν such that log( 1)x + h xνν< for each 0x > .

Lemma 5.3. If a random payoff is effective, G < +∞ .

Proof. If ( ) 1a x > and 0 1ν< < , we have log ( ) ( ( ) 1) ( )a x h a x h a xν ν

ν ν< − < . If ( ) 1a x > and 1ν ≥ , we have log ( ) ( ) ( )a x a x a x ν< ≤ . Thus, we obtain the conclusion.

We have some properties of the double sequence [ ]N nE X , as follows.

Lemma 5.4. If a random payoff is effective, ( )lim lim [ ]N n N nE X→+∞ →+∞ , ( )lim lim [ ]n N N nE X→+∞ →+∞ ,=

lim [ ]n N nN

E X→+∞ ,→+∞

= < +∞.

If a random payoff is ineffective, lim [ ]N N nE X→+∞ , = +∞ for each n , if 0t > .

Proof. For the assumption ( ) 1

( ) ( )a x

a x dF xν

>< +∞∫ , we

can assume that 1ν < . Suppose 0 2h ν< < / and ( )a x u> , we have

2

2

( ( ) 1) ( ( ) 1) log( ( ) 1)

( ( ) 1)

ha x t u t a x t u t a x t u th

h a x t u t

ν

νν

/

/

∂/ − + < / − + / − +

∂

< / − + . This guarantees that

( ( ) 1) ( ) ( ( ) 1) ( )h h

I Ia x t u t dF x a x t u t dF x

h h∂ ∂

/ − + = / − + < +∞.∂ ∂∫ ∫

Thus, since 1( ) ( ) ( )N Nf x f x a x+≤ ≤ , and lim ( )N Nf x→+∞ ( )a x= , we have

( )

( )( )0

lim lim [ ]

( ( ) 1) log( ( ) 1) ( )exp lim

( ( ) 1) ( )

exp log ( ) 1 ( )

N nn N

h

Ihh

I

I

E X

a x t u t a x t u t dF x

a x t u t dF x

a x t u t dF x

+

,→+∞ →+∞

→

/ − + / − + = / − +

= / − + < +∞.

∫∫

∫

Using Lemma 5.1, we have

( ) ( )lim lim [ ] lim lim [ ]N n N nN n n NE X E X, ,→+∞ →+∞ →+∞ →+∞

= .

Set ( ) ( )lim lim [ ] lim lim [ ]N n N n n N N nE X E Xα →+∞ →+∞ , →+∞ →+∞ ,:= = ,

NU lim [ ]n N nE X→+∞ ,:= , and lim [ ]n N N nW E X→+∞ ,:= Since

1( ) ( ) ( )N Nf x f x a x+≤ ≤ , [ ]N nE X , increases with respect to N . For any 0ε > , there exits 0N and 1n such that

NU α ε− < and nW α ε− < for each 0N N≥ and

1n n≥ . By setting 1h n:= / and applying l’Hôpital’s theorem twice, we have

( )

( )

( )

0

2

2 1

lim [ ]

exp log( ( ) 1) ( )

2( )log( ( ) 1)

0 if 1log( ( ) 1) ( )

N nh

NI

NI

NI

E Xh

f x t u t dF x

dF xf x t u tp

f x t u t dF x

+ ,→

∂∂

/ − += ×

/ − + > , < . − / − +

∫

∫

∫

This implies that [ ]N nE X , decreases with sufficiently large n ( 2 )n h h∂ / ∂ = − ∂ / ∂ . Then, there exits

2 2 0 1( )n n N n= ≥ such that

0 2 0 0 01[ ] [ ] [ ]N n N n N n NE X E X E X U, , , +> > ≥

for each 2n n≥ . Therefore, we have

0 0[ ] [ ]N N n N n nU E X E X Wα ε α ε, ,− < < ≤ ≤ < + ,

which implies lim [ ]n N nN

E X α→+∞ ,→+∞

= as a double

sequence.

If ( ) 1

( ) ( )a x

a x dF xν

>= +∞∫ (ineffective) for each 0ν > ,

using Lemma 5.2, we have

( )1

1

log ( ( ) 1) ( )lim [ ] exp

n

IN nN

n

a x t u t dF xE X ,→+∞

/ − + = = +∞

∫

for each 1n ≥ and 0t > .

Lemma 5.5. If a random payoff is effective, lim [ ] 0n N n

NV X→+∞ ,

→+∞= .

Proof. This can be proved in a manner similar to Lemma 5.4.

Lemma 5.6. If a random payoff is effective,

( )( )( )2

lim [ exp log ( ) 1 ( ) ] 0N n InN

E X a x t u t dF x,→+∞→+∞

− / − + = .∫

Proof. The equality ( ) ( )2 2[ ] [ ] [ ]N n N n N nE X c V X E X c, , ,− = + −

for each c implies the conclusion (Lemmas 5.1, 5.4, and 5.5).


47

We denote ( )exp( log ( ) 1 ( ))I

a x t u t dF x/ − +∫ by ( )uG t

on the region 0u > , 0 1t≤ ≤ , and 1 0t u tξ / − + > . We refer to this as the limit expectation of growth rate. Notice that ( )u tG is defined above Lemma 4.1 on the region max(0 ) u Eξ, < < and 0 ( )t u u ξ< < / − , or 0ξ > , 0 u ξ< ≤ , and 0t > . It is clear that ( ) ( )uu t G tG = on the intersection of the regions.

It is said that ut is the optimal proportion of investment in order to continue to invest without borrowing with respect to 0u > , if

0 11 0

( ) 1log ( ) 0 (16)lim ( ) 1ut

Iu

a x t u t dF xa x u

ρρ

ξρ ρ ρ ρ→≤ ≤/ − + >

/ − +≤

/ − +∫

for each 0 1t≤ ≤ and 1 0t u tξ / − + > . It follows that 0 1ut≤ ≤ and 1u ut u tξ / − + 0≥ . The existence and uniqueness of ut for each 0u > is proved in Theorem 5.1.

If 0 1 1 0lim t uρ ρ ξρ ρ→ , ≤ ≤ , / − + > ( )uG t exists, we extend ( )uG t by

the value if 1t u tξ / − + 0= . Suppose G < +∞ , then,

from log( ( ) ( ))u uG t G ρ/ log(( ( ) 1)I

a x t u t= / − +∫

( ( ) 1))a x uρ ρ/ / − + ( )dF x , the equation ( )u uG t

0 1 1 0sup ( )t t u t uG tξ≤ ≤ , / − + >= implies that ut is the optimal proportion. Theorem 5.1. u ut t= ( 0)u > , where ut is a continuous function defined by (7), (8) and (9) beneath Lemma 3.15.

Proof.

(1) If u E≥ , 0ut = .

It should be noted that Lemmas 3.2, 3.3 and equality (10) are valid even if u E≥ . Using ( ) 0uw t t∂ / ∂ < and

(0 ) 1 0uw E u+ = / − ≤ , we have ( ) 0uw t < for each 0 min(1t< < , ( ))u u ξ/ − . Therefore, we have

( )( ) log ( ) 0( )

t

u I

a x t ut uw t dt dF xa x u uρ ρ ρ

− += <

− +∫ ∫

for each 0 min(1tρ< < < , ( ))u u ξ/ − , and ( ) 0t

uw tρ

>∫

for each 0 t ρ< < min(1< , ( ))u u ξ/ − . This implies that 0ut = . .

(2) If 0ξ ≤ and 0 1u Hξξ< ≤ + / , ( )ut u u ξ= / − .

Notice that the condition 0ξ = , 0H = , and 0 1u Hξξ< ≤ + / is void. Using ( ( ) )uw u u ξ −/ −

(1 ) ( 1 ) 0u H H uξ ξξ ξ= − / + / − ≥ (Lemma 3.4), we have ( ) 0uw t > (Lemma 3.2) for each 0 ( )t u u ξ< < / − . The

facts that ( ) 0t

uw t dtρ

>∫ for each 0 ( )t u uρ ξ< < < / − ,

and ( ) 0t

uw t dtρ

<∫ for each 0 t< ( )u uρ ξ< < / − imply

that ( )ut u u ξ= / − .

(3) If 0ξ > and 1u Hξξ ξ< ≤ + / , 1ut = .

Since ( ) 1u u ξ/ − > , we can show that ( ) 0uw t > for each 0 1t< < as shown in (2). (4) If 0ξ > and 1 1H u Hξξ + / < ≤ / , 1ut = .

From Lemmas 3.6 and 3.8, we have 11u Ht t /≥ = for each ( 1u Hξξ∈ + / , 1 ]H/ . Therefore, from ( ) 0u uw t = and Lemma 3.2, we have ( ) 0uw t > for each 0 1 ut t< < ≤ . Thus, we obtain the conclusion as shown in (3). (5) If 0ξ > and 0 u ξ< ≤ , 1ut = .

From ( )u a xξ≤ ≤ , we have 1 ( ) 1a x uρ ρ≤ / − +

( ) 1a x t u t≤ / − + for each 0 1tρ< < < . Therefore,

( )log( ( ) 1 ( ( ) 1)) ( ) 0I

a x t u t a x u dF xρ ρ/ − + / / − + >∫ for

each 0 1tρ< < < and ( )log( ( ) 1I

a x t u t/ − +∫

( ( ) 1))a x uρ ρ/ / − + ( ) 0dF x < for each 0 1t ρ< < < . This implies 1ut = .

(6) If 0ξ > and 1 H u E/ < < , u ut t= .

It should be noted that ( ) 1u u ξ/ − > . It is sufficient to

show that ( )u

t

ut

w t dt∫

0< for each 0 1ut t< ≠ < . From

Lemmas 3.6, 3.7 and 3.8, we have 0 1ut< < . Moreover, from ( ) 0u uw t = , we have ( ) 0uw t > for each 0 ut t< < and ( ) 0uw t < for each 1u tt < < . Therefore, we obtain the conclusion.

(7) If 0ξ ≤ and max(0, 1 )H u Eξξ + / < < , u ut t= .

It should be noted that ( ) 1u u ξ/ − ≤ . It is sufficient to

show that ( )u

t

ut

w t dt∫

0< for each 0 ( )ut u ut ξ< ≠ < / − .

From Lemmas 3.2, 3.3 and 3.4, we have 0 ut< ( )u u ξ< / − . Thus, we obtain the conclusion as shown

in (6).

Hereafter, we assume that G < +∞. Thus, it is easy to verify the following corollaries.

Corollary 5.1. Suppose 0ξ ≥ and 1 H u E/ < < , or 0ξ < and max(0, 1 )H u Eξξ + / < < . Then, the optimal proportion of investment ut is uniquely determined by

( ( ) ) ( ( ) ) ( ) 0u uIa x u a x t ut u dF x− / − + =∫ , and the

maximized limit expectation of the growth rate is

( )( )exp log ( ) 1 ( )u uIa x t u t dF x/ − +∫ .


48

Corollary 5.2. Suppose 0ξ < and 0 1u Hξξ< ≤ + / . Then, the optimal proportion of investment is ( )u u ξ/ − , and the maximized limit expectation of the growth rate is exp( log( ( ) ) ( )) ( )

Ia x dF x uξ ξ− / − .∫

Corollary 5.3. Suppose 0ξ ≥ and 0 1u H< ≤ / . Then, the optimal proportion of investment is 1, and the maximized limit expectation of the growth rate is exp( log ( )

Ia x∫

( ))dF x u/ .

Corollary 5.4. Suppose u E≥ . Then, the optimal proportion of investment is 0, and the maximized limit expectation of the growth rate is 1 .

Geometic Pricing

In order to determine the price Xu of the effective random payoff ( ( ) ( ))X a x F x= , , we require the risk-free (simple or continuously compounded) interest rate 0r > for a period. As ( )u uG t is strictly decreasing (Theorems 4.1 and 5.1), the solution of the equation

( ) 1u uG t r= + (if r is simple) or ( )u uG t re= (if r is continuously compounded) is uniquely determined. The latter case ( )u uG t re= is rewritten as:

Theorem 6.1 (the geometric price of a random payoff). Consider an effective random payoff ( ( )X a x= , ( ))F x . The geometric price Xu and optimal proportion of investment,

Xut , with respect to a continuously compounded risk-free

rate 0r > , are uniquely given by the equation

( )0 1

1

sup log ( ) 1 ( )t

t t u

Ia x t u t dF x r

ξ≤ ≤≤ / +

/ − + = ,∫

except for the following two cases, where no solution exists.

(1) 0ξ < , 1 0Hξξ + / ≤ and log( ( ) 1) ( )I

r a x dF xη≥ + ,∫

where η is calculated by the equation

1 ( ( ) 1) ( ) 1I

a x dF xη/ + = .∫

(2) 0ξ < , 1 0Hξξ + / > and

log( ( ) ) ( )) log( )I

r a x dF xξ ξ≥ − − − .∫

To calculate the values Xu and Xut , we have the

following three special cases.

(3) 1Xut = and exp( log ( ) ( ) )X

Iu a x dF x r= −∫ if 0ξ >

and r ≥ log log ( ) ( )I

H a x dF x+ .∫

(4) 1Xut = and exp( log ( ) ( ) )X

Iu a x dF x r= −∫ if 0ξ = ,

H < +∞ and r ≥ log log ( ) ( )I

H a x dF x+ .∫

(5) ( )XX X

ut u u ξ= / − and exp( log( ( ) ) ( ) )X

Iu a x dF x rξ ξ= + − −∫ if

0ξ < , 1 0Hξξ + / > , and log log( ( ) ) ( )I

H a x dF x rξ ξ+ − ≤∫

log( ( ) ) ( )I

a x dF xξ< −∫ log( )ξ− − .

Apart from the five cases listed above, Xu and Xut are

uniquely determined by the simultaneous equations

( )

( ( ) ) ( ) 1

log ( ) 1 ( )I

I

u a x t ut u dF x

a x t u t dF x r

/ − + = ,

/ − + = .

∫∫

In this section we assume that 0 04r = . . It is easy to verify that the following examples are effective.

Example 6.1. Suppose that the payoff and distribution functions are given by ( )a x x= and ( ) [0F x x I= ∈ = , 1] respectively, then 0ξ = , 1 2E = / , and H = +∞ . Set

uy t u= / (0 1 2)u< < / , then the equation ( ) 0u uw t =

can be reduced to 1 (I

xy/∫ 1) 1ut dx− + = . This integral

equation has the solution ( 1) ( 1)y yut e y e= − − / − .

Therefore, we obtain

( ) exp 11 1u u y y

y yG t ye e

= − + , − −

which strictly increases from 1 to +∞ with respect to (0y∈ , )+∞ . The price u should be the solution of the

equation 0 04( )u uG t e .= . Thus, we have 0 9918y . ,≒ 0 4187u .≒ and 0 4152ut .≒ .

In the prevailing pricing theory, 0 04E u e ./ = produce

0 4804u .≒ . In this case, from 1

0( )u uu xt ut u/ − +∫ 1dx =

we have 0 1131ut . ,≒ and ( )u uG t 1 0023.≒ ( 0 04 1 0408e .< .≒ ).

Example 6.2. Suppose that the payoff a or b ( 1 )a b> > occurs with probability p or 1q p= − , respectively. Further assume that 1 1H u E/ < = < (if

0b > ) or 1u E= < (if 0b ≤ ). Then, from 1 1 1 1( 1) ( 1) 1p at t q bt t/ − + + / − + = , we obtain

1 (17)1 1

q pta b

= + ,− −

1 1( ) ( )1 1

q pq pG t a ba b

= − . − −

Samuelson (1971) deals with the case in which 2 7a = . , 0 3b = . , and 0 5p q= = . , where 0 3ξ = . , 1 5E = . ,

1 0 54H/ = . and 1 50 119 0 4202t = / .≒ . However, Samuelson (1971) may have misinterpreted the criterion to be the geometric mean 0 5 0 52 7 0 3 0 9 1. .. . = . < , instead of 0 5 0 5

1 1( ) (2 7 0 3)(0 5 1 7) (0 5 0 7)G t . .= . − . . / . . / .


49

1 1000 1. >≒ .

Example 6.3. Consider a two person matrix game with

the payoff matrix 11 12

21 22

21 39 12

a aa a

− = −

for the row

player A. Under the mixed strategies 11 12

21 22

(1 )(1 ) (1 )(1 )

p p xy x yp p x y x y

− = − − −

, the geometric price

(u u x= , )y and optimal proportion ( )u x yt t ,= of investment are uniquely determined by the simultaneous equations:

( )

( ) 1(18)

log 1

ij iji j

ij iji j

up a t ut u

p a t u t r,

,

/ − + = ,

/ − + = ,

∑

∑

if E= 0ij iji ja p

,>∑ and (x, ) (0y ∉ , 0), (0, 1), (1, 0),

(1, 1). For each 0 1x≤ ≤ there exits 0 1y≤ ≤ such

that 0E ≤ or ( )log 1 0 04ij iji jp a rη

,+ < = . .∑ Thus, the

row player A has no solution for investment (Theorem 6.1 (1)).

In the prevailing pricing theory, rE u e/ = with the mixed strategies 7 15x = / and 1 3y = / produces that

0 045 4 8039u e .= / . ,≒ 0 0084ut .≒ and ( ) 1 0002u uG t .≒ 0 04( 1 0408)e .< . .≒ Under the fixed conditions 7 15x = / , 1 3y = / , and 0 0002r = . , we have the geometric price 4 7885u .≒ with 0 0091ut . .≒

Example 6.4. In the St. Petersburg game (Bernoulli 1954), suppose that the payoff 2 j occurs with probability 1 2 j/ ( 1 2j = , , . ), then ξ 2= , E = +∞, and 1 3H = / . This game is effective, because

1 21(2 ) 2j j

j

∞ /=

/∑ 1 ( 2 1)= / − < +∞ . From Lemma 4.21

we have 1 1( )H HG t/ / 1

1 3 exp( (log 2 ) 2 )j jj

∞

== / × /∑

4 3= / .

Thus, ( )u uG t ( (3u∈ , )+∞ ) strictly decreases from 4 3/ to 1 . The equation 0 04( )u uG t e .= yields the price

5 0815u .≒ . Therefore, if the investors invest 0 1686ut .≒ of their current capital, they can maximize the limit expectation of growth rate to 0 04e . .

In the prevailing pricing theory, 0 04 0 04u E e e. .= / = +∞ / yields the infinity price.

Example 6.5. The lognormal distributed random payoff is given by

2 2( 2)22

1( ) ( ) (19)2

xr xa x Se e dF x e dx

σ

σ

πσ

+ /−= , = ,

and ( )I = −∞,+∞ . In this case, we have rE Se= ,

2

( )rH e Sσ− += / , and ( )exp log ( ) ( )I

a x dF x∫ 2 2rSe σ− /= .

When 100S = , 0 3σ = . , and 0 04r = . , we have 0ξ = , 104 0811E . ,≒ H 0 0105127. ,≒ and 1 95 1229H/ . .≒

From Lemma 4.16, ( )u uG t ( (1u H∈ / , ))E strictly

decreases from ( ) 2 2exp log ( ) ( )I

H a x dF x eσ /=∫ 1 0460.≒

to 1 . The equation 0 04( ) 1 0408u uG t e .= .≒ yields the price 95 6132u .≒ . Therefore, if the investors invest

0 9450ut .≒ of their current capital, then they can maximize the limit expectation of growth rate to 0 04e . .

In the prevailing pricing theory, rE u e/ = yields the (higher) price 100u = ( 95 6132> . ). Under this price, if the investors invest 0 4433ut .≒ of their current capital, they can maximize the limit expectation of growth rate to 1 0088. ( 0 04e .< 1 0408.≒ ). Because 2exp( 2)r σ− /

0 9950.≒ ( 1 0088< . ), the statement (Luenberger 1998) that the expected growth rate is equal to 2 2r σ− / is not necessarily true.

Example 6.6. The European put option is given by 2 2( 2)22

1( ) max( 0) ( )2

x T

TrT xa x K Se e dF x e dxT

σ

σ

π σ

+ /−= − , , = , (20)

and ( )I = −∞,+∞ . We assume that the stock price rT XY Se e= ( (X x= , ( )F x ) is lognormally distributed

with volatility Tσ , where S is the current stock price, r is the continuously compounded interest rate, K is the exercise price of the put option, and T is the exercise period. The expectation E of ( ( )a x , ( ))F x is given by

2 2( 2)22

log1 ( ) (21)2

x TKS T

rT rT xE K Se e e dxT

σ

σ

π σ

+ /− −

−∞= −∫

2 2

2 2log ( ) log ( )S SrTK Kr T r T

KN Se NT T

σ σ

σ σ

+ − + += − − − ,

where 2 2( ) 2

x tN x e dtπ− /

−∞= /∫ is the cumulative

standard normal distribution function.

When 90S = , 120K = , 2T = , 0 1σ = . , and 0 04r = . , we have 0ξ = , E 22 9848. ,≒ and H = +∞ . Therefore, from Theorems 4.1 and 5.1, ( )u uG t ( (0u∈ , )E ) strictly decreases from +∞ to 1 . The equations ( ) 0u uw t = and

0 08( ) rTu uG t e e .= = yield the price 17 8157u .≒ . With this

price, if investors continue to invest ut 0 5434.≒ of their current capital, they can maximize the limit expectation of growth rate to 0 08 1 0833e . . .≒ In the prevailing pricing theory, rTE u e/ = yields the price


50

2 2

2 2log ( ) log ( )S SrT K Kr T r T

u Ke N SNT T

σ σ

σ σ− + − + +

= − − − ,

(22)

which is the Black-Scholes formula for a European put option. Substituting the above-mentioned values for this formula, we obtain the (higher) price 21 2176u .≒ ( 17 8157> . ). With this price, if the investors continue to invest 0 2278ut .≒ of their current capital, they can maximize the limit expectation of growth rate to 1 0096. ( 1 0833< . ) .

REMARK

This paper is a revised form of the pre-print “Game pricing and double sequence of random variables”.

REFERENCES

Algoet, P. H. and Cover, T., “Asymptotic optimality and

asymptotic equipartition properties of log-optimum

investment,” The Annals of Probability, Vol.16, 1988, pp.

876-898.

Bernoulli, D., “Exposition of a new theory of the

measurement of risk, ” Econometrica, Vol. 22, 1954, pp.

23–36.

Black, F. and Scholes, M., “The pricing of options and

corporate liabilities, ” Journal of Political Economy, Vol.

81, 1973, pp. 637-654.

Browne, S. and Whitt, W., “Portfolio choice and the Bayesian

Kelly criterion, ” Advances in Applied Probability, Vol.

28, 1996, pp. 1145-1176.

Carter, M. and Brunt, B. van, “The Lebesgue Stieltjes

integral, ” Springer-Verlag, New York, 2000.

Dutka, J., “On the St. Petersburg paradox, ” Archive for

History of Exact Sciences, Vol. 39, 1998, pp. 13–39.

Feller, W., “An introduction to probability theory and its

application, ” John Wiley and Sons, New York, 1957.

Hirashita, Y., pre-print: “Game pricing and double sequence

of random variables, ”

http://arxiv.org/abs/math/0703076: Accessed 3 Mar 2007, pp.

1–29.

Hirashita, Y., “Delta hedging without the Black-Scholes

formula, ” Far East Journal of Applied Mathematics , Vol.

28, 2007, pp. 157–165.

Hirashita, Y., “Least-squares prices of coin-flipping games,”

International Journal of Applied Mathematics and

Statistics , Vol. 13, 2008, pp. 3–8.

Hirashita, Y., “Geometric price of lognormal distribution, ”

Far East Journal of Applied Mathematics, Vol. 42, 2010,

pp. 61–73.

Hirashita, Y., “Geometric pricing of Games, ” International

Journal of Applied Mathematics and Statistics , Vol. 25,

2012, pp. 35–43.

Karatzas, I. and Shreve, S. E., “Methods of mathematical

finance,” Springer-Verlag, New York, 1998.

Kelly, J. L. Jr., “A new interpretation of information rate, ”

Bell system Technical Journal , Vol. 35, 1956, pp. 917–926.

Luenberger, D. G., “A preference foundation for log mean-

variance criteria in portfolio choice problems,” Journal of

Economic Dynamics and Control, Vol. 17, 1993, pp. 887-

906.

Luenberger, D. G., “Investment science, ” Oxford University

Press, Oxford, 1998.

Robbins, H., “Recurrent games and the Petersburg paradox,”

Annals of Mathematical Statistics , Vol. 32, 1961, pp. 187–

194.

Rotando, L. M. and Thorp, E. O., “The Kelly criterion and the

stock market,” American Mathematical Monthly , Vol. 99,

1992, pp. 922-931.

Samuelson, P., “The fallacy of maximizing the geometric

mean in long sequences of investing or gambling,”

Proceedings of the National Academy of Sciences of the

United States of America, Vol. 68, 1971, pp.2493–2496.

Samuelson, P., “St. Petersburg paradoxes: defanged,

dissected, and historically described,” Journal of

Economic Literature , Vol.15, 1977, pp. 24–55.

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Game Pricing and Double Sequence of Random Variables

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