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9. Two Functions of Two RandomVariables
In the spirit of the previous lecture, let us look at animmediate generalization: Suppose X and Y are two random
variables with joint p.d.f Given two functionsand define the new random variables
How does one determine their joint p.d.f Obviouslywith in hand, the marginal p.d.fs andcan be easily determined.
(9-1)
).,( y x f XY
).,(
),(
Y X hW
Y X g Z
),( y x g
),,( y xh
(9-2)
?),( w z f ZW ),( w z f ZW )( z f Z )(w f W
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Example 9.1: Suppose X and Y are independent uniformlydistributed random variables in the intervalDefine Determine
Solution: Obviously both w and z vary in the intervalThus
We must consider two cases: and since theygive rise to different regions for (see Figs. 9.2 (a)-(b)).
).,max( ),,min( Y X W Y X Z
.0or0 if ,0),( w z w z F ZW
).,( w z f ZW
).,0(
).,0(
(9-4)
.),max( ,),min(,),( wY X z Y X P wW z Z P w z F ZW (9-5)
z w , z w
w z D ,
X
Y
w y
),( ww
),( z z
z wa
)(
X
Y
),( ww
),( z z
z wb )(Fig. 9.2 PILLAI
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For from Fig. 9.2 (a), the region is represented by the doubly shaded area. Thus
and for from Fig. 9.2 (b), we obtain
With
we obtain
Thus
, ,),(),(),(),( z w z z F z w F w z F w z F XY XY XY ZW
w z D ,
(9-6)
, z w
, z w. ,),(),( z www F w z F XY ZW (9-7)
,)()(),( 2 xy y x y F x F y x F Y X XY (9-8)
2
2 2
(2 ) / , 0 ,( , )
/ , 0 . ZW w z z z w
F z ww w z
.otherwise,0
,0,/2),(
2 w z w z f ZW
(9-9)
(9-10)
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From (9-10), we also obtain
and
If and are continuous and differentiablefunctions, then as in the case of one random variable (see (5-30)) it is possible to develop a formula to obtain the joint
p.d.f directly. Towards this, consider the equations
For a given point (z,w), equation (9-13) can have manysolutions. Let us say
,0 ,12
),()(
z z
dww z f z f z ZW Z
(9-11)
.),( ,),( w y xh z y x g (9-13)
.0 ,2
),()(
0 2
w
wdz w z f w f
w
ZW W (9-12)
),( y x g ),( y xh
),( w z f ZW
),,( , ),,( ),,( 2211 nn y x y x y x PILLAI
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represent these multiple solutions such that (see Fig. 9.3)
(9-14).),( ,),( w y xh z y x g iiii
Fig. 9.3
Consider the problem of evaluating the probability
z
(a)
w
),( w z z
www
z z
.),(,),(
,
wwY X hw z z Y X g z P
wwW w z z Z z P (9-15)
(b)
x
y1
2
i
n
),( 11 y x
),( 22 y x
),( ii y x
),( nn y x
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Using (7-9) we can rewrite (9-15) as
But to translate this probability in terms of we need
to evaluate the equivalent region for in the xy plane.Towards this referring to Fig. 9.4, we observe that the point
A with coordinates ( z ,w) gets mapped onto the point withcoordinates (as well as to other points as in Fig. 9.3(b)).As z changes to to point B in Fig. 9.4 (a), letrepresent its image in the xy plane. Similarly as w changesto to C , let represent its image in the xy plane.
(9-16)
),,( y x f XY
.),(, w z w z f wwW w z z Z z P ZW
w z
A),( ii y x
z z B
ww C
(a)
w
A z
ww
B
z z
C D
Fig. 9.4 (b)
y
Ai y B
xi x
C D
i
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As the point ( z ,w) goes to the pointthe point and the pointHence the respective x and y coordinates of are given by
and
Similarly those of are given by
The area of the parallelogram in Fig. 9.4 (b) isgiven by
,),( A y x ii ,),( Bw z z
,),( C ww z .),( Dww z z
B
,),(),( 1111 z z g x z
z g w z g w z z g i
.),(),(11
11 z z h
y z z h
w z hw z z h i
(9-20)
(9-21)
C
. , 11 w
w
h yw
w
g x ii (9-22)
DC B A
.cossinsincos
)sin(
C A B AC A B A
C A B Ai (9-23)
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But from Fig. 9.4 (b), and (9-20) - (9-22)
so that
and
The right side of (9-27) represents the Jacobian ofthe transformation in (9-19). Thus
.cos ,sin
,sin ,cos
11
11
ww
g C A z z h B A
wwh
C A z z g
B A
(9-25)
(9-24)
w z z
h
w
g
w
h
z
g i
1111 (9-26)
wh
z h
w g
z g
z
h
w
g
w
h
z
g
w z
i
11
11
1111 det (9-27)
( , ) J z w
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Next we shall illustrate the usefulness of the formula in(9-29) through various examples:
Example 9.2: Suppose X and Y are zero mean independentGaussian r.vs with common varianceDefine whereObtain
Solution: Here
Since
if is a solution pair so is From (9-33)
),/(tan , 122 X Y W Y X Z .2
).,( w z f ZW
.2
1),(
222 2/)(2
y x
XY e y x f (9-32)
,2/||),/(tan),(;),(122
w x y y xhw y x y x g z (9-33)),( 11 y x ).,( 11 y x
.tanor,tan w x yw x y
(9-34)
.2/|| w
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Substituting this into z , we get
and
Thus there are two solution sets
We can use (9-35) - (9-37) to obtain From (9-28)
so that
.cos or,sec tan1 222 w z xw xw x y x z (9-35)
.sintan w z w x y (9-36)
.sin ,cos ,sin ,cos 2211 w z yw z xw z yw z x (9-37)
).,( w z J
,cossin
sincos),( z
w z ww z w
w y
z y
w x
z x
w z J (9-38)
.|),(| z w z J (9-39) PILLAI
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We can also compute using (9-31). From (9-33),
Notice that agreeing with (9-30).Substituting (9-37) and (9-39) or (9-40) into (9-29), we get
Thus
which represents a Rayleigh r.v with parameter and
),( y x J
.11
),(22
2222
2222
z y x
y x x y x y
y x
y
y x
x
y x J (9-40)
|,),(|/1|),(| ii y x J w z J
.
2|| ,0 ,
),(),(),(22 2/
2
2211
w z e z
y x f y x f z w z f z
XY XY ZW
(9-41)
,0 ,),()(22 2/
2
2/
2/
z e z dww z f z f z ZW Z
(9-42)
,2
,2|| ,1
),()(
0
wdz w z f w f ZW W (9-43)PILLAI
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which represents a uniform r.v in the intervalMoreover by direct computation
implying that Z and W are independent. We summarize theseresults in the following statement: If X and Y are zero meanindependent Gaussian random variables with common
variance, then has a Rayleigh distribution andhas a uniform distribution. Moreover these two derived r.vsare statistically independent. Alternatively, with X and Y asindependent zero mean r.vs as in (9-32), X + jY represents acomplex Gaussian r.v. But
where Z and W are as in (9-33), except that for (9-45) to holdgood on the entire complex plane we must haveand hence it follows that the magnitude and phase of
).2/,2/(
22
Y X )/(tan 1
X Y
, jW Ze jY X (9-45)
)()(),( w f z f w z f W Z ZW (9-44)
, W PILLAI
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a complex Gaussian r.v are independent with Rayleighand uniform distributions ~ respectively. Thestatistical independence of these derived r.vs is an interesting
observation.Example 9.3: Let X and Y be independent exponentialrandom variables with common parameter .Define U = X + Y , V = X - Y . Find the joint and marginal
p.d.f of U and V .Solution: It is given that
Now since u = x + y, v = x - y, always and there isonly one solution given by
Moreover the Jacobian of the transformation is given by
.0 ,0 ,1
),( /)(2 y xe y x f y x XY
(9-46)
.2
,2
vu y
vu x
(9-47)
,|| uv
),( U
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and hence
represents the joint p.d.f of U and V . This gives
and
Notice that in this case the r.vs U and V are not independent.
As we show below, the general transformation formula in(9-29) making use of two functions can be made useful even
when only one function is specified.
2 11
1 1),( y x J
,||0 ,21),( /2 uvevu f uUV (9-48)
,0 ,2
1),()( /2
/2
ueudvedvvu f u f uu
u
uu
u UV U
(9-49)
/ | |/2 | | | |
1 1( ) ( , ) , .
2 2u v
V UV v v f v f u v du e du e v
(9-50)
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Auxiliary Variables :
Suppose
where X and Y are two random variables. To determine by making use of the above formulation in (9-29), we can
define an auxiliary variable
and the p.d.f of Z can be obtained from by proper
integration.
Example 9.4: Suppose Z = X + Y and let W = Y so that thetransformation is one-to-one and the solution is given
by
),,( Y X g Z
Y W X W or
(9-51)
(9-52)
),( w z f ZW
. , 11 w z xw y
)( z f Z
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The Jacobian of the transformation is given by
and hence
or
which agrees with (8.7). Note that (9-53) reduces to theconvolution of and if X and Y are independentrandom variables. Next, we consider a less trivial example.
Example 9.5: Let and be independent.Define
1 1 0
1 1),( y x J
),(),(),( 11 ww z f y x f y x f XY XY ZW
,),(),()( dwww z f dww z f z f XY ZW Z (9-53)
)( z f X )( z f Y
)1,0( U X )1,0( U Y
).2cos(ln2 2/1 Y X Z (9-54)PILLAI
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Find the density function of Z .Solution: We can make use of the auxiliary variable W = Y in this case. This gives the only solution to be
and using (9-28)
Substituting (9-55) - (9-57) into (9-29), we obtain
,,
1
2/)2sec(1
2
w ye x w z (9-55)
(9-56)
.)2(sec
10
)2(sec
),(
2/)2sec(2
12/)2sec(2
11
11
2
2
w z
w z
ew z
w x
ew z
w y
z y
w x
z x
w z J
(9-57)
2sec(2 ) / 22( , ) sec (2 ) ,
, 0 1,
z w ZW f z w z w e
z w
(9-58)
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and
Let so that Noticethat as w varies from 0 to 1, u varies from toUsing this in (9-59), we get
which represents a zero mean Gaussian r.v with unit
variance. Thus Equation (9-54) can be used asa practical procedure to generate Gaussian random variablesfrom two independent uniformly distributed randomsequences.
221 1 tan(2 ) / 2/ 2 2 0 0
( ) ( , ) sec (2 ) . z w z Z ZW f z f z w dw e z w e dw
tan(2 )u z w 22 sec (2 ) .du z w dw .
, ,21
2
21)( 2/
1
2/2/ 222 z eduee z f z u z Z
(9-59)
(9-60)
).1,0( N Z
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Example 9.6 : Let X and Y be independent identically distributedGeometric random variables with
(a) Show that min ( X , Y ) and X Y are independent random variables.(b) Show that min ( X , Y ) and max ( X , Y ) min ( X , Y ) are also
independent random variables.Solution: (a) Let
Z = min ( X , Y ) , and W = X Y . Note that Z takes only nonnegative values while W takes both positive, zero and negative values We have P ( Z = m, W = n) = P {min ( X , Y ) = m, X Y = n}. But
Thus
},,21,0,{ }.,21,0,{
.,2,1,0 ,)()( k pq K Y P k X P k
negative.is enonnegativis ),min(
Y X W Y X X Y X W Y X Y Y X Z
),,),(min(
),,),(min(
)}(,,),{min(),(
Y X nY X mY X P
Y X nY X mY X P
Y X Y X nY X mY X P nW m Z P
(9-61)
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,2,1,0 ,2,1,0 ,
0,0,)()(
0,0,)()(
),,(
),,(),(
||22 nmq p
nm pq pqnmY P m X P
nm pq pqmY P nm X P
Y X nmY m X P
Y X nm X mY P nW m Z P
nm
nmm
mnm
(9-63)
PILLAI
represents the joint probability mass function of the random variables Z and W . Also
Thus Z represents a Geometric random variable sinceand),1(1 2 q pq
.,2,1,0 ,)1(
)1()1(
)221(
),()(
2
222
222
||22
12
mqq p
q pqq p
qqq p
qq pnW m Z P m Z P
m
mm
m
n n
nm
(9-64)
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2 2 | |
0 0
2 | | 2 4 2 | | 12
| |1
1
( ) ( , )
(1 )
, 0, 1, 2, .
m n
m m
n n
nq
pq
P W n P Z m W n p q q
p q q q p q
q n
PILLAI
(9-65)
Note that
establishing the independence of the random variables Z and W .The independence of X Y and min ( X , Y ) when X and Y are
independent Geometric random variables is an interesting observation.(b) Let Z = min ( X , Y ) , R = max ( X , Y ) min ( X , Y ).
In this case both Z and R take nonnegative integer valuesProceeding as in (9-62)-(9-63) we get
),()(),( nW P m Z P nW m Z P (9-66)
(9-67).,21,0,
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.0 ,,2,1,0 , ,2,1 ,,2,1,0 ,2
0 ,,2,1,0 ,
,2,1 ,,2,1,0 ,
},,(),,{
},,(),,{
},),min(),max(,),{min(
},),min(),max(,),{min(},{
22
22
nmq pnmq p
nm pq pq
nm pq pq pq pq
Y X nmY m X P Y X mY nm X P
Y X nmY m X P Y X nm X mY P
Y X nY X Y X mY X P
Y X nY X Y X mY X P n Rm Z P
m
nm
mnm
nmmmnm
(9-68)
Eq. (9-68) represents the joint probability mass function of Z and R in (9-67). From (9-68),
,2,1,0 ,)1(
1)21(},{)(
2
0
22
1
22 2
mqq p
q pqq pn Rm Z P m Z P
m
n
m
n
nm pq
(9-69)PILLAI
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(9-71)
and
From (9-68)-(9-70), we get
which proves the independence of the random variables Z and R defined in (9-67) as well.
)()(),( n R P m Z P n Rm Z P
.,2,1 ,
0 ,},{)(
12
1
0 nq
nn Rm Z P n R P
nm q
p
q p
(9-70)