11

Continuous Distributions

ch4ch4

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A random variable X of the continuous type has a support or space S that is an interval(possibly unbounded) or a union of intervals, instead of a set of real numbers (discrete points).

The probability density function(p.d.f.) of X is an integrable function f(x) satisfying: f(x)>0, x S.∈ ∫Sf(x)dx= 1, P(X A) = ∫∈ Af(x)dx, where A S.⊂

Ex.4.1-1: Assume X has the p.d.f. P(X>20)=?

The distribution function of X is(cumulative distribution function)(c.d.f.)

If the derivative F’(x) exists,

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Zero Probability of Points For the random variable of the continuous type, the probability of any po

int is zero. Namely, P(X=b)=0. Thus, P(a≤X≤b) = P(a<X<b) = P(a≤X<b) = P(a<X≤b) = F(b)-F(a). For instance,

Let X be the times between calls to 911. 105 observations aremade to construct a relative frequent histogram h(x). It is compared with the exponential model in Ex4.1-1.

30 17 65 8 38 35 4 19 7 14 12 4 5 4 2 7 5 12 50 33 10 15 2 10 1 5 30 41 21 31 1 18 12 5 24 7 6 31 1 3 2 22 1 30 2 13 12 129 28 6 5063 5 17 11 23 2 46 90 13 21 55 43 5 19 47 24 4 6 27 4 6 37 16 41 68 9 5 28 42 3 42 8 52 2 11 41 4 35 21 3 17 10 16 1 68 105 45 23 5 10 12 17

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(Dis)continuity, Differentiable, Integration Ex4.1-4: Let Y be a continuous random variable

with p.d.f. g(y)=2y, 0<y<1. The distribution function of Y is P(½≤Y≤¾)=G(¾)-G(½)=5/16. P(¼≤Y<2)=G(2)-G(¼)=15/16.

Properties of a continuous random variable: The area between the p.d.f. f(x) and x-axis must equal 1.

f(x) is possibly unbounded (say, >1). f(x) can be discontinuous function (defined over a set of

intervals), However, its c.d.f. F(x) is always continuous since integration. It is possible that F’(x) does not exist at x=x0.

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Mean, Variance, Moment-generating Fn Suppose X is a continuous random variable.

The expected value of X, the mean of X is The variance of X is The standard deviation of X is The moment-generating function of X, if it exists, is

The rth moment E(Xr) exists and is finite E(X⇒ r-1), …, E(X1) do. The converse is not true.

E(etX) exists and is finite –h<t<h all the moments do.⇒ The converse is not necessarily true.

Ex4.1-5: Random variable Y with p.d.f. g(y)=2y, 0<y<1. (Ex4.1-4)

66

Percentile (π) and Quartiles Ex.4.1-6: X has p.d.f. f(x)=xe-x, 0≤x<∞.

The (100p)th percentiles a number πp s.t. the area under f(x) to the left of πp is p. The 50th percentile is called the median: m = π.5.

The 25th and 75th percentiles are called the first and third quartiles. Namely, π.25= q1, π.75= q3, and m = π.5= q2 the second quartile.

Ex4.1-7: X with p.d.f. f(x)=1-|x-1|, 0≤x<2.To find 32rd percentile π.32 is to solve F(π.32)=.32

∵F(1)=.5>.32

To find 92rd percentile π.92 is to solve F(π.92)=.92

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More Example

Ex4.1-8: X has the p.d.f. f(x)=e-x-1, -1<x<∞.

The median m = π.5 is

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Uniform Distribution Random variable X has a uniform distribution if its p.d.f. equals a con

stant on its support. If the support is the interval [a, b], then p.d.f. ⇒ This distribution, denoted as U(a, b), is also referred to as rectangular due to

the shape of f(x). The mean, variance, distribution function and moment-generating function a

re

Pseudo-random number generator: a program applies simple arithmetical operations on the seed (starting number) to deterministically generate a sequence of numbers, whose distribution follows U(0, 1). Table IX on p.665 shows an example of these (random) numbers*104. Ex.4.2-1: X has p.d.f. f(x)=1/100, 0<x<100, namely U(0,100).

The mean and variance are μ=(0+100)/2=50, σ2=10000/12. The standard deviation is , 100 times of U(0,1).

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Exponential Distribution The waiting (inter-change) times W between successive changes

whose number X in a given interval is a Poisson distribution is indeed an exponential distribution.

Such time is nonnegative the distribution function F(w)=0 for w<0.⇒ For w ≥0, F(w) = P(W≤w) = 1 -P(W>w) = 1 -P(no changes in [0, w])

= 1 –e–λw, For w>0, the p.d.f. f(w) = F’(w) = λe–λw⇒

Suppose λ=7, the mean number of changes per minute;

⇒θ= 1/7, the mean waiting time for the first (next) change.

Ex4.2-2: X has an exponential distribution with a mean of θ=20.

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Examples Ex.4.2-3: Customers arrivals follow a Poisson process of 20 per hr.

What is the probability that the shopkeeper will have to wait more than 5 minutes for the arrival of the first (next) customer?

Let X be the waiting time in minutes until the next customer arrives.

Having awaited for 6 min., what is the probability that the shopkeeper will have to wait more than 3 min. additionally for the new arrival? Memory-less, forgetfulness property!

Percentiles: To exam how close an empirical collection of data is to the exponen

tial distribution, the q-qplot (yr,πp) from the ordered statistics can be constructed, where p=r/(n+1), r=1,2,…,n.

If θis unknown, πp=-ln(1-p) can be used in the plot, instead. As the curve ≈a straight line, it matches well. (Slope: an estimate of 1/θ)

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Gamma DistributionGamma Distribution Generalizing the exponential distribution, the Gamma distribution considGeneralizing the exponential distribution, the Gamma distribution consid

ers the waiting time W until the αers the waiting time W until the αth th change occurs, α≥1.change occurs, α≥1. The distribution function F(w) of W is given byThe distribution function F(w) of W is given by

Leibnitz's rule:Leibnitz's rule:

The Gamma function is defined byThe Gamma function is defined by

F(w)

F’(w)

generalized factorial

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Fundamental CalculusFundamental Calculus There are some formula from Calculus:There are some formula from Calculus:

Generalized integration by parts (also ref. p.637):Generalized integration by parts (also ref. p.637):

Formula used in the Gamma distribution:Formula used in the Gamma distribution:

Pf:Pf:

SupposeSuppose Then,Then,

By the induction hypothesis, the equation holds!By the induction hypothesis, the equation holds!

α=1:

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Chi-square DistributionChi-square Distribution A A Gamma Gamma distribution with θ=2, α=r/2, (r N) is a ∈distribution with θ=2, α=r/2, (r N) is a ∈ Chi-Chi-

square distribution square distribution with with r degrees of freedomr degrees of freedom, denoted , denoted as χas χ22(r)(r).. The mean μ=r, and variance σThe mean μ=r, and variance σ22=2r.=2r.

The mode, the point for the maximal p.d.f., is x=r-2The mode, the point for the maximal p.d.f., is x=r-2

Ex.4.3-3: X has a chi-square distribution with r=5.Ex.4.3-3: X has a chi-square distribution with r=5. Using Table IV on p.655 Using Table IV on p.655 P(X>15.09)1-F(15.09)=1-0.99=0.01.P(X>15.09)1-F(15.09)=1-0.99=0.01.

P(1.145 X 12.83)=F(12.83)-F(1.145)=0.975-0.05=0.925.≦ ≦P(1.145 X 12.83)=F(12.83)-F(1.145)=0.975-0.05=0.925.≦ ≦ Ex.4.3-4: X is χEx.4.3-4: X is χ22(7)(7). Suppose there are two constants a & b s.t. . Suppose there are two constants a & b s.t.

P(a<X<b)=0.95. One of many possible is a=1.69 & b=16.01⇒P(a<X<b)=0.95. One of many possible is a=1.69 & b=16.01⇒

Percentiles:Percentiles: The 100(1-α) percentile isThe 100(1-α) percentile is The 100αpercentile isThe 100αpercentile is

f(x)

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Normal DistributionNormal Distribution The random variable X has a The random variable X has a normal normal

distributiondistribution, N(μ, σ, N(μ, σ22), if its p.d.f. is defined ), if its p.d.f. is defined byby where the mean μ (-∞, ∞), and variance σ∈where the mean μ (-∞, ∞), and variance σ∈ 22 (0, ∞).∈(0, ∞).∈

If Z is N(0, 1), Z is called as a standard If Z is N(0, 1), Z is called as a standard normal distribution.normal distribution.

M(t)=

M’(t)=

E(X)=

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ExamplesExamples Ex.4.4-3: If Z is N(0,1), then from Table Va on p.656,Ex.4.4-3: If Z is N(0,1), then from Table Va on p.656,

Φ(1.24)=P(Z 1.24)=0.8925, ≦Φ(1.24)=P(Z 1.24)=0.8925, ≦ P(P(1.24 Z 2.37)=Φ (2.37)-Φ (1.24)=0.9911-0.8925=0.0986≦ ≦1.24 Z 2.37)=Φ (2.37)-Φ (1.24)=0.9911-0.8925=0.0986≦ ≦ P(-2.37 Z -1.24)= P(1.24 Z 2.37)=0.0986≦ ≦ ≦ ≦P(-2.37 Z -1.24)= P(1.24 Z 2.37)=0.0986≦ ≦ ≦ ≦

From Table Vb (right-tail) on p.657, (the negative z is deriveFrom Table Vb (right-tail) on p.657, (the negative z is derived from –Z)d from –Z) P(ZP(Z>1.24)=0.1075 P(Z -2.14)=P(Z 2.14)=0.0162≦ ≧>1.24)=0.1075 P(Z -2.14)=P(Z 2.14)=0.0162≦ ≧

Ex.4.4-4: If Z is N(0,1), then find a & b from Table Va & Vb.Ex.4.4-4: If Z is N(0,1), then find a & b from Table Va & Vb. P(Z a)=0.9147 => a=1.37≦P(Z a)=0.9147 => a=1.37≦ P(Z b)=0.0526 => b=1.62≧P(Z b)=0.0526 => b=1.62≧

Percentiles:Percentiles: The 100(1-α) percentile is The 100(1-α) percentile is ZZαα, where P(Z Z≧, where P(Z Z≧ αα)=α=P(Z -Z≦)=α=P(Z -Z≦ αα) Z) Z1-α1-α=-Z=-Zαα

(the upper 100αpercent point)(the upper 100αpercent point)

The 100p percentile is πThe 100p percentile is πpp, where P(X, where P(X≦≦ππpp)=p )=p p=1-α =>πp=1-α =>πpp=Z=Z1-p1-p=-Z=-Zpp

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Conversion: N(μ,σConversion: N(μ,σ22) ) ⇒ N(0,1)) ) ⇒ N(0,1) Thm4.4-1: If X is N(μ,σThm4.4-1: If X is N(μ,σ22), then Z=(X-μ)/σis N(0,1).), then Z=(X-μ)/σis N(0,1).

Usage:Usage:

Ex.4.4-6: X is N(3,16). Ex.4.4-6: X is N(3,16).

Ex.4.4-7: X is N(25,36). Find c such that P(|X-25|≤c) Ex.4.4-7: X is N(25,36). Find c such that P(|X-25|≤c) =0.9544.=0.9544.

Within 2*σabout the mean, X and Z share the same the Within 2*σabout the mean, X and Z share the same the probability.probability.

P(4X8)

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Conversion N(μ,σConversion N(μ,σ22) ⇒ χ) ⇒ χ22(1)(1) Thm4.4-2: If X is N(μ,σThm4.4-2: If X is N(μ,σ22), then V=[(X-μ)/σ]), then V=[(X-μ)/σ]22=Z=Z22is χis χ22(1).(1).

Pf: V=ZPf: V=Z22, Z is N(0,1). Then G(v) of V is, Z is N(0,1). Then G(v) of V isG(V)G(V)

Ex.4.4-8: If Z is N(0,1), k=? So Ex.4.4-8: If Z is N(0,1), k=? So From the chi-square table with r=1, kFrom the chi-square table with r=1, k22=3.841, so k=1.96=3.841, so k=1.96..

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Q-Q PlotQ-Q Plot Given a set of observations, how to determine its distribuGiven a set of observations, how to determine its distribu

tion?tion? If its population is large, relative frequency histogram can be useIf its population is large, relative frequency histogram can be use

d.d.

For small samples, a q-q plot can be used to check.For small samples, a q-q plot can be used to check.

Q-Q plot: say against a “theoretical” normal distribution NQ-Q plot: say against a “theoretical” normal distribution N(0,1).(0,1). The mean and variance are obvious, N(0,1): the quantile zThe mean and variance are obvious, N(0,1): the quantile z1-p1-p..

The sample is suspected as N(μ,σThe sample is suspected as N(μ,σ22): the quantile q): the quantile qpp..

The ideal curve in the plot would be qThe ideal curve in the plot would be qpp=μ+σz=μ+σz1-p1-p..

If the actual curve is approximated to a straight line, the distribution If the actual curve is approximated to a straight line, the distribution of the data is verified as a normal distribution.of the data is verified as a normal distribution.

The reciprocal of the slope is σ.The reciprocal of the slope is σ.

The interception on the x-axis is μ. (Ref. Ex.4.4-9 & Fig.4.4-3 on p.2The interception on the x-axis is μ. (Ref. Ex.4.4-9 & Fig.4.4-3 on p.201)01)

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Distributions of Functions of Random Distributions of Functions of Random VariablesVariables From a known random variable X, we may consider From a known random variable X, we may consider

another Y = u(X), a function of X, and want to know another Y = u(X), a function of X, and want to know Y’s distribution function.Y’s distribution function. Distribution function technique:Distribution function technique:

We directly find G(y) = P(Y≤y) = P[u(X)≤y], and g(y)=G’(y).We directly find G(y) = P(Y≤y) = P[u(X)≤y], and g(y)=G’(y).

E.g., finding the gamma distribution from the Poisson distribution.E.g., finding the gamma distribution from the Poisson distribution.

Also, N(μ,σAlso, N(μ,σ22) N(0,1), and N(μ,σ⇒) N(0,1), and N(μ,σ⇒ 22) χ⇒) χ⇒ 22(1).(1).

It requires the knowledge of the related probability models.It requires the knowledge of the related probability models.

Change-of-variable technique:Change-of-variable technique: Find the inverse function X=v(Y) from Y=u(X), and Find the inverse function X=v(Y) from Y=u(X), and

Find the mapping: boundaries, one-to-one, two-to-one, etc.Find the mapping: boundaries, one-to-one, two-to-one, etc.

It requires the knowledge of calculus and the like.It requires the knowledge of calculus and the like.

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Distribution Function Distribution Function TechniqueTechnique Ex4.5-1: [Lognormal] X is N(μ,σEx4.5-1: [Lognormal] X is N(μ,σ22). If W=e). If W=eXX, G(w)=?, g(w)=?, G(w)=?, g(w)=?

Ex4.5-2: Let w be the smallest angle between the y-axis and the spinner,Ex4.5-2: Let w be the smallest angle between the y-axis and the spinner, and have a uniform distribution on (-π/2, π/2). and have a uniform distribution on (-π/2, π/2).

G(w)=

g(w)=

<=Cauchy p.d.f.

Both limits E(x) does not exist⇒

g(x)

(0,1)

y

w

x

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Change-of of-variable Change-of of-variable TechniqueTechnique Suppose X is a R.V. with p.d.f. f(x) with support cSuppose X is a R.V. with p.d.f. f(x) with support c11< x < c< x < c22..

Y=u(X) ⇔the inverse X=v(Y) with support dY=u(X) ⇔the inverse X=v(Y) with support d11< y < d< y < d22..

Both u and v are conti. increasing functions, and dBoth u and v are conti. increasing functions, and d11=u(c=u(c11), d), d22=u(c=u(c22).).

Suppose both u and v are conti. decreasing functions:Suppose both u and v are conti. decreasing functions:

The mapping of cThe mapping of c11< x < c< x < c2 2 would be dwould be d11> y > d> y > d22

Generally, the support mapped from.Generally, the support mapped from.

G(y)=

g(y)= Ex4.5-1 is an example.

G(y)=…=

g(y)=G’(y)

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Conversions: Any ⇔ U(0,1)Conversions: Any ⇔ U(0,1) Thm.4.5-2: X has F(x) that is strictly increasing on Sx={x: a<xThm.4.5-2: X has F(x) that is strictly increasing on Sx={x: a<x

<b}.Then R.V. Y, defined by Y=F(X), has a distribution U(0,1).<b}.Then R.V. Y, defined by Y=F(X), has a distribution U(0,1). Pf: The distribution function of Y isPf: The distribution function of Y is

The requirement that F(x) is strictly increasing can be dropped.The requirement that F(x) is strictly increasing can be dropped.

It will take tedious derivations to exclude the set of intervals as f(x)=0.It will take tedious derivations to exclude the set of intervals as f(x)=0.

The The change-of-variable techniquechange-of-variable technique can be applied to the discr can be applied to the discrete type R.V.ete type R.V. Y=u(X), X=v(Y): there exists one-to-one mapping.Y=u(X), X=v(Y): there exists one-to-one mapping.

The p.m.f. of Y is g(y)=P(Y=y)=P[u(X)=y]=P[X=v(y)]=f[v(y)], y S∈The p.m.f. of Y is g(y)=P(Y=y)=P[u(X)=y]=P[X=v(y)]=f[v(y)], y S∈ yy.. There is no term “|v’(y)|”, since f(x) presents the probability.There is no term “|v’(y)|”, since f(x) presents the probability.

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ExamplesExamples Ex.4.5-6: X is a Poisson with λ=4.Ex.4.5-6: X is a Poisson with λ=4.

If Y=XIf Y=X1/21/2, X=Y, X=Y22, then, then

When the transformation Y=u(X) is not one-to-one, say When the transformation Y=u(X) is not one-to-one, say V=ZV=Z22.. For instance, Z is N(0,1): -∞<z<∞, 0≤v<∞. [2-to-1 mapping]For instance, Z is N(0,1): -∞<z<∞, 0≤v<∞. [2-to-1 mapping]

Each interval (case) is individually consideredEach interval (case) is individually considered.. Ex.4.5-7: X has p.d.f. f(x)=xEx.4.5-7: X has p.d.f. f(x)=x22/3, -1<x<2./3, -1<x<2.

If X=YIf X=Y1/21/2, Y=X, Y=X22, then 0≤y<4:, then 0≤y<4: -1 < x-1 < x11< 0 ⇔0 ≤y< 0 ⇔0 ≤y11< 1< 1

0 ≤x0 ≤x22< 1 ⇔0 ≤y< 1 ⇔0 ≤y22< 1< 1

1 ≤x1 ≤x33< 2 ⇔1 ≤y< 2 ⇔1 ≤y33< 4< 4

G(v)=

g(v)=

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How to find E(X) & Var(X) nowHow to find E(X) & Var(X) now Ex4.6-4: Find the mean and variance of X in previous example.Ex4.6-4: Find the mean and variance of X in previous example.

Ex4.6-5: Reinsurance companies may agree to cover the wind Ex4.6-5: Reinsurance companies may agree to cover the wind damages that ranges between $2 and $10 million.damages that ranges between $2 and $10 million. X is the loss in million and has a distribution function:X is the loss in million and has a distribution function: If losses beyond $10 is set as $10, thenIf losses beyond $10 is set as $10, then

The cases (x>10) will all be attributed to x=10: P(X=10)=1/8.The cases (x>10) will all be attributed to x=10: P(X=10)=1/8.

F(x)=

μ=E(X)

σ2=E(X2)-μ2