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Handout Ch5(2)實習

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Normal Distribution• There are three reasons why normal distribution is important

– Mathematical properties of the normal distribution have simple forms

– Many random variables often have distributions that are approximately normal

– Central limit theorem tells that many sample functions have distributions which are approximately normal

• The p.d.f. of a normal distribution

2

1exp

)2(

1),|(

2

2/12

x

xf

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Properties of Normal Distribution

•

Proof: If we let y = (x–)/, then

.1),|( 2

dxxf

dyydyydxxf

)

2

1exp(

2

1 )

2

1exp(

2

1),|( 222

dyyI

)

2

1exp(Let 2

2 )

2

1exp(

)sin and cos (choose )](2

1 exp[

)2

1exp( )

2

1exp(

0

22

0

22

222

drdrr

rzrydzdyzy

dzzdyyIII

.1 ) ,|( 2

dxxf

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極座標

• 直角座標與極座標的轉換• ∫ ∫dxdy=rdrdθ

• X=r*cosθ， y=r*sinθ

• Ex: x^2+y^2 1≦(x,y)

r

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•

•

)2

1exp(

2

)]([exp

2

1)

2

1 exp()( so

2

)]([

2

1

2

)( since

2

)(exp

2

1)()(

22

2

2222

2

2222

2

2

2

2

tt

dxtx

ttt

txtt

uxtx

dxx

txeEt tX

)0()(XE

22)]0([)0()(Var X

The m.g.f. of Normal Distribution

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Properties of Normal Distribution

• If the random variables X1, …, Xk are independent and if Xi has a normal distribution with mean i and variance i

2, then the sum X1+ . . .+ Xk has a normal distribution with mean 1 + . . .+ k and variance 1

2 + . . .+ k2.

Proof:

• The variable a1x1 + . . .+ akxk+ b has a normal distribution with mean a11 + . . .+ akk + b and variance a1

212 + . . .+ ak

2k2

• Suppose that X1, …, Xn form a random sample from a normal distribution with mean and variance 2 , and let denote the sample mean. Then

has a normal distribution with mean and 2/n.

2 2 2 2

1 11 1

1 1( ) ( ) exp exp

2 2

k k k k

i i i i ii ii i

t t t t t t

nX

nX

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5.6.16

• Suppose that the joint p.d.f. of two random variables X and Y is

• Find

2 2(1/ 2)( )1( , ) for -

2 and - <y<

x yf x y e x

Pr( 2 2 2)X Y

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Solution

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Gamma Function

• For each positive number , let () be defined as:

• If , then ()= () ()

Proof: Let u= x–1, v= e – x then du= (1) x–2 and dv= e–x dx

• For every integer

0

1 dxex x

.2n

! 11 ! 12 211 1 nnnnnnnn

1 10 1

0

20

1

000

dxexex

vduuvudv

xx

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Gamma Distribution

• X has gamma distribution with parameters and ( and >0)

•

So

0for ,| 1

xexxf x

0

1

0 ,| dxexdxxf x

0

10

1 1dyeydxex yx

1 ,|0

dxxf

/ and )/( Let 11 dydxyxxy

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Gamma Distribution

• For k = 1, 2, …, we have

• XE

2

2

2

1Var

X

k

kk

xkkk

k

kk

dxexdxxfxXE

11

,| 0

1

0

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Gamma Distribution

• The m.g.f. of X can be obtained as:

• If X1, ..., Xk are independent r.v. and if Xi has a gamma distribution with parameters i and , then the sum has a gamma distribution with parameters and .

Proof:

ttt

dxexdxxfet xttx

for

,| 0

1

0

t

ttt

kk

ii for

1

1

k 1

kXX 1

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Physical Meaning of Gamma Distribution

• When is a positive integer, say = n, the gamma distribution with parameters (, ) often arises as the distribution of the amount of time one has to wait until a total of n events has occurred.

• Let Xn denote the time at which the nth event occurs, then

Notice that the number of events in [0, x] has a Poisson distribution with parameter x, and is the rate of events.

nj

j

njn j

xjxNnxNxX

!

e)(Pr)(PrPr

x

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Exponential Distribution

• A gamma distribution with parameters = 1 and is an exponential distribution.

• A random variable X has an exponential distribution with parameters has:

•

• Memoryless property of exponential distribution

)0for ( | xexf x

)for ( ,1

Var ,1

2

t

ttXXE

hXe

e

e

tX

htXtXhtX

edxetX

ht

ht

t

t

x

PrPr

Pr|Pr

Pr

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Life Test

• Suppose X1, …, Xn denote the lifetime of bulb i and form a random sample from an exponential distribution with parameter . Then the distribution of Y1=min{X1, …, Xn} will be an exponential distribution with parameter n.

Proof:

• Determine the interval of time Y2 between the failure of the first bulb and the failure of a second bulb.

– Y2 will be equal to the smallest of (n1) i.i.d. r.v., so Y2 has an exponential distribution with parameter (n1).

– Y3 will have an exponential distribution with parameter (n2).

– The final bulb has an exponential distribution with parameter .

tntt

n

n

eeetXtX

tXtXtY

1

11

PrPr

,... ,Pr Pr

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Physical Meaning of Exponential Distribution

• Following the physical meaning of gamma distribution, an exponential distribution is the time required to have for the 1st event to occur, i.e.,

where is rate of event.

• In a Poisson process, both the waiting time until an event occurs and the period of time between any two successive events will have exponential distributions.

• In a Poisson process, the waiting time until the nth occurrence with rate has a gamma distribution with parameters n and .

tetX Pr

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Example 5.9.1: Radioactive Particles

• For the Poisson process example in Example 5.4.3, suppose that we are interested in how long we have to wait until a radioactive particle strikes our target.

• Let Y1 be the time until the first particle strikes the target, and X be the number of particles that strike the target during a time period of length t. X has a Poisson distribution with mean t, where is the rate of the process.

The c.d.f. of Y1= F(y) = (for t > 0)

Taking the derivative of the c.d.f. we obtain:

The p.d.f. of Y1 = f(y) = (for t > 0)

teXXtY 10Pr11PrPr 1

te 1

te

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5.9.22

• Consider the Poisson process of radioactive particle hits in Example 5.9.1. Suppose that the rate of the Poisson process is unknown and has a gamma distribution with parameters and . Let X be the number of particles that strike the target during t time units. Prove that the conditional distribution of given X=x is a gamma distribution, and find the parameters of that gamma distribution

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Solution

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補充

• Let X be a random variable for which the p.d.f. is f and for which Pr(a<X<b)=1. Let Y=r(x), and suppose that r(x) is continuous and either strictly increasing or strictly decreasing for a<x<b. Suppose also that a<X<b if only if α<Y<β. Then the p.d.f. g of Y is specified by the relation

( )[ ( )] for

( )

0 otherwise

ds yf s y y

g y dy

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Example

• f(x)=2x,for0<x<1, 求 g(y)

• Solution:

• y=2x =>x=y/2, s(y)=y/2

• ds(y)/dy=1/2

• 所以 g(y)=f(y/2)*(1/2)=y/2 for 0<y<2