1 Introduction to Stochastic Models GSLM 54100. 2 Outline exponent distribution memoryless...

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1 Introduction to Introduction to Stochastic Models Stochastic Models GSLM 54100 GSLM 54100

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3 Memoryless Property  for X ~ exp(  ), (X  t|X > t) ~ X  P(X > t+s|X > t) = P(X > s) = e , for s, t > 0  Y  0, Y independent of X  P(X > Y+s|X > Y) = P(X > s), for s, t > 0  P(X > Y+s|X > Y) = E[P(X > Y+s|X > Y)|Y ] = P(X > s)  in this course, assume that P(X > Z+s|X > Z) = P(X > s) for any random variable Z

Transcript of 1 Introduction to Stochastic Models GSLM 54100. 2 Outline exponent distribution memoryless...

Page 1: 1 Introduction to Stochastic Models GSLM 54100. 2 Outline  exponent distribution  memoryless property  minimum of independent exp random variables.

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Introduction to Stochastic ModelsIntroduction to Stochastic ModelsGSLM 54100GSLM 54100

Page 2: 1 Introduction to Stochastic Models GSLM 54100. 2 Outline  exponent distribution  memoryless property  minimum of independent exp random variables.

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OutlineOutline exponent distribution

memoryless property minimum of independent exp random variables simple formula to find P(X > Y) for ind exp’s convolution of exponentials examples

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Memoryless Property Memoryless Property for X ~ exp(), (X t|X > t) ~ X

P(X > t+s|X > t) = P(X > s) = e, for s, t > 0

Y 0, Y independent of X P(X > Y+s|X > Y) = P(X > s), for s, t > 0 P(X > Y+s|X > Y) = E[P(X > Y+s|X > Y)|Y ]

= P(X > s)

in this course, assume that P(X > Z+s|X > Z) = P(X > s) for any random variable Z

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Minimum of Independent Minimum of Independent Exponential Random VariablesExponential Random Variables

U = min(T1, …, TM), Ti ~ exp(i), independent U ~ exp(1+…, M)

P(U > t) = P(min(T1, …, TM) > t)

= P(T1 > t, …, TM > t) = P(T1 > t) … P(TM > t)

= exp(1)…exp(M) = exp(1+…, M)

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To find To find PP((XX > > YY) for Exp ) for Exp XX & & YY X ~ exp(), Y ~ exp(), X ind. of Y P(X < Y) = E[P(X < Y)|X] = E[eX]

0x xe e dx

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Example 7.1.4Example 7.1.4 X ~ exp(), Y ~ exp(), Z ~ exp(); all

independent

P(X + Y < Z) = ???

one possibility: easiest way: by memeoryless property

{ }( ) ( ) ( )X Y Z

x y zf x f y f z dxdydz

X Y

Z

P(X < Z)

P(X+Y < Z|Z > X)

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ConvolutionConvolution X and Y: ind. continuous random variables the convolution of X and Y = the distribution of X

+ Y

fX+Y(s) = for all s

if X, Y 0, fX+Y(s) =

( ) ( )X Yf x f s x dx

0 ( ) ( )sX Yf x f s x dx

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Distribution of Distribution of XX11 + + XX22

X1, X2 ~ i.i.d. exp()

1 2( )X Xf t

1 20 ( ) ( )tX Xf x f t x dx

20 (1)tte dx

( )0t x t xe e dx

1!tte

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Distribution of Distribution of XX11 +…+ +…+ XXnn

X1, …, Xn ~ i.i.d. exp()

assume

1( )

k kS Xf t

10 ( ) ( )k

tX Sf t x f x dx

1( )( )0 ( 1)!

kxt t x xke e dx

( )!

kttke

1( )( )( 1)!k

kt

Stf t e

k

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Distribution of (Distribution of (XX||XX<<YY) for ) for Independent Exponential Random Independent Exponential Random

VariablesVariables X ~ exp(), Y ~ exp(), X independent of Y

distribution (X|X < Y) = ?

( | )P X s X Y ( )( )

P s X YP X Y

( )P X Y

( )P s X Y [ ( | )]E P s X Y X ( ) ( )Xs P x Y f x dx

x xs e e dx

( )xs e dx

( ) sxe

( )se

( )( | ) , and sP X s X Y e ( | ) ~ exp( )X X Y

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Example 7.1.6Example 7.1.6 X1, X2, X3 ~ i.i.d. exp()

L = max(X1, X2, X3)

E(L) = ??

min(X1, X2, X3) ~ exp(3)

~ exp(2)

~ exp()

L = max(X1, X2, X3) = min(X1, X2, X3) + +

X1X2X3

min(X1, X2, X3)

' '2 3min( , )X X

"3X

' '2 3min( , )X X

"3X

"3X

' '2 3min( , )X X 11

6( )E L

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Examples of RossExamples of Ross Example 5.4, 5.5, 5.8