Question

I am trying to understand why the conditional PDF f(X|X>1) is what the answer the picture indicates given that P(X>1) = 1/e. Should f(X|X>1) not equal e^(-x-1) instead given the exponential RV is e^(-x)?

Example 5.20. Let X ~ Exponential(1). (a) Find the conditional PDF and CDF of X given X > 1. (b) Find E[X|X > 1). (c) Find Var(X|X > 1). Solution: (a) Let A be the event that X > 1. Then P(A) = S, ſed 1 Thus, e-*+1 > 1 fx|X>1(x) = 0 otherwise

Answer 1

solution : Let of Xis, X~ Exponential (1) then the polf fx (2) = et a> 0 of X given X>1 is Now A be the event that x>1 , then the conditional polf given by p( x | x>1) P(x | A) = P(X, A) PCA) - P(x | x1) P(x, x>1) P(X>12 of conditional probability } So P(x, x>1) = et and P(X>1) [(810) {by the definition et dhe こ P (x71) of 271 and hence the conditional polf of a given xy 1 is given by diplom PC x1 x1) = (ne) Sto1 1 PC X1 X21) ate And hence we get exel x>1 ex+1 f (x) = XX1 EXH This the sequired conditional palf of X given XM1.

And hence the conditional CDF of a given X1 is, E (a) = f (x1x1) (de) de (X/X1) 1 06 at de 31 1 1 2 e² - exti Flx (x) = 2x71x This is the Bequided conditional CDF of x given X>1. © conditional expectation, ECx\x71) is gives blis Elx|x>1) = x fxxx (x) dx = 1 x et de 1 1 -elze'de = e lar" (6:40) e [ 2x (x+1) In el (1+1)-0 ex ex2 -ECX|X21) = 2 to find v (XIX) 1) be need to find E (X?/»>1) VOX 1x1) = (x*270)- F(x \x3D) ? and

da E(x2 ]X31) = 5x2 fx 1xza (2) da - J x2 7441 1 = e l²pt der 1 e [-e? (x2+2+2)1 -e [( 24+22+2) I) e Cel ( 1²+ 2(1) +2) -0 n - F(x? | x>1) = 5 ve xlx>1) E (x? | XX1) - E(XlX91)? che v (Xx>1)=5-2² 5-9 - 1: 2