Re: Exponential RV and Conditional Expectation Problem
From: Robert Israel (israel_at_math.ubc.ca)
Date: 10/14/04
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Date: 14 Oct 2004 17:20:50 GMT
In article <216501ce.0410140209.1edcfdb3@posting.google.com>,
bg <Stridar@gmail.com> wrote:
>I am working alone through a book (Sheldon Ross's _Probability Models
>for Computer Science_). While I have enjoyed the exposition and have
>been able to do most problems, I do not understand the math behind one
>question. Would anyone be kind enough to show me how to calculate the
>conditional expectation in the following problem?
>Problem 1.23
>Let X,Y be ind. exponential r.v.'s with rates \lambda and \mu, and let
>c >= 0.
>a) Find E( min(X,Y) | X > c )
>b) Find E( min(X,Y) | X > Y + c)
>For part (a), I have tried breaking the integral into three parts to
>remove the min function instead of finding a joint probability
>density. Is this a correct method?
Yes. You might also break it into two pieces Y <= c and Y > c, and
for the second piece use the facts that
1) the conditional density for X - c given X > c is the same as the
density for X [and similarly for Y]
2) the minimum of independent exponential r.v.'s is an exponential r.v.
>For part (b), I am lost. I thought conditioning on Y would work since
>it seems min(X,Y) should simply be Y instead of integrating over the
>exponential r.v. with parameter \mu+\lambda. However, Ross gives a
>later problem to show
> E(min(X,Y) | X > Y + c) = E(min(X,Y)| X>Y) = \frac{1, \lamda +
>\mu}
>The above technique does not evaluate to this fraction.
Yes, min(X,Y) = Y when X > Y + c. By "conditioning on Y" I hope you mean
E[Y | X > Y + c] =
int_0^infty y P{X > y + c} f_Y(y) dy / int_0^infty P(X > y + c) f_Y(y) dy
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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