Re: Exponential RV and Conditional Expectation Problem
From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 10/14/04
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Date: 14 Oct 2004 11:31:00 -0700
Stridar@gmail.com (bg) wrote in message news:<216501ce.0410140209.1edcfdb3@posting.google.com>...
> Dear all,
>
> 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)
>
>
> Thank you,
> BG
>
> P.S.
> 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?
If done correctly. What are your three parts?
I'd do it this way:
Think in terms of a new rv, Z = min(X,Y). You can easily
reason about the cumulative distribution of Z.
Prob(Z<z|X>c) = 1 - P(Z>=z | X>c). I find this latter
probability easier to reason about. Z>=z iff both X>=z and
Y>=z. Thus
P(Z>=z | X>c) = P(Y>=z & X>=z|X>c)
= P(Y>=z|X>c) P(X>=z|X>c)
You know how to evaluate those expressions.
This is in general a useful trick for dealing with the
minimum.
> For part (b), I am lost. I thought conditioning on Y would work since
> it seems min(X,Y) should simply be Y
That seems correct.
> 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)
I have to think about this one a little.
> = \frac{1, \lamda +
> \mu}
While it may work out to this, I don't think it's because
there is an exponential rv with parameter lambda +mu.
- Randy
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