Chaitanya's Random Pages

April 17, 2011

The maximum of two exponential random variables

Filed under: mathematics — ckrao @ 12:21 am

Suppose X and Y are independent exponential random variables with mean 1. It is well known that the minimum of X and Y is also exponential with mean 1/2. A lesser known fact is that the maximum of X and Y has the same distribution as X + Y/2!

To prove this curiosity, we simply show their distributions are equal. Since X and Y have pdf and cdf given by $e^{-x}$ and $1-e^{-x}$ respectively for $x \geq 0$, we have

$\begin{array}{lcl} {\rm Pr}(X + Y/2 \leq a) &=& \int_0^a e^{-x} {\rm Pr}\left(Y \leq 2(a-x)\right)\ dx\\&=&\int_0^a e^{-x} \left(1-e^{-2(a-x)}\right)\ dx\\&=& \int_0^a e^{-x}\ dx - e^{-2a}\int_0^a e^x\ dx\\& = & 1 - e^{-a} - e^{-2a}(e^a-1)\\ &=& 1 - 2e^{-a} + e^{-2a}\\&=& \left(1-e^{-a}\right)^2\\ &=& {\rm Pr}(X \leq a, Y \leq a)\\&=& {\rm Pr}(\max\{X , Y\} \leq a),\end{array}$

as required.

More generally, if X and Y have respective means $1/\lambda_1$ and $1/\lambda_2$, then max{X,Y} has cdf given by $(1-e^{-\lambda_1})(1-e^{-\lambda_2})$, the product of the respective cdfs of X and Y. However it is only for the special case of $\lambda_1 = \lambda_2$ that this may be expressed as the cdf of a linear combination of X and Y.

Reference: G. Grimmett and D. Stirzaker, Probability and Random Processes, Oxford University Press, 2001, p142.

1. How would you determine the variance of max{X,Y}?

I think the expected value would be
for V = max{X,Y}
E[V] = E[X] + 1/2E[Y]
E[V] = 1 + 1/2
E[V] = 1.5

but I don’t think Var[V] = Var[X] + Var[Y]…

Comment by Dan Ciborowski — November 9, 2011 @ 1:47 pm

• Nice question! You are correct about E[V] = 1.5. Since V = max{X,Y} has the same distribution as X + Y/2, its variance would be

var[V] = var[X] + var[Y/2] (by independence)
var[V] = var[X] + var[Y]/4 (var[kY] = k^2 var[Y] for any constant k)
var[V] = 1 + 1/4
var[V] = 5/4

You can check this answer by the harder way of finding var[V] from first principles: from the blog post the cdf of V is (1-exp(-a))^2, so taking derivatives, its pdf is 2(1-exp(-a))exp(-a). Then by working out two integrals you will find that

var[V] = E[X^2] – (E[X])^2 = 7/2 – (3/2)^2 = 5/4.

Knowing that the distribution is the same as X + Y/2 makes things somewhat easier!

Comment by ckrao — November 9, 2011 @ 9:01 pm

2. I understood CDF of the maximum of X and Y.
But how can I find the PDF of the maximum of X and Y?

Comment by Frin — March 13, 2016 @ 3:42 pm

• To find the PDF differentiate the CDF expression with respect to a.

Comment by ckrao — March 15, 2016 @ 1:29 am

3. The result holds if lambda1=lambda2. Not only in the case where both are 1.

Comment by John Lawrence — March 8, 2018 @ 6:37 pm

• Well spotted, now fixed. Thanks!

Comment by ckrao — March 8, 2018 @ 9:09 pm

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