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 and respectively for , we have

as required.

More generally, if X and Y have respective means and, then max{X,Y} has cdf given by , the product of the respective cdfs of X and Y. However it is only for the special case of 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.

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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 |

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 |