Problem of the Week #78 - November 25th, 2013

[Chris L T521]

Here's this week's problem.

-----

Problem: Let $X_1,X_2,\ldots X_n$ be i.i.d. exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is $n/2^{n-1}$. That is, if $M=\max\limits_j X_j$ then show that
\[P\left\{ M > \sum_{i=i}^n X_i - M\right\} = \frac{n}{2^{n-1}}\]

-----

Hint:

Spoiler

What is $\displaystyle P\left(X_1 > \sum_{i=2}^n X_i\right)$?

Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Chris L T521]

No one answered this week's problem. You can find the solution below.

Spoiler

We first note that
\[\begin{aligned} P\left\{X_1 > \sum_{i=2}^n X_i\right\} = &P\{X_1>X_2\} P\{X_1-X_2>X_3\mid X_1>X_2\} \\ &\times P\{X_1-X_2-X_3>X_4\mid X_1>X_2+X_3\}\\ &\times \ldots\\ &\times P\{X_1-X_2 - \ldots - X_{n-1} > X_n \mid X_1> X_2 + \ldots + X_{n-1}\} \\ = & (1/2)^{n-1}\end{aligned}\]
due to the memoryless property of the exponential distribution. Therefore, we now see that
\[P\left\{M > \sum_{i=1}^n X_i - M\right\} = \sum_{i=1}^nP\left\{X_1 > \sum_{j\neq i}^n X_i\right\} = \sum_{i=1}^n \frac{1}{2^{n-1}} = \frac{n}{2^{n-1}}\]