For a set of n Exponential iid RV, what is the probability that the maximum exceeds the sum of the others? My question stems from self-study of question 91 in page 372 of Ross's Introduction to probability models, 12th edition.
The answer is given in the textbook but I am trying to understand a specific step that I suspect depends on Lebesgue integration.
For clarity I present the entire problem.
Let $X_1, ..., X_n$ be iid Exponential RV with rate $\lambda$ and $M=\max_j X_j$. Show
$$
P\left \{ M > \sum_{i=1}^n X_i - M \right \} = \frac{n}{2^{n-1}}
$$
A hint is given: What is $P\{X_1 > \sum_{i=2}^n X_i \}$?
The hint suggests the following approach:
$$
P\left \{ M > \sum_{i=1}^n X_i - M \right \} = \sum_{k=1}^n P\left \{ M > \sum_{i=1}^n X_i - M \bigg| M = X_k \right  \} P\left \{M=X_k \right \}
$$
and
$$
\begin{split}
P\left \{X_1 > \sum_{i=2}^n X_i \right \} &= \int_0^\infty \left ( P\{X_1 > Y  | X_1 = x \} f_{X_1}(x) \right ) dx \\
&= \int_0^\infty \left ( P \{x > Y \} f_{X_1}(x) \right ) dx 
\end{split}
$$
Where $Y \sim Gamma[n-1,\lambda]$.
This leads to
$$
P\left \{X_1 > \sum_{i=2}^n X_i \right \} = \frac{1}{2^{n-1}}
$$
How do you compute $P\{M=X_k\}$? or am I setting this wrong? The book answer is the sum as if $P\{M=X_k\}=1$. I suspect there is some special Lebesgue integration.
Anyone can shed some light on the issue?
Thanks!
GMercier
 A: You seem to have proved that
$$ P\left \{ M > \sum_{i=1}^n X_i - M \bigg| M = X_k \right  \} = P\left \{X_1 > \sum_{i=2}^n X_i \right \} = \frac{1}{2^{n-1}}$$
Assuming that's right, then you don't need to compute $P\{M=X_k\}$, because your conditional is a constant, hence it goes outside the summation  - and of course you must have $\sum_k P\{M=X_k\}=1$ (your guess $P\{M=X_k\}=1$ is clearly absurd, both because of the above, and also by intuition).
The problem is that the first equality above is wrong.
When you condition on the event $M=X_1$, it's true that the event $ M > \sum_{i=1}^n X_i - M $ can be expressed $X_1 >\sum_{i=2}^n X_i  $... but you can't forget the condition. And $P(X_1 >\sum_{i=2}^n X_i) \ne P(X_1 >\sum_{i=2}^n X_i | X_1 = M) $
What we know (why?) is this: $ X_1 >\sum_{i=2}^n X_i  \implies X_1 = M $
Then $$P(X_1 >\sum_{i=2}^n X_i \mid  X_1 = M) = \frac{P(X_1 >\sum_{i=2}^n X_i \cap X_1 = M)}{P( X_1 = M)}=\frac{P(X_1 >\sum_{i=2}^n X_i) }{P( X_1 = M)}$$
Can you go on from here?
( Granted,  this implies that we didn't need to use conditionals, only total probability:)
$$ P( M > \sum_{i=1}^n X_i - M)\\ =\sum_k  P( M > \sum_{i=1}^n X_i - M \cap M = X_k) \\ =\sum_k P(X_k >\sum_{i\ne k} X_i) \\= n P(X_1 >\sum_{i=2}^n X_i)$$
A: Since all the $X_i$ are non-negative, the statement $X_1 >  \sum\limits_{i=2}^n X_i$ implies $X_1$ must be the maximum, with $X_1 >  \sum\limits_{i=1}^n X_i - X_1$, so your $\frac{1}{2^{n-1}}$ is the joint probability that $X_1=M$ is the maximum and $M >  \sum\limits_{i=1}^n X_i -M$.
The same is true for each of the other $X_j$, i.e. $\frac{1}{2^{n-1}}$ is the joint probability that $X_j=M$ is the maximum and $M >  \sum\limits_{i=1}^n X_i -M$. Summing over the $n$ possible values of $j$ gives
$$\mathbb P\left(M >  \sum\limits_{i=1}^n X_i -M\right) =\frac{n}{2^{n-1}}$$
