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

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
$$\mathbb P\left(M >  \sum\limits_{i=1}^n X_i -M\right) =\frac{n}{2^{n-1}}$$
