Probability of max of exponentially distributed random variables greater than sum of the others First of all, hi. I am new here.
Let$$X_1,\dots, X_n$$
be i.i.d. exponential random variables.
$$
Pr({\max X_n}>{(\sum X_n-\max X_n ) }) = ?
$$
I think we should take integrals on exponential distribution functions over corresponding intervals but I could not make it work.
Thanks.
edit: my initial work was similar to and more generalized version of problem 6.45 in http://www.stat.washington.edu/~hoytak/teaching/current/stat395/
instead of 1s as in uniform distribution I tried to put exponential pdf. But nothing good seemed to come out of it.
2nd edit: thanks a lot for all of you, especially sinbadh and BGM. you spent much time. I could not put my initial work thoroughly because I just start getting used to LATEX format. Hopefully after earning some repition I start giving thumbs up.
 A: First of all, you want to find probability for $\max\{X_i\}\ge\sum\mbox{ of the other $X_i$}$, not only $P(\max\{X_i\}\ge \sum X_i)$. Now, by total probability:
On the other hand, supposing independence for random variables,
$\begin{eqnarray}
&&P(\max\{X_i\}\ge\sum\mbox{ of the other $X_i$})\\
&=&\sum_{k=1}^nP(\max\{X_i\}\ge\sum\mbox{ of the other $X_i$}|\max\{X_i\}=X_k)P(\max\{X_i\}=X_k)\\
&=&\sum_{k=1}^nP(X_k\ge\sum\mbox{ of the other $X_i$}|\max\{X_i\}=X_k)P(\max\{X_i\}=X_k)\\
&=&\sum_{k=1}^nP(X_k\ge\sum\mbox{ of the other $X_i$})\\
&=&nP(X_1\ge\sum_{k=2}^nX_k)...(*)
\end{eqnarray}$ 
I believe you did all this steps.
Now, let $S=X_2+...+X_n$. Since $X_i\sim Exp(\lambda)$ are iid, then $S\sim\Gamma(n-1,\lambda)$.
Moreover, $S$ and $X_1$ are independent. Then, setting $X=X_1$,
$\begin{eqnarray}
P(X\ge S)&=&\int_0^{\infty}\int_s^\infty f_{X,S}(x,s)\,\mathrm{d}x\mathrm{d}s\\
&=&\int_0^{\infty}\int_s^\infty f_X(x)f_S(s)\,\mathrm{d}x\mathrm{d}s\\
&=&\int_0^{\infty}f_S(s)\int_s^\infty f_X(x)\,\mathrm{d}x\mathrm{d}s\\
&=&\int_0^{\infty}f_S(s)\int_s^\infty \lambda e^{-\lambda x}\,\mathrm{d}x\mathrm{d}s\\
&=&\int_0^{\infty}f_S(s)e^{-\lambda s}\mathrm{d}s\\
\end{eqnarray}$
Now, the last integral is $\frac{1}{2^{n-1}}$. It would be calculated in several ways. For example, noting that it is the moment generating function evaluated in $t=-2\lambda$ for a gamma distribution $S\sim\Gamma(n-1,\lambda)$, or writing explicitely $f_S(s)$ and "complete it" to a $\Gamma(n-1,2\lambda)$.
Finally, in $(*)$ we have $P(\max\{X_i\}\ge\sum\mbox{ of the other $X_i$})=\frac{n}{2^{n-1}}$ (please, note independence of parameter $\lambda$). 
A: I just missed sinbadah's simple approach using the conditional argument. Here I provide another approach, which is more complicated, and just for a reference.
One special thing about the order statistics of exponential is the spacings: Let 
$$W_i = X_{(i+1)} - X_{(i)}, i = 1, 2, \ldots, n - 1$$
be the spacings, with $W_0 = X_{(1)}$, then 
$$ W_i \sim \text{Exp}\left(\frac {\theta} {n - i}\right) $$
where $\theta$ is the mean of the original exponential sample. Another good thing is that $W_i$ are independent. Hence,
$$ \begin{align} 
&~ \Pr\left\{2X_{(n)} > \sum_{i=1}^n X_i\right\} \\
=&~ \Pr\left\{2\sum_{i=0}^{n-1}W_i > \sum_{i=1}^n X_{(i)}\right\}\\
=&~ \Pr\left\{2\sum_{i=0}^{n-1}W_i > \sum_{i=1}^n \sum_{j=0}^{i-1} W_j\right\}\\
=&~ \Pr\left\{2\sum_{i=0}^{n-1}W_i > \sum_{i=0}^{n-1} (n - i)W_j\right\}\\
=&~ \Pr\left\{W_{n-1} > \sum_{i=0}^{n-3} (n - i - 2)W_j\right\} \\
=&~ \int_0^{+\infty}\cdots\int_0^{+\infty} \int_{\sum_{i=0}^{n-3} (n - i - 2)w_i}^{+\infty}
\frac {n!} {2\theta^{n-1}}\exp\left\{-\frac {1} {\theta} 
\sum_{i = 0}^{n-3}(n - i)w_i - \frac {w_{n-1}} {\theta}\right\}
dw_{n-1}dw_{n-3} \ldots dw_0 \\
=&~ \int_0^{+\infty}\cdots\int_0^{+\infty} \frac {n!} {2\theta^{n-2}}
\exp\left\{-\frac {1} {\theta} \sum_{i = 0}^{n-3}(n - i)w_i\right\}
\exp\left\{-\frac {1} {\theta} \sum_{i = 0}^{n-3} (n - i - 2)w_i\right\}
dw_{n-3} \ldots dw_0 \\
=&~ \frac {n!} {2\theta^{n-2}} \int_0^{+\infty}\cdots\int_0^{+\infty} 
\exp\left\{-\frac {2} {\theta} \sum_{i = 0}^{n-3}(n - i - 1)w_i\right\}
dw_{n-3} \ldots dw_0 \\
=&~ \frac {n!} {2\theta^{n-2}} \prod_{i=0}^{n-3} \int_0^{+\infty} \exp\left\{-\frac {2(n-i-1)} {\theta} w_i\right\} dw_i \\
=&~ \frac {n!} {2\theta^{n-2}} \prod_{i=0}^{n-3} \frac {\theta} {2(n - i - 1)} \\
=&~ \frac {n!} {2\theta^{n-2}} \frac {\theta^{n-2}} {(n-1)!2^{n-2} } \\
=&~ \frac {n} {2^{n-1}}
\end{align}$$
Finally it agrees with the answer of sindbah.
