Probability of max of exponentially distributed random variables greater than sum of the others

Question

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.

Answer 1

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 $\lambda$).

Answer 2

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.