# Find $P\left( 2M > X_1 + \cdots + X_n\right)$

Let $X_1,\cdots,X_n$ be mutually independent and identical distributed exponential random variables. Let $M = \max(X_1, \cdots, X_n)$. Find $P\left( X_1 + \cdots + X_n < 2M\right)$.

I have the following. First, we can do

\begin{align} P\left( X_1 + \cdots + X_n < 2 M \right) = 1 - P\left( X_1 + \cdots + X_n > 2M\right) \end{align}

Then

\begin{align} P\left(X_1 + \cdots + X_n > 2M\right) &= P\left(X_1 + \cdots + X_n>2 \max(X_1,\cdots,X_n)\right) \\ &=P\left( X_2 + \cdots + X_n > X_1 \cap \cdots \cap X_1 + \cdots + X_{n-1} > X_n\right) \end{align}

I think I can somehow find

$$P\left(\sum_{i\ne j} X_i > X_j\right)$$

by using symmetry argument (e.g. $P(X_1 > X_2 + X_3 + X_4) = P(X_2 > X_1 + X_3 + X_r)$). But the problem now is that I don't know if I can turn the expression into product of probabilities. Besides, I haven't used the fact that $X_i$'s are exponential random variables. Any hint would be great.

I assume that $X_i\overset{iid}{\sim}exp(1)$. Let $S=\sum_{i=1}^n X_i$ and $S_{-i}=\sum_{j\ne i}X_i$. Then, noticing that $\mathsf{P}(X_i=X_j)=0$ for $i\ne j$, we have \begin{align} \mathsf{P}(S<2M)&=\sum_{i=1}^n\mathsf{P}(S_{-i}<M, X_i=M) \\ &=\sum_{i=1}^n\mathsf{P}(S_{-i}<X_i), \end{align} where \begin{align} \mathsf{P}(S_{-i}<X_i)&=\int_0^\infty \mathsf{P}(S_{-i}<x)e^{-x}dx \\ &=\int_0^\infty\frac{\gamma(n-1,x)}{\Gamma(n-1)}e^{-x}dx=\frac{1}{2^{n-1}} \end{align} because $S_{-i}\sim \Gamma(n-1,1)$. Hence, $$\mathsf{P}(S<2M)=\frac{n}{2^{n-1}}.$$
The last integral can be evaluated using the power series expansion of the lower incomplete gamma function, i.e. $$\int_0^\infty\frac{\gamma(n-1,x)}{\Gamma(n-1)}e^{-x}dx=\sum_{k=0}^\infty\frac{1}{\Gamma(n+k)}\int_0^{\infty}e^{-2x}x^kdx=\sum_{k=0}^\infty \frac{1}{2^{n+k}}$$
• The summation is over independent events, so you could just write: $$P(S<2M) = n P(S_{-n} < M, X_n=M)$$ – adfriedman Oct 11 '17 at 1:21
• I am a little bit confused by $M$... I should be looking at $\sum_{i=1}^n P\left( \sum_{j\ne i} X_j < X_i\right)$, right? – 3x89g2 Oct 11 '17 at 3:52
• I simplified the argument. Now $$\mathsf{P}(S_{-i}<X_i)=\mathsf{E}[\mathsf{P}(S_{-i}<X_i)\mid X_i]=\int_0^\infty \mathsf{P}(S_{-i}<X_i\mid X_i=x)e^{-x}dx.$$ – d.k.o. Oct 11 '17 at 3:59