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suppose that $X_{1},\cdots, X_{k+1}$ are i.i.d. random variables obeying the exponential distribution. My question is how to calculate $P(X_{1}+\cdots+X_{k}\leq x_{0}, where $x_{0}$ is a positive number.

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Let $a$ denote the parameter of the exponential random variables. The maximal index $k$ such that $X_1+\cdots+X_k\le x_0$ is the value at time $x_0$ of a Poisson process of intensity $a$, see here. Hence the probability of the event you are interested in is also the probability that a Poisson random variable of parameter $ax_0$ is $k$, which is $ \mathrm{e}^{-ax_0}\frac{(ax_0)^k}{k!}. $

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The sum of $k$ i.i.d. exponential r.v.'s is an Erlang r.v. of shape $k$, say $Y_k$. Let's consider all possible values of $Y_k$ from 0 to $x_0$:

$\mathrm P ( Y_k \leq x_0 < Y_k + X_{k+1} ) = \mathrm \int_0^{x_0} \mathrm P( x_0 < t+X_{k+1} ) f_{Y_k}(t) \mathrm d t$

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The sum of $k$ i.i.d exponential random variables is a gamma distribution with cdf $F(x) = 1-e^{\lambda x} \sum_{k=0}^{r-1} \frac{(\lambda x)^{k}}{k!}$