Suppose that $X_1,X_2,\ldots$ are independent and exponentially distributed random variables with the parameter $\lambda$. Let us denote $S_n=X_1+\ldots+X_n$. I am interested in the distribution of the random variable $N=\min\{n\ge1:S_n\ge c\}$ with $c>0$, i.e. $N$ is a random variable that gives the smallest $n$ such that $S_n\ge c$.

I am trying to calculate the probabilities $P(N=k)$, where $k\ge1$. For $k=1$, $$ P(N=1)=P(S_1\ge c)=P(X_1\ge c)=e^{-\lambda c}. $$ For a general $k\ge1$ with a convention that $S_0=0$, $$ P(N=k)=P(\{S_k\ge c\}\cap\{S_{k-1}<c\})=P(c-X_k\le S_{k-1}<c), $$ but I have no idea how to evaluate this probability. How can I evaluate this probability?

I know that $S_n\sim\mathrm{Gamma}(n,\lambda^{-1})$, but I am not sure if that is useful.

Any help is much appreciated!

  • $\begingroup$ I would say it is $P(S_{k-1}+X_{k}\ge c\mid S_{k-1}<c) = P(c-X_k\leq S_{k-1}< c$. Then, conditioning on $X_k$, this can be evaluated from the CDF of the Gamma distribution. Then, you de-condition by integrating over the PDF of $X_k$ which is the exponential distribution. $\endgroup$
    – user164550
    Mar 16, 2016 at 16:10
  • 1
    $\begingroup$ $N_c\sim 1+Poisson(c)$. $\endgroup$
    – A.S.
    Mar 16, 2016 at 17:20

2 Answers 2


This is a Poisson process of rate $\lambda$. The given event $$ \{S_k\ge c\}\cap\{S_{k-1}<c\} $$ is equivalent to saying that the number $N_c$ of arrivals within the time interval $(0,c)$ is equal to $k-1$.

In the Poisson process, $N_c$ follows Poisson distribution of parameter $\lambda c$. Therefore, $$P(\{S_k\ge c\}\cap\{S_{k-1}<c\})=P(N_c=k-1)=e^{-\lambda c} \frac{(\lambda c)^{k-1}}{(k-1)!}.$$


\begin{align} P(S_{k−1}+X_k≥c\mid S_k−1<c)&=P(c−X_k≤S_{k−1}<c)\\ &=\mathbb{E}_{X_k}[P(c−X_k≤S_{k−1}<c)]\\ &=\mathbb{E}_{X_k}\left[\frac{1}{\Gamma(k-1)} \gamma\left(k-1,\, \frac{c}{\theta}\right) - \frac{1}{\Gamma(k-1)} \gamma\left(k-1,\, \frac{c−X_k}{\theta}\right)\right]\\ &=\frac{1}{\Gamma(k-1)} \gamma\left(k-1,\, c \lambda\right) - \mathbb{E}_{X_k}\left[\frac{1}{\Gamma(k-1)} \gamma\left(k-1,\, \lambda(c−X_k)\right)\right]\\ &=\frac{1}{\Gamma(k-1)} \gamma\left(k-1,\, \frac{c}{\theta}\right) - \frac{1}{\Gamma(k-1)} \int_0^\infty \gamma\left(k-1,\, \lambda c− \lambda t\right) \lambda e^{-\lambda t} dt \end{align}


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