# $P(\{S_k\ge c\}\cap\{S_{k-1}<c\})$, where $S_k$ is the sum of iid exponential random variables

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!

• 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.
– user164550
Mar 16, 2016 at 16:10
• $N_c\sim 1+Poisson(c)$.
– A.S.
Mar 16, 2016 at 17:20

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, $$P(\{S_k\ge c\}\cap\{S_{k-1}<c\})=P(N_c=k-1)=e^{-\lambda c} \frac{(\lambda c)^{k-1}}{(k-1)!}.$$