How to calculate the probability of $\sum_{i=1}^{n}X_{i}\leq t,\;\sum_{i=1}^{n+1}X_{i}>t$?

Question

$\{X_{i}\;,i=1,2,3,...\}$ are i.i.d. exponential distributions with $E(X_{i})=\lambda^{-1}$

How to calculate the probability of the event $$\{\sum_{i=1}^{n}X_{i}\leq t,\;\sum_{i=1}^{n+1}X_{i}>t\}$$ where $t$ is given?

Answer 1

Using theory about Poisson processes, this event is equivalent to "there are exactly $n$ arrivals of a Poisson process with rate $\lambda$ during the time interval $[0,t]$." Since the number of arrivals in $[0, t]$ follows a $\text{Poisson}(\lambda t)$ distribution, the probability of this event is $e^{-\lambda t}\frac{(\lambda t)^n}{n!}$.

Answer 2

Here is a way to solve the problem with direct calculations (i.e. no knowledge from the theory of Poisson processes is required):

1. Set $S_n := \sum_{i=1}^n X_i$ then $$p:=\mathbb{P} \left( \sum_{i=1}^n X_i \leq t, \sum_{i=1}^{n+1} X_i > t \right) = \mathbb{P}(S_n \leq t, S_n+X_{n+1}>t).$$
2. Since $S_n$ and $X_{n+1}$ are independent, it follows from the tower property of the conditional expectation that \begin{align*} p&= \mathbb{E} \big[ \mathbb{E}(1_{\{S_n \leq t\}} 1_{\{S_n+X_{n+1} >t\}} \mid S_n) \big] =\mathbb{E}(1_{\{S_n \leq t\}} f(S_n))\end{align*} where $$f(s) := \mathbb{P}(s+X_{n+1}>t), \qquad s \geq 0.$$

3. From $X_{n+1} \sim \text{Exp}(\lambda)$, it follows that $$f(s) = e^{-\lambda (t-s)}$$ for all $s \leq t$. Hence, by Step 2, $$p = e^{-\lambda t} \mathbb{E}(1_{\{S_n \leq t\}} e^{\lambda S_n}).$$

4. Since $S_n = X_1+\ldots+X_n$ for iid random variables $X_i$, we have

\begin{align*} \mathbb{E}(1_{\{S_n \leq t\}} e^{\lambda S_n}) &= \int_{\mathbb{R}} \ldots \int_{\mathbb{R}} 1_{\{x_1+\ldots+x_n \leq t\}} e^{\lambda(x_1+\ldots+x_n)} d\mathbb{P}_{X_1}(x_1) \ldots \, d\mathbb{P}_{X_n}(x_n) \end{align*} where $\mathbb{P}_{X_i}$ denotes the distribution of $X_i$. As $X_i \sim \text{Exp}(\lambda)$, this shows

\begin{align*} \mathbb{E}(1_{\{S_n \leq t\}} e^{\lambda S_n}) &= \lambda^n\int_{(0,\infty)} \ldots \int_{(0,\infty)} 1_{\{x_1+\ldots+x_n \leq t\}} e^{\lambda(x_1+\ldots+x_n)} e^{-\lambda x_1} \cdots e^{-\lambda x_n} \, dx_1 \ldots dx_n \\ &= \lambda^n \int_{(0,\infty)} \ldots \int_{(0,\infty)} 1_{\{x_1+\ldots+x_n \leq t\}} \, dx_1 \ldots dx_n.\end{align*}

Note that the integral on the right-hand side is the volume of an $n$-simplex with side length $t$; it equals $t^n/n!$. In summary,

$$p = e^{-\lambda t} \frac{(\lambda t)^n}{n!}$$