Understanding the Problem

We are given $n$ independent and identically distributed (i.i.d.) random variables, $X_1, \dots, X_n$, following an exponential distribution with mean $E[X_i] = \frac{1}{\lambda}$. The sum is $S_n = X_1 + \dots + X_n$. We define $N$ as the smallest integer $n \ge 1$ such that $S_n > 1$. We need to find the variance of $N$, denoted as $Var(N)$.

Relating to Poisson Process

The sum of $n$ i.i.d. exponential random variables with rate $\lambda$ corresponds to the time of the $n$-th event in a Poisson process with rate $\lambda$. Let $T_n$ be the time of the $n$-th event in such a process. Then $S_n = T_n$. The definition of $N$ becomes $N = \inf\{n \ge 1: T_n > 1\}$.

This implies that the $(N-1)$-th event occurs at or before time 1 ($T_{N-1} \le 1$), and the $N$-th event occurs after time 1 ($T_N > 1$).

Let $M(t)$ be the number of events occurring in the Poisson process up to time $t$. The condition $T_{N-1} \le 1 < T_N$ means that exactly $N-1$ events have occurred by time $t=1$. Therefore, $N-1 = M(1)$.

Calculating Variance

For a Poisson process with rate $\lambda$, the number of events $M(t)$ in the time interval $[0, t]$ follows a Poisson distribution with mean $\lambda t$ and variance $\lambda t$. Thus:

• $E[M(1)] = \lambda \times 1 = \lambda$
• $Var(M(1)) = \lambda \times 1 = \lambda$

Since $N-1 = M(1)$, we can find $E[N]$ and $Var(N)$:

• $E[N-1] = E[M(1)] = \lambda \implies E[N] = \lambda + 1$.
• $Var(N-1) = Var(M(1)) = \lambda$.

The variance is unchanged by subtracting a constant:

• $Var(N) = Var(N-1) = \lambda$.

Conclusion

The variance of $N$ is $\lambda$. This corresponds to Option B.