Expected waiting time of last item for a set of m exponential random variables

Question

So I've been mulling over a question: If I have a type of object that breaks after a waiting time $T \sim Exp(\lambda)$. Now I'm looking at $m$ of these same objects and I want to know the expected waiting time of the last object to break i.e. $E(T_m)$. My intuition tells me this is a Gamma distribution and so $E(T_m)=\frac{m}{\lambda}$ with a $Var(T_m)=\frac{m}{\lambda^2}$. That being said I'd like to know if I'm wrong and if so how to think about the problem differently. Thanks!

Answer 1

This is the maximum of a set of i.i.d. exponentially distributed r.v.s $X_1, \dots, X_m$, each with cumulative distribution function $F(x) = 1 - e^{-\lambda x}$ for $x > 0$. Therefore the c.d.f. of $T_m$ is $$ P(T_m \le x) = P(\forall i \, X_i \le x) = \prod_i P(X_i \le x) = F(x)^m = \left(1 - e^{-\lambda x} \right)^m $$ and the probability density function is the derivative of this expression. That is not a gamma distribution. In particular, $E(T_m) = \lambda^{-1}c_m$, where $$ \begin{aligned} c_m &= \int_0^\infty m x e^{-x}\left(1-e^{-x}\right)^{m-1} dx = \sum_{k=0}^{m-1} (-1)^k m \binom{m-1}{k}\int_0^\infty x e^{-x-kx} dx \\ &= \sum_{k=0}^{m-1} (-1)^k \binom{m}{k+1} \frac{1}{k+1} \\ &=\psi^{0}(1+m) + \gamma \end{aligned} $$ where $\psi^{(0)}(z) = \frac{\Gamma'(z)}{\Gamma(z)}$ is the polygamma function of order $0$ and $\gamma \approx 0.577\dots$ is Euler's constant. This means that $E(T_m)\sim (\log m + \gamma)/\lambda$ for large $m$