We know that that given $X_1,…,X_n$ independent random variables which are distributed exponentially, their minimum is also distributed exponentially. Is the converse true? Given $X = \min\{X_1, …, X_n\}$ distributed exponentially, and $X_1, …, X_n$ are independent, is it true that $X_i$ is distributed exponentially for all $i = 1,…,n$?
Edit: I don't assumme the $X_i$'s are identically distributed. I believe it is true that if $X \sim \exp(\lambda)$, and if $p_i \equiv \mathbb{P}(X_i = X)$ then $X_i \sim \exp(\lambda p_i)$. However, I couldn't find a reference and didn't manage to prove this by myself.
Further edit: The result is true if you assume $\{X = X_i\}$ is an event independent of $X$. This is proven in the following manner:
- Construct a Poisson process $N_t$ whose waiting times are IID copies of $X$.
- Split this Poisson process to $n$ counting processes – the process $N_t^i$ will count the number of occurences where $X = X_i$.
- Using the fact that $\{X = X_i\}$ is an event independent of $X$, and that $X_1,…,X_n$ are independent random variables, one can show that $N_t^i$ is a Poisson process with intensity $\lambda p_i$.
- The waiting times of $N_t^i$ will be distributed as $\exp(\lambda p_i)$, but on the other hand, they will be equal to $X_i$.
Best Answer
Example: Let $\lambda_1:[0,\infty)\to(0,1)$ be a continuous function, and then define $\lambda_2(t):=1-\lambda_1(t)$. Also define $\Lambda_k(t):=\int_0^t\lambda_k(s)\,ds$ for $t\ge 0$ and $k=1,2$. Then $F_k(t) :=1-\exp(-\Lambda_k(t))$, $t\ge 0$, are distribution functions of non-negative random variables, say $X_1$ and $X_2$. Their minimum $X:=\min(X_1,X_2)$ has cdf $1-(1-F_1(t))\cdot(1-F_2(t))=1-e^{-t}$, $t\ge 0$. Thus $X$ is exponentially distributed, but the $X_k$ need not be.