# Why does $\min(X_1, X_2 ,\ldots, X_n) \sim \mathrm{Expo}(X_1 + X_2 + \cdots + X_n)$?

Let $X_1, X_2,\ldots,X_n$ be identical and independent random variables that are distributed exponentially with rate value $\lambda$. Then, does $\min(X_1, X_2 ,\ldots, X_n) \sim \mathrm{Expo}(X_1 + X_2 + \cdots X_n)$?

I think I have a proof: Let $M$ be $\min(X_1, X_2 ,\ldots, X_n)$. That means all of $X_1, X_2 ,\ldots, X_n$ must be $\ge M$. Hence,

\begin{align} P(M \le m) & = 1 - P(M \ge m) \\ & = 1 - e^{-\lambda}e^{-\lambda}e^{-\lambda} \cdots \\ & = 1 - e^{-\lambda n} \end{align} which matches the cumulative distribution function of an exponential distribution.

However, I am wondering why $\min(X_1, X_2 ,\ldots, X_n)$ is the same thing as saying that all of the values $X_1, X_2 ,\ldots, X_n$ must be greater than or equal to a certain value $m$. What if none of $X_1, X_2 ,\ldots, X_n$ equal that value $m$? How are these two statements equivalent?

• I think my answer is at least simpler than the others. – Michael Hardy Nov 26 '12 at 3:45
• You have $\min\sim\operatorname{Expo}(X_1+\cdots+X_n)$ where you need $\min\sim\operatorname{Expo}(\lambda + \cdots + \lambda). \qquad$ – Michael Hardy Apr 16 '18 at 14:51

We do the problem in a somewhat more inefficient way, so that the logic will be clear.

Let $Y=\min(X_1, X_2,\dots,X_n)$. We want to find the cumulative distribution function $F_Y(y)$ of $Y$. By definition, $$F_Y(y)=\Pr(Y\le y).$$ It is clear that $F_Y(y)=0$ if $y\le 0$. We now find $F_Y(y)$ for $y\gt 0$.

The minimum of $X_1,X_2,\dots,X_n$ is $\le y$ if and only if at least one of the $X_i$ is $\le y$. This is straightforward to prove. If at least one of the $X_i$ is $\le y$, then the minimum of the $X_i$ must be $\le y$. And if the minimum is $\le y$, than at least one of the $X_i$ must be $\le y$.

To find the probability that at least one of the $X_i$ is $\le y$, we first find the probability of the complementary event that they are all $\gt y$.

Since each $X_i$ is exponentially distributed with parameter $\lambda$, $\Pr(X_i\gt y)=e^{-\lambda y}$. Since the $X_i$ are independent, the probability that all of them are $\gt y$ is $\left(e^{-\lambda y}\right)^n$, which is $e^{-n\lambda y}$.

So the probability that at least one of the $X_i$ is $\le y$ is $1-e^{-n\lambda y}$. We conclude that $$F_Y(y)=1-e^{-n\lambda y}$$ if $y \gt 0$.

We may recognize this as the cumulative distribution function of an exponentially distributed random variable with parameter $n\lambda$. Or we can take the derivative of the cdf $F_Y(y)$, and find that $Y$ has density function $n\lambda e^{-n\lambda y}$ (for $y\gt 0$), which we recognize as the density function of an exponentially distributed random variable with parameter $n\lambda$.

The statement you make is somewhat unclear, i.e. the meaning of $\mathrm{Expo}(X_1+X_2+\cdots+X_n)$. What you proved is that the $\min(X_1,X_2,\ldots,X_n) \sim \exp(n\lambda)$, given $X_i \sim \exp(\lambda)$. Concerning your second question - the minimum is not the same thing as a certain value of $m$, since the later is not a random variable.

$\min(X_1,\ldots,X_n)$ is not the same thing as saying all of the values are greater than $m$.

$\min(X_1,\ldots,X_n)>m$ is the same thing as saying all of the values are greater than $m$.