2
$\begingroup$

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?

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

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 $n\lambda$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$\min(X_1,\ldots,X_n)$ 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 $m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.