I have a question regarding the probability distribution of the difference of two random variables.

Suppose we have $n$ exponential random variables, and they are all independent and identically distributed with parameter $\lambda$. Then, suppose we define random variable $Y_i$ as the $i$-th smallest $X_k$. For example, $Y_1 = \min(X_1, X_2, \ldots, X_n)$. $Y_2$ is the second smallest $X$.

I understand the following two facts: 1) Exponential random variables are memoryless. That is, if $X$ is an exponential random variable, $P(X \geq t + t_0 | X \geq t_0) = P(X \geq t)$.

and 2) If $X_1, X_2, \ldots, X_k$ are independent, each with parameter $\lambda$, then the random variable $Y_1 = \min(X_1, X_2, \ldots, X_k)$ has the probability density function: $f(y_1) = k \lambda e^{-k \lambda y_1}$

Then, my question is: how do I get the probability density function of $Y_2 - Y_1$? $Y_2 - Y_1$ is the second smallest exponential random variable minus the smallest exponential random variable. I understand that the expected value of $(n-1)$ exponential random variables with the same parameter $\lambda$ is $\frac{1}{(n-1)\lambda}$ (since we can use fact 2 above and set the number of random variables to $k-1$), but I do not see how $E(Y_2-Y_1)$ equals that quantity.

I believe strongly that fact 1 is involved, but I do not see how $Y_2 - Y_1$ is related to $(Y_2 - Y_1 | Y_1 = y_1)$.

Thanks everyone.

  • 2
    $\begingroup$ Hint: A. Renyi, On the theory of order statistics, Acta Mathematica Hungarica, vol. 4, nos. 3-4, 191-231. (See also, e.g., S. Resnick, 1992.) $\endgroup$
    – cardinal
    Commented Jun 2, 2011 at 19:52
  • $\begingroup$ @cardinal: Well done. $\endgroup$
    – Did
    Commented Jun 2, 2011 at 20:40
  • $\begingroup$ This shows up, for example, (oftentimes in slightly disguised form) in standard proofs of the limiting distribution of the Kolmogorov-Smirnov statistic. $\endgroup$
    – cardinal
    Commented Jun 2, 2011 at 23:20

1 Answer 1


A simple method is to compute the distribution of $(Y_1,Y_2)$. Pick some positive $y_1<y_2$. What does it take to know that $Y_1\in(y_1,y_1+\mathrm{d}y_1)$ and $Y_2\in(y_2,y_2+\mathrm{d}y_2)$? You have to choose:

(1) the rank of the smallest value, for which there exists $n$ possibilities,

(2) the rank of the second smallest value, for which there exists $n-1$ possibilities,

(3) that these two random variables are indeed in the intervals $(y_1,y_1+\mathrm{d}y_1)$ and $(y_2,y_2+\mathrm{d}y_2)$, which happens with respective probabilities $\lambda \mathrm{e}^{-\lambda y_1}\mathrm{d}y_1$ and $\lambda \mathrm{e}^{-\lambda y_2}\mathrm{d}y_2$,

and (4) finally that the $n-2$ other random variables are greater than $y_2$, which happens with probability $\mathrm{e}^{-(n-2)\lambda y_2}$.

All this yields the probability density function of $(Y_1,Y_2)$ as $$ n(n-1)\lambda^2 \mathrm{e}^{-\lambda y_1-(n-1)\lambda y_2}\mathbf{1}_{0<y_1<y_2}. $$ From this, to compute the value of $E(Y_2-Y_1)$ is straightforward. For example, introducing $Z_2=Y_2-Y_1$, one sees that the probability density function of $(Y_1,Z_2)$ is $$ n(n-1)\lambda^2 \mathrm{e}^{-n\lambda y_1-(n-1)\lambda z_2}\mathbf{1}_{y_1>0}\mathbf{1}_{z_2>0}. $$ This is a product distribution hence $Y_1$ and $Z_2$ are independent with well known distributions, exponential with parameters $n\lambda$ and $(n-1)\lambda$, respectively.

  • $\begingroup$ Hi Didier Piau. My understanding is the steps were: 1) Create the joint pdf of (Y1,Y2). 2) Do a change of variables. 3) In the new joint pdf, it is apparent it is the product of two known pdf's. Hence, the new joint pdf can be separated. 4) In one of the two separations ($Z_2$), we can see our object of interest and it turns out to have an exponential $(n-1) \lambda$ distribution. $\endgroup$
    – jrand
    Commented Jun 3, 2011 at 8:33
  • $\begingroup$ Can I do step (2) (Change of variables) to any p.d.f. or does the substitution have to be linear? So if I have $f(x,y)$, can I create $f(x,z)$, with $z = g(x,y)$, $g()$ being any arbitrary function? $\endgroup$
    – jrand
    Commented Jun 3, 2011 at 8:38

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