Show equal distribution of the max and average of two exponential i.i.d.s

Question

I've been trying to prove: $Z \overset{d}{=} W$

Where:

• $Z = \frac{X+Y}{2}$,
• $W = \max(X,Y)$,
• X and Y are independent and identically exponantially distributed r.v:s with $\lambda = 1$

My idea was to calculate and compare the characteristic functions. For $Z$ we can derive its c.f., $\gamma_Z(t)$, quite easily by the following:

• $\gamma_Z(t)=\gamma_{X+Y}(\frac{t}{2}) = (\gamma_{X}(\frac{t}{2}))^2 = \frac{1}{(1-it/2)^2}\hspace{15mm}$ (1)

which is the c.f. of a $\Gamma(2,2)$-distributed r.v. While the c.f. of $W$ is harder to derive. Using the answer to this question:

• https://stats.stackexchange.com/questions/18433/how-do-you-calculate-the-probability-density-function-of-the-maximum-of-a-sample

First we can see that the c.d.f. of $W$ is:

• $F_W(w) = F_X(w)F_Y(w) = (1 - e^{-w})^2 = 1 - 2e^{-w} + e^{-2w}$

And by deriving it w.r.t. $w$ we get:

• $f_W(w) = \frac{dF_W(w)}{dw} = 2(e^{-w} - e^{-2w})$

Then:

• $\gamma_W(t) = E[e^{itw}] = \int_0^{\inf}e^{itw}f_W(w)dw = 2\int_0^{\inf}e^{itw}(e^{-w} - e^{-2w})dw$

• ... $\hspace{8mm}= 2\int_0^{\inf}e^{-w(1-it)} - e^{-w(2-it)}dw $

• ... $\hspace{8mm}= 2[\frac{e^{-w(1-it)}}{it-1} - \frac{e^{-w(2-it)}}{it-2}]_0^{\inf} $

• ... $\hspace{8mm}= 2[0 - (\frac{1}{it-1} - \frac{1}{it-2})] $

• $\gamma_W(t) = \frac{1}{1-3/2it-t^2/2} $

Which does not match the results in (1) and makes me doubt that my approach is the correct one to solve this problem. Any hints or corrections would be greatly appreciated!

Thanks!

Answer 1

The result you've stated is actually false, and @drhab gave a fantastic quick argument as to why.

My best guess as to what's going on is that you may have mis-copied the question. Is it possible that $Z$ should be $X + \frac Y 2$ instead? That variable has the same distribution as $\max\{X, Y\}$, and a good explanation for why can be found here (copied below).

\begin{align} {\rm Pr}(X + Y/2 \leq a) &= \int_0^a e^{-x} {\rm Pr}\left(Y \leq 2(a-x)\right)\ dx\\&=\int_0^a e^{-x} \left(1-e^{-2(a-x)}\right)\ dx\\&= \int_0^a e^{-x}\ dx - e^{-2a}\int_0^a e^x\ dx\\& = 1 - e^{-a} - e^{-2a}(e^a-1)\\ &= 1 - 2e^{-a} + e^{-2a}\\&= \left(1-e^{-a}\right)^2\\ &= {\rm Pr}(X \leq a, Y \leq a)\\&= {\rm Pr}(\max\{X , Y\} \leq a)\end{align}

Answer 2

They are not equally distributed since $\mathbb{E} Z = 1$ and $\mathbb{E}W=1+1+\tfrac{1}{2} = \tfrac{3}{2}$, see here why.