3
$\begingroup$

Let $X_{1},X_{2},X_{3}$ be random variates from $U(0,1)$. It is required to compute $E(\frac{X_{1}+X_{2}}{X_{1}+X_{2} + X_{3}})$.

Here is what I did..

$E(\frac{X_{1}+X_{2}}{X_{1}+X_{2} + X_{3}}) = E(1 - \frac{X_{3}}{X_{1}+X_{2} + X_{3}}) = 1 - E(\frac{X_{3}}{X_{1}+X_{2} + X_{3}})$

From the exchangeability property of random variates, we can write:

$E(\frac{X_{3}}{X_{1}+X_{2} + X_{3}}) = E(\frac{X_{2}}{X_{1}+X_{2} + X_{3}}) = E(\frac{X_{3}}{X_{1}+X_{2} + X_{1}})$

Now, adding all of the above three, we get 1 and consequently, the value of the above expectation will be $\frac{1}{3}$

Hence, from the first equation, we get $\frac{2}{3}$.

I just wanter to know whether my approch of this problem correct? If yes, can you suggest some alternative methods also.

$\endgroup$
1
  • 3
    $\begingroup$ The similar thread at stats.stackexchange.com/questions/374989 validates your approach. Because distributions of sums of uniform variables are complicated, alternative approaches are unattractive. You could shorten your analysis by observing that the exchangeability of the $X_i$ implies $$E\left[\frac{X_1+X_2}{X_1+X_2+X_3}\right] \\= \frac{1}{3}\left(E\left[\frac{X_1+X_2}{X_1+X_2+X_3}\right] + E\left[\frac{X_2+X_3}{X_2+X_3+X_1}\right] + E\left[\frac{X_3+X_1}{X_3+X_1+X_2}\right] \right) = \frac{1}{3}E[2] = \frac{2}{3}.$$ $\endgroup$
    – whuber
    Feb 3 at 23:07

Your Answer

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

Browse other questions tagged or ask your own question.