Let $X$ and $Y$ be independent random variables, with known moment generating functions $M_X(t)$ and $M_Y (t)$ and $I$ be such that $P(I = 1) = 1 − P(I = 0) = p \in (0, 1)$. Compute the moment generating function of the random variable $S = IX + (1−I)Y $.

I am given the hint that taking condition on I may help but still I have no idea how to compute. Any help would be appreciated.


The idea is to use the double expectation theorem, that is $$ \DeclareMathOperator{\E}{\mathbb{E}} \E X = \E \E X \mid Y $$ and we use this with $I$ as the conditioning variable.

$$ M_S(t) = \E e^{tS} = \E e^{t(IX+(1-I)Y)} \\ = \E \left\{ \E e^{tIX +t(1-I)Y}\mid I \right\} \\ = p\E e^{tX+0} + (1-p)\E e^{tY+0} \\ = pM_X(t) + (1-p)M_Y(t) $$ which, indeed, is simpler that the other answer (and correct).

  • 2
    $\begingroup$ Unfortunately I did not get pinged by the downvote nor the top comment in your answer so I have not seen my erroneous answer until now. You are correct, my answer was wrong and I have now deleted it as your answer is the same method but correctly applied. $\endgroup$ – Therkel Oct 12 '17 at 16:47

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