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Here $X$, $Y$, and $U$ are independent random variables, $X$ and $Y$ have moment generating functions $M_{X}(t)$ and $M_{Y}(t)$ respectively, and $U$ has the uniform distribution on $[0,1]$. I can't seem to find any function of $X$, $Y$, and $U$ that gives the moment generating function required, $$\frac{M_{X}(t) +M_{Y}(t)}{2}.$$ Any ideas?

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More generally, let $X_1,\ldots,X_n$ be independent random variables and set $$V_i=\mathsf 1_{U^{-1}\left(\frac{i-1}n,\frac in\right)},$$ that is, $$ V_i=\begin{cases} 1,& \frac{i-1}n<U<\frac in\\ 0,& \text{otherwise}. \end{cases} $$ Set $Z=\sum_{i=1}^n V_iX_i$, then as the sets $\{V_i = 1:1\leqslant i\leqslant n\}$ are disjoint and $$\mathbb P\left(\bigcup_{i=1}^n\{V_i=1\}\right)=1,$$ by conditioning on $V_i$ we obtain \begin{align} \mathbb E\left[e^{tZ}\right] &= \sum_{i=1}^n\mathbb E\left[e^{tZ}\mid V_i=1 \right]\mathbb P(V_i=1)\\ &=\sum_{i=1}^n p_i\mathbb E\left[e^{tX_i}\right]. \end{align}

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  • $\begingroup$ Does this answer not make use of a discrete uniform distribution, which you've written as $Y$, rather than a continuous uniform distribution on $[0,1]$ (which I've written as $U$)? $\endgroup$ – Exp5LogMingus Dec 14 '15 at 14:39
  • $\begingroup$ Good point, let me revise it. $\endgroup$ – Math1000 Dec 14 '15 at 22:57

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