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Let $\{ Y_j: 1\leq j \leq K \}$ be a collection of i.i.d. random variables. Suppose we have two random variables $W$ and $W'$ that have the same distribution function, where $W'$ is given by: $$W'=\sum_{j=1}^{K} Y_j,$$ where $K$ is a random number. Then:

$M_W(s)=G_K(M_Y(s))$, where $M_Y$ is the moment generating function of $Y$.

Since $W$ and $W'$ have the same distribution, they have the same moment generating function: $$M_W(s)=M_{W'}(s).$$

By definition of the probability generating function, the RHS is given by: $$G_K(M_Y(s))=\mathbb{E}[(M_Y(s))^K]=\mathbb{E}[(\mathbb{E}[e^{Ys}])^K].$$ To compute the LHS, I condition on $K$:

$$M_{W'}(s)=\mathbb{E}[e^{W' s}]=\mathbb{E}[\mathbb{E}[e^{W' s} \vert K]]=\mathbb{E}[\mathbb{E}[e^{s\sum_{j=1}^{K}Y_j } \vert K]]=\mathbb{E}[\mathbb{E}[e^{sY_1s+...Y_Ks} \vert K]]=\mathbb{E}[\mathbb{E}[e^{Y_1s}\vert K]...\mathbb{E}[e^{Y_Ks}\vert K] ]=\mathbb{E}[\mathbb{E}[e^{Ys}\vert K]^K ].$$

So is $\mathbb{E}[(\mathbb{E}[e^{Ys}])^K]=\mathbb{E}[\mathbb{E}[e^{Ys}\vert K]^K ]$?

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  • $\begingroup$ Is the notation $G_K$ standard? What does it mean? $\endgroup$ Aug 29, 2014 at 9:10
  • $\begingroup$ It is the probability generating function. $G_K(s):=\mathbb{E}[s^K]$. $\endgroup$
    – mr_T
    Aug 29, 2014 at 9:22
  • $\begingroup$ Is K random or deterministic? Depending on this, some parts of the proof may need to be revised. $\endgroup$
    – Did
    Aug 29, 2014 at 10:05
  • $\begingroup$ It is random (it gives the number of some points). $\endgroup$
    – mr_T
    Aug 29, 2014 at 11:44
  • $\begingroup$ So I have to condition on $K$, isn't it? I edited my post. $\endgroup$
    – mr_T
    Aug 29, 2014 at 14:13

1 Answer 1

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If the random variables $Y$ and $K$ are independent, then $ \Bbb E[e^{Ys}\mid K] = \Bbb E[e^{Ys}]$ a.s., hence $$ \Bbb E[\Bbb E[e^{Ys} \mid K]^K] = \Bbb E[\Bbb E [e^{Ys}]^K]. $$

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