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

  • $\begingroup$ 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


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


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.