Let $\{X_n:n\ge 1\}$ be a sequence of i.i.d. Bernoulli random variables with probability of success $0<p<1$, i.e, $$P\{X_1=1\}=1-P\{X_1=0\}=p.$$ The random variable $Y$ is independent of the sequence $\{X_n:n\ge 1\}$ and has a Poisson distribution with parameter $\lambda>0.$ Find the characteristic function of the random variable $$Z=\sum_{k=1}^{Y+1} X_k.$$

So I know the characteristic function is $\phi_Z(t)=E\left[e^{itZ}\right]=E\left[e^{it\sum_{k=1}^{Y+1} X_k}\right]$ But I am not sure how to expand the $E\left[e^{it\sum_{k=1}^{Y+1} X_k}\right]$ term because whether or not the event $X_k$ for $k>1$ occurs is Poisson distributed.

My attempt:

The probability of $P(Y=k)= \frac{\lambda^ke^{-\lambda}}{k!}$.

$E\left[e^{it\sum_{k=1}^{Y+1} X_k}\right]=\Pi^{\infty}_{k=0}E\left[P(Y=k)e^{itX_k}\right]=\Pi^{\infty}_{k=0}P(Y=k)\left(1-p+pe^{it}\right)=\Pi^{\infty}_{k=0}\frac{\lambda^ke^{-\lambda}}{k!}\left(1-p+pe^{it}\right)$

Am I on the right track? I am not sure because I have never seen a random variable defined like $Z$ before.

  • $\begingroup$ I think that if Y is independent of Xi then E[Z] = p*E[(Y+1)]. That might simplify things $\endgroup$
    – MikeP
    Commented Mar 14, 2016 at 12:30

1 Answer 1


The idea here is to use conditioning:

\begin{align} \text{E} \left ( e^{it \sum_{k=1}^{Y+1} X_k} \right ) &= \text{E} \left [ \text{E} \left ( e^{it \sum_{k=1}^{Y+1} X_k} \mid Y \right ) \right ] \\ &= \text{E} \left [ \prod_{k=1}^{Y+1} \text{E} \left ( e^{it X_k} \right ) \right ] \\ &= \text{E} \left [ \text{E} \left ( e^{it X_1} \right )^{Y+1} \right ] \\ &= \text{E} \left [ (pe^{it} + 1 - p)^{Y+1} \right ] \\ &= \sum_{k=0}^{\infty} e^{- \lambda} \frac{\lambda^k}{k!} (pe^{it} + 1 - p)^{k+1} . \end{align}


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