Preparation for an Exam in Probabilities I am preparing for an exam in Probability and I have the two following questions.
First: let $\{X_{n,k}:k\in\mathbb{N}\}$ be a collection of identically distributed and independent random variables that have a variance $\sigma^2>0$ and a mean $\mu >0$. 
Let $p_0=\mathbb{P}(X_{1,1}=0)$ and define $Z_0,Z_1,...$ by: $Z_0=1$ and $Z_n=X_{n,1}+X_{n,2}+...+X_{n,Z_{n-1}}$, for $n\in \mathbb{N}$. 
Now I need to show that: $\mathbb{E}(Z_{n+1}^2|\mathcal{F}_n)=\mu^2Z_{n}^2+\sigma^2Z_n$, for $\mathcal{F}_n=\sigma(Z_0,Z_1,...,Z_n)$. 
I have already shown that $M_n=Z_n/\mu^n$ is an $(\mathcal{F}_n)-martingale$, since: $\mathbb{E}(Z_n/\mu^n|\mathcal{F}_{n-1})=1/\mu^n\mathbb{E}(Z_n|\mathcal{F}_{n-1})=1/\mu^n(Z_{n-1})\mu=M_{n-1}$.
The second exercise asks me to show that if $X_\lambda$ is a Poisson random variable with parameter $\theta\lambda$, then $X_\lambda/\lambda$ converges in probability as $\lambda\rightarrow\infty$ and to determine the limit.
I have already calculated the characteristic function of $X_\lambda$ to be:
$\phi_{X_\lambda/\lambda}=\mathbb{E}e^{iuX}=\sum_{k=0}^{\infty}(\theta\lambda)^k/k!e^{-\theta\lambda}e^{iuk}=e^{\theta\lambda(e^{iu}-1)}$.
 A: Doing your computations more carefully, one gets $\phi_{X_\lambda/\lambda}(u)=\exp(\theta\lambda(\mathrm e^{\mathrm iu/\lambda}-1))$. One knows that $\mathrm e^z=1+z+o(z)$ when $z\to0$, hence, using this for $z=\mathrm iu/\lambda$, one sees that $\phi_{X_\lambda/\lambda}(u)\to\mathrm e^{\mathrm iu\theta}$ for every fixed $u$, when $\lambda\to\infty$. This proves that $X_\lambda/\lambda\to\theta$ in distribution as $\lambda\to\infty$. The limit is deterministic hence $X_\lambda/\lambda\to\theta$ in probability.
A: Here is a solution to the first part of your question. Since you have deduced that 
$$
\mathbb{E}\left[Z_{n+1}{}\left|\mathcal{F}_n \right.\right]{}={}Z_{n}\mu\,,
$$
then, observe that, because $Z_{n}$ is determined by the sigma-algebra $\mathcal{F}_n$ and the $X_{n,k}$ are independent and identically distributed, we have
$$
\begin{eqnarray*}
\mathbb{V}ar\left(Z_{n+1}\left|\mathcal{F}_n \right.\right)&{}={}&\mathbb{V}ar\left(\left. \sum\limits_{i=1}^{Z_{n}}X_{n+1,i}\right|\mathcal{F}_n \right)\newline
&{}={}&\sum\limits_{i=1}^{Z_{n}}\mathbb{V}ar\left(X_{n+1,i}\left|\mathcal{F}_n \right.\right)\newline
&{}={}&\sum\limits_{i=1}^{Z_{n}}\mathbb{V}ar\left(X_{n+1,i}\right)\newline
&{}={}&Z_{n}\sigma^2\,.
\end{eqnarray*}
$$
Which means that,
$$
\mathbb{E}\left[Z^2_{n+1}{}\left|\mathcal{F}_n \right.\right]{}={}\mathbb{V}ar\left(Z_{n+1}\left|\mathcal{F}_n \right.\right){}+{}\left(\mathbb{E}\left[Z_{n+1}{}\left|\mathcal{F}_n \right.\right]\right)^2{}={}Z_{n}\sigma^2{}+{}Z^2_n\mu^2.
$$
