Say we have a binomial random variable $B \sim \operatorname{Bin}(n,p)$ and $n$ i.i.d. Normal random variables $N_i \sim N(p, 100)$. What is the distribution of $\sum_{i=1}^B N_i$?
I'm not really sure how to go about this. I know that I can calculate the expectation through Wald's Lemma, but in general, I'm not sure how to deal with random sums like this.
I'm not sure how to deal with random sums like this
Let $X_{1}, X_{2}, X_{3},\cdots$ be independent and identically distributed random variables and let $N$ be a non-negative integer valued random variable and is independent of the random variable $X$. If $$S_{N}=X_{1}+X_{2}+\cdots +X_{N}$$ be the sum of a random number of independent and identically distributed random variables, then \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E[S_{N}]&=&E[N]\cdot E[X]\\ Var[S_{N}]&=& E[N]\cdot Var[X] + Var[N]\cdot \left\{E[X]\right\}^{2} \end{eqnarray*} We first find the conditional expectation of $S_{N}$ given the value of $N$ as \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E\left[S_{N}|N=n\right] &=& E\left[X_{1}+X_{2}+\cdots +X_{N}|N=n\right] \\ &=& E\left[X_{1}+X_{2}+\cdots +X_{n}\right]\\ & &\quad \mbox{ since X and N are independent RV's}\\ &=& n E[X], \quad\quad\mbox{ since $X_{i}'s$ are iid RV's}\\ E\left[S_{N}\right] &=& \sum_{n}\underbrace{E\left(S_{N}|N=n\right)}_{n\cdot E[X]} P\left\{N=n\right\}\\ &=& \sum_{n} n\cdot E[X]\cdot P\left\{N=n\right\}\\ &=&E(X)\cdot \sum_{n}n\cdot P\left\{N=n\right\}\\ E\left[S_{N}\right] &=& E[X]\cdot E[N] \end{eqnarray*} In order to find an expression for $Var[S_{N}]$, we first find the second conditional moment of $S_{N}$. \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E(S_{N}^{2}|N=n) &=& E[(X_{1}+X_{2}+\cdots +X_{N})^{2}|N=n] \\ &=& E[(X_{1}+X_{2}+\cdots +X_{n})^{2}]\\ &=& E\left\{\sum_{i=1}^{n}X_{i}^{2}+2\sum_{i<j}X_{i}X_{j}\right\}\\ &=& \sum_{i=1}^{n}E\left(X_{i}^{2}\right) + 2 \sum_{i<j} E\left(X_{i}\right)E\left(X_{j}\right)\\ &=& n\cdot E\left(X^{2}\right) + 2\cdot \frac{n(n-1)}{2}E\left(X\right)E\left(X\right)\\ \end{eqnarray*} \begin{eqnarray*} E(S_{N}^{2}|N=n) &=& n\cdot E\left(X^{2}\right) + n(n-1)\left(E(X)\right)^{2}\\ &=& n[E(X^{2})-(E(X))^{2}] + n^{2}(E(X))^{2}\\ E(S_{N}^{2}|N=n) &=& n Var(X) + n^{2}(E(X))^{2} \end{eqnarray*} Now \begin{eqnarray*} % \nonumber to remove numbering (before each equation) Var(S_{N}) &=& E(S_{N}^{2}) - \left[E(S_{N})\right]^{2} \\ &=& \sum_{n}E(S_{N}^{2}|N=n)P\{N=n\}-[E(X)E(N)]^{2}\\ &=&\sum_{n}\left[nVar\left(X\right)+n^{2}\left(E\left(X\right)\right)^{2}\right] P\left\{N=n\right\}\\ & &\quad\quad\quad -\left[E(X)E(N)\right]^{2}\\ &=& Var(X) \sum_{n}nP\left\{N=n\right\}\\ & & + \left(E\left(X\right)\right)^{2}\sum_{n}n^{2}P\left\{N=n\right\} - \left[E(X)E(N)\right]^{2}\\ &=&E(N)Var(X) + \left(E\left(X\right)\right)^{2} \left[E(N^{2})-(E(N))^{2}\right]\\ Var(S_{N}) &=& E(N)Var(X) + Var[N](E(X))^{2} \end{eqnarray*}
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