# Binomial Sum of Normals

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.

• What have you tried? Do you have a general strategy for approaching problems like this? Is this a homework problem? If so, you should add the self-study tag and read the corresponding wiki
– jld
Oct 25 '17 at 0:57
• Thank you for the feedback. I edited my response to answer these questions. Where can I find the corresponding wiki? Oct 25 '17 at 1:00
• thanks for the edits. You can just click on self-study and then click on the learn more
– jld
Oct 25 '17 at 1:02
• Let $Z = \sum_{i=1}^B N_i$. You want $P(Z \leq z)$ for $z \in \mathbb{R}$. Expand $P(Z)$ using the tower law, then you have many different sums of iid normal RVs which are still normal.
– Alex
Oct 25 '17 at 1:03
• The random variable $\sum_{i=1}^{B}N_{i}$ is called a compound random variable and its mean is $E(B)E(N)$ and variance $E[B]\cdot Var[N] + Var[B]\cdot \left\{E[N]\right\}^{2}$. Oct 25 '17 at 1:03

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*}