Let $Y=B_1 + B_2 +...+B_n$ where the R.V. $B_1...B_n$ are i.i.d. and each has a Bern(p) distribution. Of course, we know that each of these R.V. $B_i$ has a mean $p$ and a variance $p(1-p)$ and that the MGF of a Bernoulli R.V. is $1-p+pe^t$.
I was told to find the following: $E(Y)$, $Var(Y)$ and MGF of $Y$. Here is my work:
For the $E(Y)$ we have $\sum_{i=1}^{n}$ $Y$$*$$pmf$ is the way to get the expectation. Thus in this case, we get $\sum_{i=1}^{n}$ $(B_1 + B_2 + ... B_n)*pmf$ and we come to the conclusion that the expectation is $n(1-p)$.
Now the real question is for the $Var(Y)$. I believe there are 2 ways that I know of to get the $Var(Y)$. The first, is to use this equation $Var(Y)=E(Y^2)-[E(Y)]^2$ and the second is to find $M_Y(t)$ and then figure out the first and second moments and use the equation above. Unfortunately, I am not sure how to proceed. Here is what I would do in each case:
In the first case: Would the $E(Y^2)$ simply be $\sum_{i=1}^{n}$ $(B_1^2 + B_2^2 + ... B_i^2) * pmf$?
In the second case using the MGF:
$M_Y(t)$ $=$ $E(e^{tY})$ = $\sum_{i=1}^{n}$ $e^{tY} * pmf$ is this correct? Then we can simply this to $M_Y(t)$ $=$ $\sum_{i=1}^{n}$ $[e^{tB_1} * e^{tB_2} * ... e^{tB_n}] * pmf$ If this is correct, I am unsure of the next step and help would be greatly appreciated!
Thanks so much!
By the linearity of expectation, $${\rm E}[Y] = {\rm E}\sum_{i=1}^n B_i = \sum_{i=1}^n {\rm E}[B_i] = \sum_{i=1}^n p = np.$$ By the independence of each of the $B_i$'s, the variance of $Y$ is the sum of the variances of each $B_i$: $${\rm Var}[Y] = \sum_{i=1}^n {\rm Var}[B_i] = \sum_{i=1}^n p(1-p) = np(1-p).$$ Again, using the independence of the $B_i$'s (hence the $\exp(B_i t)$'s are also independent), the expectation of their product is the product of the expectations, so the MGF of $Y$ is given by $$\begin{align*} M_Y(t) &= {\rm E}[e^{Yt}] = {\rm E}\left[\exp \sum_{i=1}^n B_i t \right] \\ &= {\rm E}\left[ \prod_{i=1}^n e^{B_i t} \right] \\ &= \prod_{i=1}^n {\rm E}[e^{B_i t}] \\ &= \prod_{i=1}^n M_{B_i}(t) \\ &= (1 - p + pe^t)^n. \end{align*}$$