There are $n$ students and you pick a student replacement so each pick is independent . $p$ is the portion of the student body that will vote for A instead of B for student council president.
Let $X = # of the students who'll vote for A in n$
Let $Y = # of students who'll vote for B$
- $E(X-Y)$
- $Var(X-Y)$
Express X in terms of Y
How do I approach this? From definition, I'm assuming:$$E(X) = \sum_{n=0}^{pn}aP(X=a)$$ Which means $$E(Y) = \sum_{n=0}^{(1-p)n}bP(Y=b)$$
And since $E(X-Y) = E(X)-E(Y)$, I subtract one from the other. Here I get stuck...but I know if I can solve for $E(X)$ and $E(Y)$, I can find $Var(X-Y)$
We assume that there are only two candidates. Then $Y=n-X$, so $X-Y=2X-n$.
The expectation of $2X-n$ is $2E(X)-E(n)$. We have $E(n)=n$. The random variable $X$ has Binomial distribution, parameters $n$ and $p$, so $E(X)=np$.
Thus $E(X-Y)=2np-n$.
For the variance of $X-Y$, we want $\text{Var}(2X-n)$. This is $2^2\text{Var}(X)$.
The variance of $X$ is $np(1-p)$, by standard information about the Binomial. Thus $X-Y$ has variance $4np(1-p)$.