Let $x$ represent the number of successes and $y$ the number of failures of $n$ independent Bernouli trials with $p$ representing the probability of success in any one trial. Find the distribution of $z = x - y$ . Show that $E[z] = n(2p − 1)$, $Var(z) = 4np(1 − p)$.
I had a solution by myself but I'm not sure about it, my solution is that the $x$ and $y$ are bionimial distribution, am I right?
You are correct about $X \sim Bin(n, p)$ and $Y \sim Bin(n, 1- p)$. This is because $1-p$ is the probability of failure and then we get this by using the definition of the binomial random variable.
Now to find the expected value apply LOTUS on $z$, by $E(z)=E(x)-E(y)=np-n(1-p)=n(2p-1)$
Now since $x$ and $y$ are not independent (i.e. $x$ and $y$ are dependent) we cannot simply add their variances. Notice however that $x+y = n$, hence $y = n−x$.
Therefore, $V(z) = V(x − y) = V(x − n + x) = V(2x − n) = 4 V(x) = 4np(1 − p)$