I am given that X is a random variable with a mean of two and variance of 4, as per the question:
So, I try to solve the problem by expanding $E[x+2]^2$
$$E[x+2]^2 = E[x+2] * E[x+2]$$ $$E[x+2] = E[x] + E[2] = 2 + 2 = 4$$ $$E[x+2]^2 = 4*4=16$$
However, the answer is 16. I think I am missing something obvious, but I haven't been able to figure it out. Does anyone know why?
$$E(X+2)^2=E(X^2+4X+4)=E(X^2)+4E(X)+4$$ and $E(X^2)=Var(X)+(E(X))^2=4+2^2=8$ so the answer is 20.
In statistical calculation,usually, when we write $E(X+2)^2$ we means $E\bigg((X+2)^2\bigg)$ and $E^2(X+2)=\bigg(E(X+2)\bigg)^2$