Variance of a linear transformed standard normal r.v.

Question

I have the sequence $X_i$ that converges to $N(0, 1)$ in distribution and $$Y = 2X_i + 1$$

I was able to find $E(Y) = 1$, but I am struggling at $Var(Y)$.

From the general variance formula $E[(X-\mu)^2]$ I get $E[Y^2]$, since $\mu = 0$ by standard normal definition of $X_i$: $$E[Y^2] = E[(2X_i+1)^2] = E[4X_i^2+4X_i+1] = 0+0+1 = 1 $$ since the expectation of $X_i$ is $0$.

But this answer is incorrect. Where is my mistake?

Answer 1

How do you get $E(4X_i^{2})=0$? The correct value is $4EX_i^{2}=4$ since $EX_i^{2}$ is the variance of the standard normal distribution.

$EY^{2}=5$ and $EY=1$ so $var (Y)=5-(1)^{2}=4$.