# expectation & variance of square of non-standard normal distribution

Let $$A \sim Normal(k, 1)$$ (ie. with mean $$k$$ and variance $$1$$). Let $$B = A^2$$. The first step is to express the CDF of $$B$$ in terms of the CDF $$S_X(x)$$ of the standard normal distribution (with mean $$0$$ and variance $$1$$).

That part I think is fine. I said the CDF of $$B$$ is expressed by: $$P(B \leq b) = P(A^2 \leq b) = P(-\sqrt{b} \leq A \leq \sqrt{b}) = S_X(\sqrt{b} - k) - S_X(-\sqrt{b} - k)$$.

The next part is to find the expectation and variance of $$B$$. I am not seeing how to do that short of evaluating some unappealing integrals. How would one go about doing this?

My class has not yet introduced the Chi-squared distribution, though I am aware this question is related to that. If possible, answers not using the Chi-squared distribution would be more helpful.

Edit: haven't covered moment-generating functions yet, either.

Note that $$E(B) = E((Z+k)^2)$$, where $$Z$$ ~ $$N(0, 1)$$. Do you know $$E(Z^4)$$?

• I think we can also solve for $Var(A) = E(B) - E(A)^2$, which doesn't involve $A^4$, and that would lead us to an answer of $1 + k^2$ for $E(B)$. However, I think the variance calculation for $B$ might involve $E(Z^4)$ as you suggested. would you mind elaborating? Thank you! – 0k33 Nov 6 '18 at 14:44
• Ah, thats even faster and yields $E(B) = 1+k^2$. Now $Var(B) = E(B^2) - E(B)^2 = E((Z+k)^4) - (1+k^2)^2$ and using $E(Z)=0=E(Z^3)$ we obtain $E((Z+k)^4) = E(Z^4) + 6k^2E(Z^2) + k^4 = 3+6k^2+k^4$. – Stockfish Nov 6 '18 at 14:51
• I see now, thank you so much! Also, last question -- why is $E(B)=E((Z+k)^2)$, and not $E(B)=E((Z-k)^2)$? I'm probably just messing up some translating someplace? – 0k33 Nov 6 '18 at 14:58
• We have $E(Z+k) = k$, so it cannot be $Z-k$ :) as $A$ has mean $k$, too, and $Z-k$ has mean $-k$. – Stockfish Nov 6 '18 at 15:00
• perfect! Thank you very much! @Stockfish – 0k33 Nov 6 '18 at 15:02