0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – 0k33 Nov 6 '18 at 14:44
  • 1
    $\begingroup$ 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 $E((Z+k)^4) = E(Z^4) + 6k^2E(Z^2) + k^4 = 3+6k^2+k^4$. $\endgroup$ – Stockfish Nov 6 '18 at 14:51
  • 1
    $\begingroup$ 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? $\endgroup$ – 0k33 Nov 6 '18 at 14:58
  • $\begingroup$ We have $E(Z+k) = k$, so it cannot be $Z-k$ :) as $A$ has mean $k$, too, and $Z-k$ has mean $-k$. $\endgroup$ – Stockfish Nov 6 '18 at 15:00
  • $\begingroup$ perfect! Thank you very much! @Stockfish $\endgroup$ – 0k33 Nov 6 '18 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.