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Let $X$ be a random variable that is normally distribted with mean $\mu$ and variance $\sigma^2$. Compute

  1. $E(X^2)$
  2. $E(X^4)$
  3. $Var(X^2)$

They also hint that for a standard normally distributed random variable $Z$ it follows that $E(Z^4)=3$.

I don't see how to use the hint given

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  • $\begingroup$ How would you write $X$ in term of $Z$? $X = . \times Z + . $ $\endgroup$
    – Frostic
    Apr 26, 2018 at 17:51
  • $\begingroup$ Z-score I suppose? $X=Z \sigma + \mu?$ $\endgroup$
    – Parseval
    Apr 26, 2018 at 18:01
  • $\begingroup$ Yea exactly now try to plug this equation in your expectation and work with the linearity of the expectation $\endgroup$
    – Frostic
    Apr 26, 2018 at 18:02
  • $\begingroup$ I get to $$E(X^2)=E((Z\sigma+\mu)^2)=E(Z^2\sigma^2+2Z\sigma+\mu^2)=\sigma^2E(Z^2)+2\sigma\mu E(Z)+\mu^2.$$ Is it generally true that $E(Z^k)=k-1?$ In that case then the answer should be $$E(X^2)=\sigma^2+\mu^2?$$ $\endgroup$
    – Parseval
    Apr 26, 2018 at 19:02
  • $\begingroup$ @Parseval : It is not generally true that $\operatorname E(Z^k) = k-1,$ but your conclusion that $\operatorname E(X^2) = \mu^2+\sigma^2$ is correct. $\qquad$ $\endgroup$
    – Michael Hardy
    Apr 26, 2018 at 19:06

1 Answer 1

Reset to default
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$\newcommand{\e}{\operatorname{E}}$\begin{align} \e(X^4) & = \e((\mu+\sigma Z)^4) \\[10pt] & = \e(\mu^4 + 4\mu^3\sigma Z + 6\mu^2\sigma^2 Z^2 + 4 \mu\sigma^3 Z^3 + \sigma^4 Z^4) \\[10pt] & = \mu^4 + 4\mu^3\sigma \e(Z) + 6\mu^2\sigma^2 \e(Z^2) + 4\mu \sigma^3 \e(Z^3) + \sigma^4 \e(Z^4). \end{align} Now apply what you were given about the expected values of powers of $Z.$

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  • $\begingroup$ Thanks! Can you plese check if my answer to a comment above is correct? $\endgroup$
    – Parseval
    Apr 26, 2018 at 19:06
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    $\begingroup$ +1 for your comments above ! $\endgroup$
    – Frostic
    Apr 26, 2018 at 20:25

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