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I have $X_1,...,X_n$, i.i.d. $N(\mu,\sigma^2)$ and I would like to calculate $\text{var}[(X_i−\mu)^2]$.

I know that the solution is $2\sigma^4$.

However, I can't derive it. Any suggestions?

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  • $\begingroup$ Notice that if you write 3\text{var}(X) you see $3\text{var}(X)$ whereas if you write 3\operatorname{var}(X) you see $3\operatorname{var}(X).$ And the difference between \text{var}X and \operatorname{var}X is seen here: $$\begin{align} & \text{var}X \\ \\ & \operatorname{var}X \end{align}$$ So with \operatorname{} the spacing depends on the context in a way that's build in to the software. That's why \operatorname{} is standard. I edited the question accordingly. $\endgroup$ – Michael Hardy Dec 3 '18 at 0:55
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Quite simple, just see: $$V\left[(X - \mu)^2\right] = \sigma^4 V\left[\left(\frac{X - \mu}{\sigma}\right)^2\right]$$ $$=\sigma^4 V(Z^2)$$ where, $ Z \sim N(0, 1)$

So, $Z^2 \sim \chi^2_1$, $V(Z^2) = 2$ (Variance of chi-square is 2$v$, where $v$ is degree of freedom. ).

It complete our proof, $$V\left[(X - \mu)^2\right] = 2 \sigma^4$$

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    $\begingroup$ To avoid confusion, I'd write $V\left(\lbrack X-\mu \rbrack ^2\right)$ instead of $V\left(X-\mu\right)^2$, as the latter can be misunderstood. I'd also make similar chages to the $\sigma^4V(\frac{X-\mu}{\sigma})^2$ part. $\endgroup$ – Phil Nov 28 '18 at 8:44
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    $\begingroup$ So in parentheses you write "(Variance of chi-square is $2v$, where $v$ is degree[s] of freedom.)" I thought the question was how to prove that. $\endgroup$ – Michael Hardy Dec 1 '18 at 20:17
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The question is how to prove $\operatorname{var}\left( \left( \dfrac{X-\mu} \sigma \right)^2 \right) = 2$ when $X\sim\operatorname N(\mu,\sigma^2).$

We have $Z=\dfrac{X-\mu}\sigma \sim \operatorname N(0,1),$ so the problem is how to prove $\operatorname{var}(Z^2) = 2$ when $Z\sim\operatorname N(0,1).$

First note that $$ \operatorname E(Z^2) = \operatorname E((Z-0)^2) = \operatorname E((Z-\operatorname E(Z))^2) = \operatorname{var}(Z) = 1. $$ Then recall that $$ \operatorname{var}(Z^2) = \operatorname E \Big( \big( Z^2\big)^2 \Big) - \left( \operatorname E(Z^2) \right)^2. $$ Therefore $$ \operatorname{var}(Z^2) = \operatorname E\left( Z^4\right) - 1. $$ The problem then is to show that $\operatorname E(Z^4) = 3.$

\begin{align} \operatorname E(Z^4) = {} & \int_{-\infty}^{+\infty} z^4 \varphi(z)\, dz \\ & \text{where $\varphi$ is the standard normal density} \\[10pt] = {} & 2\int_0^{+\infty} z^4\varphi(z)\,dz \\[10pt] = {} & 2\int_0^{+\infty} z^4 \cdot \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz \\[10pt] = {} & \sqrt{\frac 2 \pi} \int_0^{+\infty} z^3 e^{-z^2/2} (z\,dz) \\[10pt] = {} & \sqrt{\frac 2 \pi} \int_0^{+\infty} (\sqrt{2u\,\,}\,)^3 e^{-u} \, du \\[10pt] = {} & \frac 4 {\sqrt\pi} \int_0^{+\infty} u^{3/2} e^{-u}\, du \\[10pt] = {} & \frac 4 {\sqrt\pi} \Gamma\left( \frac 5 2\right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) \tag 1 \\[10pt] = {} & \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot\frac 1 2 \cdot \sqrt\pi \tag 2 \\[10pt] = {} & 3. \end{align} The work done on the line labeled $(1)$ can be justified by integrating by parts. What is done on line $(2)$ can be justified by an argument involving polar coordinates. Maybe I'll wait to see if there's popular demand before addressing those.

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  • $\begingroup$ @OliviaRoberts : I'm glad you liked it. $\endgroup$ – Michael Hardy Dec 3 '18 at 1:05
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Guide:

Some possible ingredients to solve the problem.

\begin{align} \operatorname{Var}(Z) &= \operatorname E(Z^2)-E(Z)^2\\ \operatorname E((X_i-\mu)^p) &= \begin{cases} 0 & ,p\text{ is odd} \\ \sigma^p(p-1)!! & ,p\text{ is odd}\end{cases} \end{align}

where $n!!$ denotes the double factorial, that is, the product of all numbers $n$ to $1$ that have the same parity as $n$. The moment formula is obtained from the wikipedia page.

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