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Okay, so I am interested if there is a way to derive the variance for a Chi-Square distribution using the property that it is the sum of independent unit normal distributions squared.

For example, if $X$ is a Chi-Square random variable with $n$ degrees of freedom, it has the distribution: $\displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ where $Z$ is Normal$(0,1)$

I know that $Var(X)= E(X^{2}) -[E(X)]^{2}$

To start finding $E(X)$ I begin with the fact that each $Z$ has $E(Z)=0$ and $Var(Z)=1$. This implies that $E(Z^2)=1$ since $Var(Z)=E(Z^{2})-[E(Z)]^2$

Since $X = \displaystyle\operatorname{}\left(\sum_{i=1}^n Z_i^2\right)$ then $E(X)=\displaystyle\operatorname{}\left(\sum_{i=1}^n 1\right)=n$

I am lost on what the next step would be. I have this start, but don't where to go next. Any thoughts on how to find $E(X^{2})$?

$\begin{align} X^2 =& (\sum_{i=1}^nZ_i^2)^2\\ =& \sum_{i=1}^nZ_i^4+\sum_{i \neq j}^nZ_i^2Z_j^2 \end{align}$

I can't see any way this sum not getting nasty. Note: I can solve this using integration of the PDF for the Chi-Square distribution, but I was wondering if there is any way to do it using the property that Chi-Squared is sum of Squared Normal.

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    $\begingroup$ Use the fact that the variance of a sum of independent random variables is the sum of the variances, then compute the variance of $Z^2$. $\endgroup$ – carmichael561 Jul 20 '17 at 1:45
  • $\begingroup$ You're almost there, just take expectations from there. $Z_i$ and $Z_j$ are independent for $i \neq j$ so the second sum isn't too bad. Do you know what $E[Z^4]$ is? $\endgroup$ – Marcus M Jul 20 '17 at 1:46
  • $\begingroup$ No I wasn't sure how to proceed with $E[Z^{4}]$ @carmicheal561 I am aware of the sum of the variances, but how does this help me with taking the variance of $Z^{2}$ because this would be the product of a random variable? $\endgroup$ – user345 Jul 20 '17 at 1:53
  • $\begingroup$ The variance of $Z^2$ is just $E[Z^4]-E[Z^2]^2$, which should be fairly straightforward to compute. $\endgroup$ – carmichael561 Jul 20 '17 at 1:58
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    $\begingroup$ @ap2010, In my opinion, the best way to find the moments of $Z$ are to find the moment generating function and taking derivatives. You can find the moment generating function of a standard normal just by straight up integrating. It's a good exercise. You'll end up getting that the m.g.f. is $e^{t^2 / 2}$. Taking the fourth derivative and setting $t = 0$ gives $E[Z^4] = 3$. $\endgroup$ – Marcus M Jul 20 '17 at 2:00
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Since $X_1^2,\ldots,X_n^2$ are independent, you have $$ \operatorname{var}(X_1^2+\cdots+X_n^2) = n \operatorname{var}(X_1^2). $$ And this is $\operatorname{var}(X_1^2) = \operatorname{E}((X_1^2)^2) - (\operatorname{E}(X_1^2))^2$. If you know that $\operatorname{E}(X_1^2) =1,$ then it remains only to find $\operatorname{E}(X_1^4).$ \begin{align} \operatorname{E}(X_1^4) & = \int_{-\infty}^\infty x^4 \varphi(x)\, dx \text{ where $\varphi$ is the standard normal density} \\[10pt] & = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2} \,dx \\[10pt] & = \sqrt{\frac 2 \pi} \int_0^\infty x^4 e^{-x^2/2} \,dx\text{ by symmetry} \\[10pt] & = \sqrt{\frac 2 \pi} \int_0^\infty x^3 e^{-x^2/2} \Big( x\,dx\Big) \\[10pt] & = \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du = \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 \Gamma\left( \frac 3 2 \right) = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \Gamma\left( \frac 1 2 \right) \\[10pt] & = \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \sqrt \pi \\[10pt] & = 3. \end{align}

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