Given that $Y$ is a degree $8$ chi square distribution i.e $Y \sim \chi_{8}^2$
I want to find $E[Y^2]$ and $Var[Y^2]$
I only know that $E[Y]$ of it should be $8$, but I don't know how to find $E[Y^2]$ from it?
Thank you
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2 Answers
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2 $$\chi_{8}^2=Gamma(4;1/2)$$
Where $1/2$ is the rate parameter. Now it is not difficult to calculate any moment you want solving the integrals
Alternative method:
Given that $Y\sim \chi_8^2$,
$$Y=Z_1^2+\dots +Z_8^2$$
Where $Z_i$ are iid Standard gaussian. Moments of the standard gaussian are well known thus...
$$\mathbb{E}[Y]=\underbrace{1+1+\dots +1}_{\text{8 times}}=8$$
$$\mathbb{V}[Y]=8\Bigg[\frac{4!}{2^2\cdot 2!}-1\Bigg]=16$$
- $\begingroup$ For the alternative method, I don't quite get it, does it means that I can decompose it to E[Y^2] = E[(x1^2 + x2^2+...+x8^2)(x1^2 + x2^2+...+x8^2)] = 8*8 =64? $\endgroup$newbie_cser– newbie_cser2021-02-03 08:38:11 +00:00Commented Feb 3, 2021 at 8:38
- $\begingroup$ @newbie_cser : the alternative method directly calculate $V(Y)=V(Z_1^2+...+Z_8^2)=8V(Z_1^2)$. If you have to calculate $V(Y^2)$ you need $E(Y^4)$ that is easy with the integral $\endgroup$tommik– tommik2021-02-03 08:40:49 +00:00Commented Feb 3, 2021 at 8:40
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$X\sim \chi_{n}^2$
$E(X)=n,V(X)=2n$
$E(X^2)=V(X)+E(X)^2$
Above mentioned stuff is sufficient to solve this problem
