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For $\alpha,\beta>0$, the probability density function of a Gamma$(\alpha,\beta)$ random variable is given by $$f(x)=\frac{x^{\alpha-1}e^{\frac{-x}{\beta}}}{\Gamma(\alpha)\beta^\alpha} \ \ \ \ \ \ \ x>0$$ For $\sigma>0$, the probability density function of a $N(0,\sigma^2)$ random variable is given by $$f(x)=\frac{e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}} \ \ \ \ \ \ \ x\in\mathbb{R}$$

Now for $G\sim\text{Gamma}(\alpha,\beta)$, I have shown that for $k\in\mathbb{N}^+$ $$\mathbb{E}(G^k)=\frac{\beta^k\Gamma(\alpha+k)}{\Gamma(\alpha)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ by manipulating the integrand $g^kf(g)$

Let $Y\sim N(0,\sigma^2)$, using the result $(1)$, or otherwise, show that for $n\in\mathbb{N}^+$ $$\mathbb{E}(Y^n)=\ \begin{cases} \frac{(2\sigma^2)^{\frac{n}{2}}\Gamma(\frac{1}{2}+\frac{n}{2})}{\Gamma(\frac{1}{2})} & \text{n is even} \\ 0 &\text{n is odd} \\ \end{cases} $$

I decide to first tackle the odd case first, as this seemed most obvious.

For odd $n$ $$ \mathbb{E}(Y^n)=\int_{-\infty}^{\infty} y^n\frac{e^{-\frac{y^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}} dy=0 $$ as the integrand is an odd function when $n$ is odd.

For the even case, I have seen a solution that actually solved the expectation by definition. I wanted to use the result $(1)$, as I thought this could speed things up. But i'm unsure if my solution is valid.

For even $n$

We first start with \begin{align} &Y\sim N(0,\sigma^2) \\ &\Rightarrow \frac{Y}{\sigma}\sim N(0,1) \\ &\Rightarrow \frac{Y^2}{\sigma^2}\sim \chi^2_1 \\ &\Rightarrow \frac{Y^2}{\sigma^2}\sim \text{Gamma}\Big(\frac{1}{2},2\Big)\\ &\Rightarrow Y^2\sim\text{Gamma}\Big(\frac{1}{2},2\sigma^2\Big) \\ \end{align} Now let $$Z=Y^2\Rightarrow Y=Z^\frac{1}{2}$$ Hence $$\mathbb{E}(Y^n)=\mathbb{E}(Z^{\frac{n}{2}})=\frac{(2\sigma^2)^{\frac{n}{2}}\Gamma(\frac{1}{2}+\frac{n}{2})}{\Gamma(\frac{1}{2})}$$

I am not entirely sure that my logic for the even case, where I derived a gamma distribution from the normal distribution, is correct. Is this a sufficient answer?

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The only problem is the equality $$Z = Y^2 \implies Y = Z^{1/2}$$ since due to that $Y \sim \mathcal{N}(0, \sigma^2)$ (i.e., $Y$ can take on negative values), it follows that $$Z^{1/2} = \sqrt{Y^2} = |Y|\text{.}$$ Instead, take advantage of the following: since $Y^2 \sim \text{Gamma}\left( \dfrac{1}{2}, 2\sigma^2\right)$, you know that for every $k \in \mathbb{N}^{+}$ that $$\mathbb{E}[(Y^2)^k] = \mathbb{E}[Y^{2k}] = \dfrac{(2\sigma^2)^{k}\Gamma(1/2+k)}{\Gamma(1/2)}$$ and we may set $n:= 2k$ (since $2k$ is obviously an even number) to obtain $$\mathbb{E}[Y^n]=\dfrac{(2\sigma^2)^{n/2}\Gamma(1/2+n/2)}{\Gamma(1/2)}\text{.}$$ The reason why your answer coincides with this one is because $$|Y|^n = Y^n$$ when $n$ is even.

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  • $\begingroup$ Thnaks! If I can just ask as well, I was just having another look at this step \begin{align} &\Rightarrow \frac{Y^2}{\sigma^2}\sim \text{Gamma}\Big(\frac{1}{2},2\Big)\\ &\Rightarrow Y^2\sim\text{Gamma}\Big(\frac{1}{2},2\sigma^2\Big) \\ \end{align} and i don't quite understand how this is true. Doesn't $$cY\sim\text{Gamma}(a,cb) \ \ \text{for} \ \ Y\sim\text{Gamma}(a,b)?$$ and hence shouldn't \begin{align}&\Rightarrow \frac{Y^2}{\sigma^2}\sim \text{Gamma}\Big(\frac{1}{2},2\Big)\\ &\Rightarrow Y^2\sim\text{Gamma}\Big(\frac{1}{2},\frac{2}{\sigma^2}\Big)?? \\ \end{align} $\endgroup$ – user557493 Jul 2 '18 at 23:36
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    $\begingroup$ @Bell Since $\dfrac{Y^2}{\sigma^2} \sim \text{Gamma}\left(\dfrac{1}{2}, 2\right)$, it follows that $$\dfrac{Y^2}{\sigma^2} \cdot \sigma^2 = Y^2 \sim \text{Gamma}\left(\dfrac{1}{2}, 2\sigma^2\right)$$ $\endgroup$ – Clarinetist Jul 3 '18 at 11:42
  • $\begingroup$ Cool, thank you! If I can ask, what is the reason for this? For example, if $$X\sim\text{Gamma}(a,b)$$ why does $$c\cdot X\sim\text{Gamma}(a,c\cdot b)$$ and why not $$c\cdot X\sim\text{Gamma}(c\cdot a,b)$$ $\endgroup$ – user557493 Jul 4 '18 at 3:41
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    $\begingroup$ @Bell How much do you know about finding PDFs of transformations? Let $Y = cX$. This means that $X = Y/c$, so you can use the formula $$f_{Y}(y) = f_{X}(y/c) \cdot \left|\dfrac{\text{d}}{\text{d}y}[y/c]\right|$$ and when you work out the details from here, you'll see why this is true. $\endgroup$ – Clarinetist Jul 4 '18 at 17:46
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    $\begingroup$ @Bell Another way to do this would be to use the MGF of $X$ and the formula $$M_{cX}(t) = M_{X}(ct) = \left[1-b(ct) \right]^{-\alpha} = \left[1-(bc)t \right]^{-\alpha} $$ which is, indeed, the MGF of a $\text{Gamma}(a, bc)$ random variable. $\endgroup$ – Clarinetist Jul 4 '18 at 17:51

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