I am trying to find $\mathbb{E}(\text{ln}(X))^2$ where $Y=\text{ln}(X)$ and $Y\sim N(0,\sigma^2)$. $$f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{y^2}{2\sigma^2}}$$

Squaring a normal r.v, I am expecting to arrive at a gamma distribution.

I let $$Z=Y^2=(\text{ln}(X))^2\Rightarrow Y=\sqrt{Z} \ \ \ \ \text{for} \ \ y>0$$ Hence $$f_Z(z)=f_Y(\sqrt{z})\Bigg|\frac{dy}{dz}\Bigg|$$ $$f_Z(z)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{z}{2\sigma^2}}\frac{1}{2\sqrt{z}}$$ $$f_Z(z)=\frac{1}{\sigma 2\sqrt{2} \ \Gamma(\frac{1}{2})}e^{-\frac{z}{2\sigma^2}}z^{-\frac{1}{2}}$$ Now, removing the 2 infront of the $\sqrt{2}$ term, this looks like $$Z\sim\text{Gamma}\Big(\frac{1}{2},2\sigma^2\Big)$$ Hence, $\mathbb{E}(Z)=\sigma^2$, which is exactly what I need to prove a result for the next part of this question.

Have I made a mistake somewhere?

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    $\begingroup$ $\mathsf{E}Y^2=Var(Y)=\sigma^2$... $\endgroup$ – d.k.o. Jul 8 '18 at 2:09
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    $\begingroup$ You know the distribution for $Y$, $N(0,\sigma^2)$ so you can readily get $E(Y^2)=\sigma^2$.. $Y=ln(X)$ seems entirely irrelevant. $\endgroup$ – herb steinberg Jul 8 '18 at 2:15
  • $\begingroup$ You are correct, I did not consider this. Thank you! For completeness/general interest though, I would like to know how to find distribution of $Y^2$. $\endgroup$ – user557493 Jul 8 '18 at 2:20
  • $\begingroup$ If $Z \sim \mathsf{Norm}(0, 1),$ then $Z^2 \sim \mathsf{Chisq}(1),$ so $Y^2$ would have a multiple of a chi-squared distribution. $\endgroup$ – BruceET Jul 8 '18 at 2:24

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