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Let $x$ be a random gaussian variable with mean=0 and sd=1, which is then squared (thus a chi-squared variable), so $y=x^2$. I understand that the expected value of $y^2$ is actually the variance of $y$, so $\text{E}(y^2)=\text{var}(y)=2$. What would be the variance of $y^2$ as $\text{var}(y^2)$?

kjetil b halvorsen
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    It's simple to apply the formula, because $\operatorname{Var}(y^2)=E(y^4)-E(y^2)^2 = E[x^8]-E[x^4]^2$ and all those moments are well-known (and easy to compute directly: see https://stats.stackexchange.com/a/176814/919). – whuber Feb 25 '21 at 21:15
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    "I understand that the expected value of $y^2$ is actually the variance of $y$." Actually no. $Y^2$ is the square of a $\chi^2$ RV. The mean of $Y$ is 1 not 0. Always remember: for any $X$, $\text{Var}(X) = E[X^2] - E[X]^2$ – AdamO Feb 25 '21 at 21:50

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