Can anyone help to prove this equation for any distribution $$ E(z^4)=1+\operatorname{Var}(z^2) $$ where $z$ is a random variable with the standard normal distribution
$$z=\frac{x−μ}σ$$
2026-04-07 03:23:03.1775532183
Variance of squared random variable
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What we have is $$\operatorname{Var}\left(z^2\right)=\mathbb E\left[\left(z^2\right)^2\right]-\left(\mathbb E\left[z^2\right]\right)^2=\mathbb E\left[z^4\right]-\left(\mathbb E\left[z^2\right]\right)^2,$$ so the formula in the opening post is true if and only if $\mathbb E\left[z^2\right]=1$.