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Given $x \sim N(\mu ,\sigma ^{2})$, we know $\frac{(n-1)s^{2}}{\sigma ^{2}} \sim x^{2}_{n-1}$, where $s$ is sample standard deviation.

But to find sample error of $\sigma$, I took the standard deviation of both sides, treating $\sigma$ and $n-1$ as constants, and got $Var(s^{2}) = Var(\frac{x^{2}_{n-1}} {\sigma ^{2} (n-1)}）$. Since Variance of chi-squared with $n-1$ degrees of freedom is $2(n-1)$, the right hand side expression reduces to $Var(\frac {2} {\sigma ^{2}})$, which is not the correct answer of $\frac {\sigma ^{4}} {n-1}$. Where did my math break down?

Thanks!

user101998
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1 Answers1

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You made an algebra error. When you move $(n-1)/\sigma^2$ to the RHS you should get: $$ s^2=\frac{\sigma^2}{n-1}\chi^2_{n-1}.$$ For the next step, use the fact that $$ \operatorname{Var}(cX)=c^2\operatorname{Var}(X) $$ for any constant $c$.

Note: The quantity $2\sigma^4/(n-1)$ is the variance of the sample variance, not the variance of the sample standard deviation (which is quite difficult to obtain: see here)

grand_chat
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