0
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.