I currently have a standard result regarding variances and it looks like:

$$\frac{(2n-1)s^2}{\sigma^2} {\sim} \chi^2_{2n-1}$$

that is, it is approximately a chi-squared distribution with $2n-1$ degrees of freedom.

Now, can I take the variance of both sides so that I get:

$\frac{(2n-1)^2 var(s^2)}{\sigma^4} = 2(2n-1)$

(since the variance of chi-square is just the degrees of freedom times 2).

Would this be a valid step? Thank you!

I currently have a standard result regarding variances and it looks like:

$$\frac{(2n-1)s^2}{\sigma^2} \sim \chi^2_{2n-1}$$

that is, it is approximately a chi-squared distribution with $2n-1$ degrees of freedom.

Now, can I take the variance of both sides so that I get:

$$\frac{(2n-1)^2 \operatorname{var}(s^2)}{\sigma^4} = 2(2n-1)$$

(since the variance of chi-square is just the degrees of freedom times $2$).

Would this be a valid step? Thank you!

I currently have a standard result regarding variances and it looks like:

$\frac{(2n-1)s^2}{\sigma^2} \overset{}{\sim} \chi^2_{2n-1}$

that is, it is approximately a chi-squared distribution with $2n-1$ degrees of freedom.

Now, can I take the variance of both sides so that I get:

$\frac{(2n-1)^2 var(s^2)}{\sigma^4} = 2(2n-1)$

(since the variance of chi-square is just the degrees of freedom times 2).

Would this be a valid step? Thank you!

I currently have a standard result regarding variances and it looks like:

$$\frac{(2n-1)s^2}{\sigma^2} {\sim} \chi^2_{2n-1}$$

that is, it is approximately a chi-squared distribution with $2n-1$ degrees of freedom.

Now, can I take the variance of both sides so that I get:

$\frac{(2n-1)^2 var(s^2)}{\sigma^4} = 2(2n-1)$

(since the variance of chi-square is just the degrees of freedom times 2).

Would this be a valid step? Thank you!