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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!

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    $\begingroup$ This question is baffling because it is so utterly trivial: when a random variable has a given distribution, then of course the variance of that RV is given by the variance of that distribution! This leads me to suspect that some complications might be lurking behind your understanding of "$\sim$" and "approximately" and "take the variance of." Could you explain what the difficulty is? $\endgroup$
    – whuber
    Oct 8, 2014 at 16:57
  • $\begingroup$ This is fine if the sample is normally distributed, and the result will hold under the conditions that the chi-squared distribution holds. Note that you can't take the variance of $~\chi^2_{2n-1}$ because it isn't a random variable it's a statement of a distribution. $\endgroup$ Oct 8, 2014 at 17:00
  • $\begingroup$ If your question is "if two things have the same distribution, do they have the same variance?" then the answer is "obviously, yes" (which is what I believe whuber is referring to when he says "utterly trivial"). If you're asking something else, you should definitely clarify. $\endgroup$
    – Glen_b
    Oct 8, 2014 at 22:08

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The symbol $\sim$ does not mean "approximately" in this context (use $\approx$ instead, for "equals approximately"). It means "follows exactly the distribution of a..." or other verbal transcription to that effect.

So if we assume that

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

we have that the random variable $Q$ follows a chi-square with $2n-1$ degrees of freedom.

It is then perfectly valid to write

$$s^2 =\frac{\sigma^2}{2n-1}Q$$

This makes the random variable $s^2$ to follow a Gamma distribution,

$$s^2 \sim \Gamma_{d}\left(\frac {2n-1}{2},2\frac{\sigma^2}{2n-1}\right)$$ where we have used the "shape-scale" parametrization. Then

$$\text{Var}(s^2) = \frac {2n-1}{2}\left(2\frac{\sigma^2}{2n-1}\right)^2 = \frac {2\sigma^4}{2n-1}$$

which is what you indeed found -but it is advisable to go through the above procedure, specifically, to use the equality symbol together with a variable symbol (like the $Q$ I used), before performing mathematical manipulations.

If on the other hand what we assume is that $\frac{(2n-1)s^2}{\sigma^2}$ follows a chi-square distribution only "approximately",

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

then still, the above calculations are not invalid, but, the accuracy of the obtained expression for the variance of $s^2$ should be under questioning and investigation (since $=$ should be everywhere changed to $\approx$).

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  • $\begingroup$ (1) The correctness of your final remarks depends on what "approximately" means. For instance, if it means the two distribution functions are uniformly within $\varepsilon$ of each other, then even for extremely small positive $\varepsilon$ you can only put a lower bound on the variance of $s^2$ (its variance could be infinite). (2) I do not understand the point of introducing the superfluous variable "$Q$" and the unnecessary calculation of the Gamma parameters. The manipulations shown in the question are valid--but the descriptions of them are strange. $\endgroup$
    – whuber
    Oct 8, 2014 at 19:29
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    $\begingroup$ @whuber On (2): It is explicitly stated in the question that the OP confused the "equals approximately" with "is distributed according to". The purpose of the $Q$ variable was exactly to permit me to write in one line both concepts/symbols (the last equation in my answer) so that it will drive the message home visually -never fails, according to my teaching experience. On (1): I believe that the way I have "put a question mark" on these last remarks covers adequately all cases, since after all, we do not know what kind of deviation exists. $\endgroup$ Oct 8, 2014 at 21:34
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    $\begingroup$ In re-reading the question and your answer, I see the wisdom of introducing the variable $Q$, because it helps dispel a possible misunderstanding (+1). $\endgroup$
    – whuber
    Mar 24, 2015 at 20:41
  • $\begingroup$ @whuber Thanks, I appreciate the gesture. I was going through my list of answers to find an older one in order to link it to some other recent question, and saw that this one had a "math processing error" in the title. I edited to correct. This, I guess, sent the thread to the top of the "active" list, and suddenly, so much commotion here! $\endgroup$ Mar 24, 2015 at 20:52

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