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I know: Let be $X$ a random variable and $c\in\mathbb{R}$. Then is

$$Var(cX)=c^2Var(X).$$

But is it true that

$$Var(cX)=\underbrace{Var[Var[\ldots Var}_{c \text{ times}}[X]\ldots]]?$$

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    $\begingroup$ I am not sure I understand what you are asking. The first equation is right, the second is not (as far as I can tell) $\endgroup$ – Repmat Feb 1 '16 at 17:18
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    $\begingroup$ I don't understand how anybody would come to that second equation. Were you possibly intending to ask something like is Var(cX) = Var(X+X+...+X) where there are "c" Xs in the sum? If so then that is true but only really makes sense when c is a whole number. $\endgroup$ – Dason Feb 1 '16 at 19:21
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No.

If $X$ is your only random variable being modeled, $\mathrm{Var}[X]$ is just a number, not a random quantity. So $\mathrm{Var}[\mathrm{Var}[X]]$ is 0, as is repeating it any more times.

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    $\begingroup$ I agree with the answer, but I would just like to point out that empirical variance (an estimator of population variance based on sample data) will be a random variable and the expression $\widehat{\text{Var}}(\widehat{\text{Var}}(X))$ will make a lot of sense and will not equal zero. $\endgroup$ – Richard Hardy Feb 1 '16 at 19:53
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    $\begingroup$ @RichardHardy I see you edited your comment to address my notational complaint as I was writing it. :) In any case, $\widehat{\mathrm{Var}}\left( \widehat{\mathrm{Var}}(X) \right)$ is a very different beast from $\mathrm{Var}(2 X)$. $\endgroup$ – Dougal Feb 1 '16 at 19:57
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    $\begingroup$ Yes, my original comment included at least two mistakes: I used "estimate" in place of "estimator" and forgot both hats. I agree with your response. $\endgroup$ – Richard Hardy Feb 1 '16 at 20:02

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