$I_n$ is the indicator rv where $I_n=1$ if $A_n$ occurs and 0 otherwise.

Then define $\eta_n = \sum_{k=1}^n I_k$

I'm given this statement:

enter image description here

But in another post about the same proof, I have this:

enter image description here

$X_n$ is the same r.v. as $\eta_n$

But in the first case, $\sigma^2(\eta_n)$ is the standard deviation but in the second case, $V(X_n)$ is variance.

Are variance and standard deviations supposed to be equal?

This is the link to the first case page 8

This is the link to the second case enter link description here


Usually, the standard deviation is denoted with $\sigma$ and it is the square root of the variance. It seems to me that there is a mistake in the upper statement: the standard deviation should be $\sigma$ (without the square); so they should either have written "we denote the standard deviation by $\sigma(X)$" or "we denote the variance by $\sigma^2(X)$". That is the only way the equation is correct anyway.

  • $\begingroup$ thanks for the clarification! $\endgroup$ – user8290579 Mar 3 at 23:45

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