I was wondering the following problem:

If $X_n$ is a sequence of independent random variables, $|X_n|\leq b, (b>0)$ and $X_n$ converges to $X$ almost surely. Using dominated convergence theorem, we have $\mathbb{E}[\lim\limits_{n\rightarrow \infty}X_n] = \lim\limits_{n\to\infty}\mathbb{E}[X_n] = \mathbb{E}[X]$. Similarly, we have $\mathbb{E}[\lim\limits_{n\rightarrow \infty}X_n^2] = \lim\limits_{n\to\infty}\mathbb{E}[X_n^2] = \mathbb{E}[X^2]$. Therefore, we have $Var(\lim\limits_{n\rightarrow \infty}X_n) = \lim\limits_{n\to\infty}Var(X_n) = Var(X)$

Then, using cesaro sum we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1} X_i}{n} = X$ almost surely. So, $Var(\lim\limits_{n\to\infty}\frac{\sum^n_{i=1} X_i}{n}) = Var(X) = \lim\limits_{n\to\infty}Var(\frac{\sum^n_{i=1} X_i}{n}) = \lim\limits_{n\to\infty}\frac{1}{n^2}Var(\sum^n_{i=1} X_i) = \lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n^2} $.

On the other hand, using cesaro sum for the sequence $Var(X_n)$, we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n} = Var(X)$. Combine with the previous results, we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n} = Var(X)= \lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n^2}$.

This is obviously not correct, but I don't know where went wrong. Any help would be greatly appreciated!

  • $\begingroup$ Consider $X_n$ identically distributed. Now, it can be seen that the claim about the Cesaro means for the random variables is false. The issue is that when we are taking the sum, the probability space to the product space. $\endgroup$ – Abhishek Shetty Nov 23 '17 at 5:41
  • $\begingroup$ @AbhishekShetty Could you elaborate is a bit? why does it turn into a product space? $\endgroup$ – metricspace Nov 23 '17 at 20:09
  • $\begingroup$ @AbhishekShetty is the cesaro means for r.v. not true or is the variance part not true? $\endgroup$ – metricspace Nov 23 '17 at 20:10
  • $\begingroup$ @AbhishekShetty ok, i see. So in general we don't have $\frac{\sum^n_1X_i(\omega)}{n} = \frac{(\sum^n_1X_i)(\omega))}{n} $ $\endgroup$ – metricspace Nov 23 '17 at 20:16
  • $\begingroup$ As an illustrative example, take $X_n \sim Bern(0.5)$. With this you can see that how the probabilities behave. Think of how $\sum X_i $ is "generated". What we need is to "draw" $\omega_i$ and then evaluate $X_i( \omega_i) $ and sum as opposed to "draw" $ \omega$ and then evaluate $\sum X_i ( \omega)$. $\endgroup$ – Abhishek Shetty Nov 24 '17 at 15:29

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