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Can you please help verify if what I have done is correct for the question below? I applied Chebyshev's theorem in the first step, but I am worried if there are any mathematical errors or misapplied theorems in my solution. Thanks for the help.

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$\ni^2$ is definitely a new one for me... –  cardinal Oct 17 '11 at 23:36
All of the $\ni$'s should probably have been $\varepsilon$. –  Henning Makholm Oct 17 '11 at 23:45
I don't know why you have $-1$ in the numerator after $\operatorname{var}(S_n)$. All you really need to know about $\operatorname{var}(S_n)$ is that it is finite. That can be deduced from the fact that $E(X^4)<\infty$. You seem to have done everything else. –  Michael Hardy Oct 18 '11 at 0:48
This application of Chebyshev doesn't quite work since the mean of $S_n/\sigma$ is not equal to 1. You should follow Mike's lead below, and consider $S_n^2/\sigma^2$ instead. Then you can use the result in math.stackexchange.com/questions/72975/… to conclude. –  Byron Schmuland Oct 18 '11 at 1:33
$S_n^2$ is an unbiased estimator for $\sigma^2$. $S_n$ is not an unbiased estimator for $\sigma$. By the way, you make the variance go to zero using math.stackexchange.com/questions/72975/… –  Byron Schmuland Oct 18 '11 at 2:00

1 Answer 1

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One way to do this is show that $\frac{S_n^2}{\sigma^2}$ converges in probability to 1 using the same method you tried (invoking Chebyshev's Inequality), and then noting (or proving if you didn't know already) that if $X_1, X_2, \ldots$ converge in probability to $X$ and $g$ is a continuous function, then $g(X_{1}), g(X_{2}), \ldots$ converge in probability to $g(X)$.

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