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$n$ random variables $X_1,\ldots,X_n$ are an i.i.d. sample. $\bar X_n$ is the sample mean. $\mu$ is the expectation of distribution. Doesn't guarantee a finite variance.

Does this always hold?

$$E[(\bar X_n-\mu)^2]\rightarrow0 \qquad (n\rightarrow \infty)$$

If yes, in which sense does this hold? (ie. almost surely / in probability / in distribution)

If not, under what condition does this hold? What if we add that $\sigma^2<\infty$ is the variance?

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  • $\begingroup$ It is not correct to say $X_1,\ldots,X_n$ are i.i.d. samples; rather one should say the $n$ random variable $X_1,\ldots,X_n$ are an i.i.d. sample. $\endgroup$ – Michael Hardy Sep 29 '18 at 16:03
  • $\begingroup$ The quantity $E[(\bar X_n-\mu)^2]$ is finite only when $E[X_1^2]$ is finite, hence the question does not make sense unless the variance is finite. If the variance is finite, surely you can compute $E[(\bar X_n-\mu)^2]$ in terms of $n$ and $\sigma^2=E[(X_1-\mu)^2]$? $\endgroup$ – Did Sep 29 '18 at 16:11
  • $\begingroup$ Showing that $\operatorname E\left( \left( \overline X_n - \mu\right)^2 \right)$ is infinite if $\operatorname E\left( \left( X_1 - \mu \right)^2\right)$ is infinite is something I'm not sure I ever thought of before. Everybody's seen the converse of that proposition. $\qquad$ $\endgroup$ – Michael Hardy Sep 29 '18 at 16:16
  • $\begingroup$ It might be instructive to play with the distribution $P[X=n]=\frac4{n(n+1)(n+2)}$, which has $E[X]=2$, but $E\!\left[X^2\right]=\infty$ $\endgroup$ – robjohn Sep 29 '18 at 17:30
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If $\operatorname{var}(X_1)<\infty$ then $\operatorname E\left(\left(\overline X_n - \mu\right)^2\right) = \dfrac{\operatorname{var}\left(X_1\right)} n \to 0$ as $n\to\infty.$

However, if $\displaystyle \Pr(X_1\in A) = \int_A \frac{du}{\pi(1+u^2)}$ for every measureable set $A,$ i.e. if $X_1$ has a standard Cauchy distribution, then the distribution of $\overline X_n = (X_1+\cdots+X_n)/n$ is actually that same Cauchy distribution. Its interquartile range is still from $-1$ to $+1$ no matter how big $n$ is.

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  • $\begingroup$ Thanks! It is clear. $\endgroup$ – Zhenduo Cao Sep 29 '18 at 16:25
  • $\begingroup$ The Cauchy distribution doesn't have a mean, so is not a "counterexample" as far as the OP's question goes. $\endgroup$ – user10354138 Sep 29 '18 at 17:23

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