# Does variance of sample mean converge to zero?

$$n$$ 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?

• 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. – Michael Hardy Sep 29 '18 at 16:03
• 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 $\sigma^2=E[(X_1-\mu)^2]$? – Did Sep 29 '18 at 16:11
• 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. $\qquad$ – Michael Hardy Sep 29 '18 at 16:16
• It might be instructive to play with the distribution $P[X=n]=\frac4{n(n+1)(n+2)}$, which has $E[X]=2$, but $E\!\left[X^2\right]=\infty$ – robjohn Sep 29 '18 at 17:30

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.