# Finding the limit probability of sample mean greater than a number and its distribution

Given that $$X_{1}, X_{2}, X_{3}...$$ is squence of independent random variable, given that $$E[X_{n}] = \frac{20}{17}$$ and $$E[|X_{n}|^3] < 12$$, let $$\overline{X_{n}} = \frac{X_{1} + X_{2} + ... + X_{n}}{n}$$

(a) compute $$\lim_{n \rightarrow \infty} P(\overline{X_{n}} > 16)$$

(b) show that exists $$X$$ such that $$\overline{X_{n}}$$ converge to $$X$$ in $$L_{2}$$ as $$n \rightarrow \infty$$, specify the distribution of $$X$$

The question only says $$X_{i}$$ independent but need not to be identically distributed(if $$i.i.d$$ is satisfied then I think CLT can be used here), then what should we do and which theorem should we use? Thanks.

• I suspect where it says $E[|X|^3]$ you mean $E[|X_n|^3]$? – joriki Dec 5 '19 at 22:51
• A variant of the CLT can also be used for independent, but non-identical random variables: sjsu.edu/faculty/watkins/CLText.htm Only constraint is that the means and variances of the individual distributions should exist and be bounded. – S Prasanth Dec 5 '19 at 23:10
• The existence of $E[|X_n|^3]$ implies the existence of $E[|X_n|^2]$ (and hence the variance) for each $n$, as mentioned here: en.wikipedia.org/wiki/Moment_(mathematics) (search for "moment about any point exists" on that page). Not only does the variance exist, it is also bounded: Theorem 3.1 equation 3.3 here: kurims.kyoto-u.ac.jp/EMIS/journals/JIPAM/images/040_04_JIPAM/… – S Prasanth Dec 5 '19 at 23:10

Answer for b): $$EX_n^{2} \leq (E|X_n|^{3})^{2/3} \leq (12)^{2/3}$$. Hence $$var(\overline {X_n})=\frac 1 {n^{2}} \sum\limits_{k=1}^{n} var(X_k) \leq \frac 1 n ((12)^{2/3}-(\frac {20} {17})^{2}) \to 0$$. Thus $$E(\overline {X_n}-\frac {20} {17})^{2} \to 0$$. Hence $$\overline {X_n} \to \frac {20} {17}$$ in $$L^{2}$$.
Answer for a): By b) $$\overline {X_n} \to \frac {20} {17}$$ in probability so the limit is $$0$$.
• Yes, it converges to $\frac {20} {17}$. @TrueWarrior09 – Kavi Rama Murthy Dec 6 '19 at 5:19