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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.

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    $\begingroup$ I suspect where it says $E[|X|^3]$ you mean $E[|X_n|^3]$? $\endgroup$ – joriki Dec 5 '19 at 22:51
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    $\begingroup$ 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. $\endgroup$ – S Prasanth Dec 5 '19 at 23:10
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    $\begingroup$ 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/… $\endgroup$ – S Prasanth Dec 5 '19 at 23:10
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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 $0$.

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  • $\begingroup$ So we shall say the sample mean converges to a constant distribution? $\endgroup$ – TrueWarrior09 Dec 6 '19 at 1:51
  • $\begingroup$ Yes, it converges to $\frac {20} {17}$. @TrueWarrior09 $\endgroup$ – Kavi Rama Murthy Dec 6 '19 at 5:19
  • $\begingroup$ Thank you. @Kabo Murphy $\endgroup$ – TrueWarrior09 Dec 6 '19 at 5:20

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