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If $X_1, X_2, X_3, \cdots$ is a sequence of independent identically distributed random variables with $E[X_i] < \infty$ and $Var(X_i)< \infty$ such that the sequence $Y_n = 3 \frac{X_1 + X_2 + \cdots + X_n }{\sqrt{n}}$ converges in law to a standard Normal distribution, compute $E[X_i]$ and $var(X_i)$.

As it is used in this question, what does it mean for this sequence to converge in law to a standard normal? I know law is synonymous with distribution but I still don't fully understand the implications of a convergence in law.

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It means that the CDF of $Y_n$ converges pointwise to the CDF of a standard normal at all points of continuity of a standard normal CDF (which is on all of $\mathbb{R}$).

Convergence in law is also sometimes called convergence in distribution or weak convergence.

In this case, note that if $X_1,\ldots,X_n$ are i.i.d. with mean $\mu$ and variance $\sigma^2<\infty$, then defining $S_n = \frac{1}{n} \sum_{i=1}^n X_i$, we have $\frac{\sqrt{n} (S_n - \mu)}{\sigma} \implies N(0,1)$ where $\implies$ denotes convergence in distribution (this is a Central Limit Theorem).

You're supposed to match the form of $Y_n$ to $\frac{\sqrt{n} (S_n - \mu)}{\sigma} $ and read off $\mu$ and $\sigma$ to solve this problem.

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  • $\begingroup$ Could we also just use the fact that a convergence in law implies that the expectations are identical but the expectation of a standard normal is just 0 implying the expectation of $Y_n$ should also be 0? $\endgroup$ – David South Nov 6 '15 at 20:00
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    $\begingroup$ Convergence in distribution doesn't mean that the means converge -- the identity function is continuous, but not bounded. Try the sequence of random variables $X_1,X_2,\ldots$ with $X_n$ taking $0$ w.p. $1-1/n$ and taking $2^n$ otherwise as a counterexample to that. $\endgroup$ – Batman Nov 6 '15 at 20:11
  • $\begingroup$ hello @Batman, what does the abreviation w.p. please ? can't find anything on the internet $\endgroup$ – Marine Galantin Mar 25 at 10:27
  • $\begingroup$ With probability. It’s fairly standard in probability texts, at least in English. So, each $X_n$ is $2^n*Bernoulli(1/n)$ in my comment above $\endgroup$ – Batman Mar 25 at 18:26

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