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Let $X_1,X_2,…$ be a sequence of independent and identically distributed random variables with $P(X_1=1)=\frac{1}{4}$ and $P(X_1=2)=\frac{3}{4}$. If $\bar{X_n}=\frac{1}{n}\sum_{i=1}^{n}X_i$, for $n=1,2,\ldots$ then $\lim_{n\to \infty} P(\bar{X_n}\leq 1.8)$ is ?

My work:

We can write $P(\bar{X_n}\leq1.8)=1-P(\bar{X_n}\geq1.8)$,

so $\lim_{n\to \infty}P(\bar{X_n}\leq1.8)=1-\lim_{n\to \infty}P(\bar{X_n}\geq1.8)=1-P \{\lim_{n\to \infty}\bar{X_n}\geq 1.8\}$.

Now I have a feeling that we can apply Weak law of large numbers, but I don't know the mean. If it were the case then we could conclude that required probability is $1$. So how should I proceed next? Is this the right path? Help please. Thanks.

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  • $\begingroup$ They gave you enough information to calculate the mean. But to answer your question, yeah you're on the right path. $\endgroup$ – snarfblaat Jul 22 '16 at 3:32
  • $\begingroup$ What do you mean by "I don't know the mean"? Do you know how to compute the expectation of a random variable? $\endgroup$ – Omnomnomnom Jul 22 '16 at 3:32
  • $\begingroup$ @Omnomnomnom yeah I do know how to calculate the expectation of a r.v. What I am saying is how to use the facts $P(X_1=1)$ and $P(X_1=2)$ to calculate it? $\endgroup$ – Harry Potter Jul 22 '16 at 3:42
  • $\begingroup$ What is $\mathbb{E}(X_1)$? Note that the sums of probabilities is $P(X_1=1)+P(X_1=2)=1$. $\endgroup$ – i707107 Jul 22 '16 at 3:54
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    $\begingroup$ @HarryPotter what formula are you used to using? What do you think is "missing" here? $\endgroup$ – Omnomnomnom Jul 22 '16 at 3:58
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Since $\overline{X}_n=\frac74$ and $\mathrm{Var}\left(X_n\right)=\frac3{16n}$, we have by Chebyshev's Inequality, $$ P\left(\left|X_n-\tfrac74\right|\ge\lambda\right)\le\frac3{16n\lambda^2} $$ Plugging in $\lambda=\frac1{20}$ yields $$ P\left(X_n\ge1.8\right)\le\frac{75}{n} $$ Therefore, $$ P\left(X_n\le1.8\right)\ge1-\frac{75}{n} $$

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  • $\begingroup$ Ok I understood everything but one point, how $P\left(\left|X_n-\tfrac74\right|\ge\lambda\right)$ become $P(X_n\geq 1.8)$ after putting $\lambda=1/20$? $\endgroup$ – Harry Potter Jul 22 '16 at 15:00
  • $\begingroup$ Got it thanks . $\endgroup$ – Harry Potter Jul 22 '16 at 15:02
  • $\begingroup$ $\left\{x:x\ge1.8\right\} \subset\left\{x:\left|x-\frac74\right|\ge\frac1{20}\right\}$ $\endgroup$ – robjohn Jul 22 '16 at 15:04

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