# Find the limit of $P(\bar{X_n}\leq 1.8)$ for i.i.d random variables $X_i$s of known distribution

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.

• They gave you enough information to calculate the mean. But to answer your question, yeah you're on the right path. – snarfblaat Jul 22 '16 at 3:32
• What do you mean by "I don't know the mean"? Do you know how to compute the expectation of a random variable? – Omnomnomnom Jul 22 '16 at 3:32
• @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? – Harry Potter Jul 22 '16 at 3:42
• What is $\mathbb{E}(X_1)$? Note that the sums of probabilities is $P(X_1=1)+P(X_1=2)=1$. – i707107 Jul 22 '16 at 3:54
• @HarryPotter what formula are you used to using? What do you think is "missing" here? – Omnomnomnom Jul 22 '16 at 3:58

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, $$P\left(X_n\le1.8\right)\ge1-\frac{75}{n}$$
• 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 $\lambda=1/20$? – Harry Potter Jul 22 '16 at 15:00
• $\left\{x:x\ge1.8\right\} \subset\left\{x:\left|x-\frac74\right|\ge\frac1{20}\right\}$ – robjohn Jul 22 '16 at 15:04