Say a fair dice is thrown $n$ times. Showing using the Chebyshev Inequality that the probability that the number of sixes thrown lies between $\frac{1}{6}n-\sqrt{n}$ and $\frac{1}{6}n+\sqrt{n}$ is at least $\frac{31}{36}$.
Idea:
The at least leads me to believe that I be doing it over the complement of $\{\frac{1}{6}n-\sqrt{n}< X<\frac{1}{6}n+\sqrt{n}\}=\{-\sqrt{n}< X-\frac{1}{6}n<\sqrt{n}\}$ and note $\mathbb E[X]=\frac{n}{6}$, where $X$ is number of sixes thrown
This looks similar to Chebyshev.
So, for $\epsilon > 0$
$P(|X-\frac{1}{6}n|< \sqrt{n})=1-P(|X-\frac{1}{6}n|\geq \sqrt{n})\leq1-\frac{\operatorname{Var}{X}}{\epsilon^2}\iff P(|X-\frac{1}{6}n|\geq\sqrt{n})\leq\frac{\operatorname{Var}{X}}{\epsilon^2}$
But how can I calculate $\operatorname{Var}{X}$? Do I have to explicitly define the Random Variable and then find an appropriate density funtion? All I have is $\operatorname{Var}{X}=\mathbb E[(X-\mathbb E[X])^2]=\int X-\frac{1}{6}n\operatorname{dP}$
Any help is greatly appreciated.
Best Answer
You're sort of on the right track, but your inequality in $1-P\left(|X-\frac{1}{6}n|\geq \sqrt{n}\right)\leq1-\frac{\operatorname{Var}{X}}{\epsilon^2}$ is the wrong way round ( it should be $1-P\left(|X-\frac{1}{6}n|\geq \sqrt{n}\right)\geq1-\frac{\operatorname{Var}{X}}{\epsilon^2}$ ), you haven't identified what $\epsilon$ is, and a power of 2 is missing from your integral for the variance. The latter should be $\int \left( X - \frac{1}{6} n \right)^2 dP$ .
I presume the form of Chebyshev's inequality you're using is $P(|X-\frac{1}{6}n|\geq \epsilon)\leq\frac{\operatorname{Var}{X}}{\epsilon^2}$ , in which case your $\epsilon$ is just $\sqrt{n}$ , and your inequality becomes $P(|X-\frac{1}{6}n|\geq \sqrt{n})\leq\frac{\operatorname{Var}{X}}{n}$
You could evaluate the integral for the variance by working out what the distribution $F_X$ of $X$ is (Hint: $F_X\left(j\right) = P\left(X=j\right)$ is the probability of getting $j$ sixes with $n$ independent throws of a fair die), but there's also a simpler way of calculating it.
If $X_i$ is the number of sixes you get on the $i^\mbox{th}$ throw, then $P\left( X_i = 1 \right) = \frac{1}{6}$ , $P\left( X_i = 0 \right) = \frac{5}{6}$ , and $X_1, X_2, \dots , X_n$ are independent, identically distributed random variables with $X = X_1 + X_2 + \dots + X_n$. Now there's a theorem which tells us that the variance of a sum of $n$ independent identically distributed random variables is just $n$ times the common variance of the summands. That is, $\mbox{Var}\left(X\right) = n \mbox{Var}\left(X_1\right)$ , so you can prove your result just by calculating the variance of the simple two-valued random variable $X_1$ .
Elaboration of hint about $F_X$:
Since $X$ can only take on one of the values $0, 1, \dots , n$ , the sample space (call it $\Omega$) can be partitioned into a union of the disjoint events $\ E_j = \left\{ \omega \in \Omega\ |\ X\left(\omega\right) = j\ \right\} \mbox{ for } j=0, 1, \dots , n $ . The integral $\int \left( X - \frac{1}{6} n \right)^2 dP$ can then be written as $\int_{\bigcup_{j=0}^n E_j}\left( X - \frac{1}{6} n \right)^2 dP = \sum_{j=0}^n \int_{E_j}\left( X - \frac{1}{6} n \right)^2 dP $ . Since $X$ has the fixed value $j$ everywhere in $E_j$ , then $\int_{E_j}\left( X - \frac{1}{6} n \right)^2 dP = $ $ \left( j- \frac{1}{6} n \right)^2 \int_{E_j} dP = \left( j- \frac{1}{6} n \right)^2 P\left(E_j\right) = \left( j- \frac{1}{6} n \right)^2 F_X\left(j\right)$ . So $Var\left(X\right) = \sum_{j=0}^n \left( j- \frac{1}{6} n \right)^2 F_X\left(j\right)$ .
As callculus noted in his answer, $X$ is $\ n, \frac{1}{6}$-binomially distributed, which gives you the expression for $F_X\left(j\right)$ as a function of $n$ and $j$ . If you don't know this expression, you will find it (as well as its variance!) in any good text on elementary probability theory (such as Volume 1 of William Feller's classic, An Introduction to Probability Theory and Its Applications—3rd Edition—where you will find the material on pp.147-8 and p.230) .