Answer to Question #159983 in Statistics and Probability for Treesa

Question

If X is the number obtained when a die is thrown, show that Tchebycheff's inequality gives P(|X-3.5|>2.5)<0.47. Compare this with actual probability

Answer 1

We compute the mean ($\mu$) and the standard deviation ($\sigma$) for $X$. We get:

$\mu=E[X]=\frac16(1+2+3+4+5+6)=3.5$

$\sigma=E[X^2]-(E[X])^2=\frac16(1^2+2^2+3^2+4^2+5^2+6^2)-(E[X])^2=\frac{91}6-\frac{49}4=\frac{35}{12}$

We remind that Chebyshev inequality states that $P(|X-\mu|\geq k\sigma)\leq\frac{1}{k^2}$. Another version of Chebyshev inequality is: $P(|X-\mu|\geq k)\leq\frac{\sigma^2}{k^2}$

We consider the second version of the inequality. It can be obtained from Markov inequality: $P(X\geq c)\leq\frac{E[X]}{c}$. We apply Markov inequality to random variable: $(X-\mu)^2$ and receive: $P(|X-\mu|\geq k)=P((X-\mu)^2\geq k^2)\leq\frac{E[(X-\mu)^2]}{k^2}=\frac{E[X^2]-2\mu E[X]+\mu^2}{k^2}=\frac{\sigma^2}{k^2}$

In a similar way one can obtain the first version.

In both cases $k$ is a strictly positive real number.

We set $\mu=3.5$ and $\sigma=\frac{35}{12}\approx2.92$ and receive: $P(|X-3.5|\geq\,2.92\,k)\leq\frac{1}{k^2}$ and

$P(|X-3.5|\geq k)\leq\frac{(2.92)^2}{k^2}$. We set $k\approx0.856$ in the first inequality and $k=2.5$ in the second one. We receive: $P(|X-3.5|\geq\,2.5)\leq1.168$ from the first inequality and $P(|X-3.5|\geq k)\leq1.364$ from the second one. Now we shall calculate the probabilities: $P(|X-3.5|\geq\,2.5)$ and $P(|X-3.5|>\,2.5)$. We receive:

$P(|X-3.5|\geq\,2.5)=P((X=5)\lor(X=6))=\frac{1}{3}$ and

$P(|X-3.5|>\,2.5)=P(X=6)=\frac{1}{6}$.

As it can be observed, two versions of Chebyshev inequality provide estimates 1.168 and 1.364, whereas direct calculations yield $\frac{1}{3}$ (or $\frac{1}{6}$ if the inequality is strict).