Chapter 3.3, Problem 44E

A result called Chebyshev’s inequality states that for any probability distribution of an rv X and any number k that is at least 1, P(|X - μ| ≥ kσ) ≤ 1/k². In words, the probability that the value of X lies at least k standard deviations from its mean is at most 1/x².

1. a. What is the value of the upper bound for k = 2? k = 3? k = 4? k = 5? k = 10?
2. b. Compute μ and σ for the distribution of Exercise 13. Then evaluate P(|X - μ| ≥ kσ) for the values of k given in part (a). What does this suggest about the upper bound relative to the corresponding probability?
3. c. Let X have possible values −1,0, and 1, with probabilities $\frac{1}{18},\frac{8}{9}$ and $\frac{1}{18}$, respectively What is P( |X − μ| ≥ 3σ), and how does it compare to the corresponding bound?
4. d. Give a distribution for which P(|X – μ| ≥ 5σ) = .04.

To determine

Find the value of the upper bound for k=2, 3, 4, 5 and 10.

Answer to Problem 44E

The values are:

k	Upper bound
2	0.25
3	0.11
4	0.06
5	0.04
10	0.01

Explanation of Solution

Given info:

The Chebyshev’s inequality states that for any probability distribution of an rvX and any numberk that is at least 1, then c $P\left(\left|X-\mu \right|\ge k\sigma \right)\le \frac{1}{k^2}$

Calculation:

The values are:

k	Upper bound
2	$\begin{array}{c}\frac{1}{k^2}=\frac{1}{2^2}\\ =0.25\end{array}$
3	$\begin{array}{c}\frac{1}{k^2}=\frac{1}{3^2}\\ =0.11\end{array}$
4	$\begin{array}{c}\frac{1}{k^2}=\frac{1}{4^2}\\ =0.06\end{array}$
5	$\begin{array}{c}\frac{1}{k^2}=\frac{1}{5^2}\\ =0.04\end{array}$
10	$\begin{array}{c}\frac{1}{k^2}=\frac{1}{{10}^2}\\ =0.01\end{array}$

To determine

Find $P\left(\left|X-\mu \right|\ge k\sigma \right)$ for different values of k.

Comment on the upper bound comparing with the corresponding probability.

Answer to Problem 44E

For k=2, the probability value is 0.04.

The upper bound is very broad comparing with the value 0.04.

For k=3, 4 and 5 the probability value can’t be found.

Explanation of Solution

Calculation:

From problem 13,

$\begin{array}{c}E\left(X\right)={\displaystyle \sum _{allx}xp\left(x\right)}\\ =\left\{\begin{array}{l}0\times 0.10+1\times 0.15+2\times 0.20+\\ 3\times 0.25+4\times 0.20+5\times 0.06+6\times 0.04\end{array}\right\}\\ =2.64\end{array}$

$\begin{array}{c}E\left(X^2\right)={\displaystyle \sum _{allx}{x}^{2}p\left(x\right)}\\ =\left\{\begin{array}{l}{0}^2\times 0.10+1^2\times 0.15+2^2\times 0.20+\\ 3^2\times 0.25+4^2\times 0.20+5^2\times 0.06+6^2\times 0.04\end{array}\right\}\\ =9.34\end{array}$

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2\\ =9.34-{2.64}^2\\ =2.3704\end{array}$

Then $\sigma =\sqrt{2.3704}=1.54$

Now for k=2

$\begin{array}{c}P\left(\left|X-2.64\right|\ge 2\times 1.54\right)=P\left(\left|X-2.64\right|\ge 3.08\right)\\ =P\left(X\ge 2.64+3.08\right)orP\left(X\le 2.64-3.08\right)\\ =P\left(X\ge 5.72orX\le -0.44\right)\\ =P\left(X=6\right)\\ =0.04\end{array}$

From part a, the upper bound for k=2 is 0.25. Hence, the upper bound is very broad comparing with the value 0.04.

Fork=3

$\begin{array}{c}P\left(\left|X-2.64\right|\ge 3\times 1.54\right)=P\left(\left|X-2.64\right|\ge 4.62\right)\\ =P\left(X\ge 2.64+4.62\right)orP\left(X\le 2.64-4.62\right)\\ =P\left(X\ge 7.26orX\le -1.98\right)\end{array}$

This probability is not possible because the value of x is up to 6.

Similarly for k=4 and 5 the probability value can’t be found.

Thus, for k=3, 4 and 5 the probability value can’t be found.

To determine

Find $P\left(\left|X-\mu \right|\ge k\times \sigma \right)$ for the given value of k.

Give some suggestion for the upper bound for the corresponding probability..

Answer to Problem 44E

$\underset{\_}{P\left(\left|X-\mu \right|\ge 3\times \sigma \right)=\frac{1}{9}}$.

The probability and the upper bound is same for k = 3.

Explanation of Solution

Given info:

X have three possible values –1, 0, 1 with probability $\frac{1}{18},\frac{8}{9},\frac{1}{18}$ respectively.

Calculation:

The value of  $E\left(X\right)andV\left(X\right)$:

$\begin{array}{c}E\left(X\right)={\displaystyle \sum _{allx}xp\left(x\right)}\\ =\left\{\left(-1\right)\times \frac{1}{18}+0\times \frac{8}{9}+1\times \frac{1}{18}\right\}\\ =0\end{array}$

$\begin{array}{c}E\left(X^2\right)={\displaystyle \sum _{allx}{x}^{2}p\left(x\right)}\\ =\left\{{\left(-1\right)}^2\times \frac{1}{18}+{\left(0\right)}^2\times \frac{8}{9}+{\left(1\right)}^2\times \frac{1}{18}\right\}\\ =\frac{1}{9}\end{array}$

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2\\ =\frac{1}{9}-0\\ =\frac{1}{9}\end{array}$

Hence,$\sigma =\sqrt{\frac{1}{9}}=\frac{1}{3}$

Hence,

$\begin{array}{c}P\left(\left|X-0\right|\ge 3\times \frac{1}{3}\right)=P\left(\left|X\right|\ge 1\right)\\ =P\left(X=1orX=-1\right)\\ =\frac{1}{18}+\frac{1}{18}\\ =\frac{1}{9}\end{array}$

From the Chebyshev’s inequality,the upper bound for k=3 is $\frac{1}{3^2}=\frac{1}{9}$.

Hence, the probability and the upper bound is same for k=3.

To determine

Give a distribution for which $P\left(\left|X-\mu \right|\ge 5\times \sigma \right)=0.04$.

Answer to Problem 44E

The distribution of X is

$\underset{\_}{p\left(x\right)=\{\begin{array}{l}\frac{1}{50},\text{If}x=-1\\ \frac{24}{25},\text{If}x=0\\ \frac{1}{50},\text{If}x=1\end{array}}$

Explanation of Solution

Given info:

$P\left(\left|X-\mu \right|\ge 5\times \sigma \right)=0.04$

Calculation:

Let X have three possible values –1, 0, 1 with probability $\frac{1}{50},\frac{24}{25},\frac{1}{50}$ respectively.

The value of  $E\left(X\right)\text{and}V\left(X\right)$:

$\begin{array}{c}E\left(X\right)={\displaystyle \sum _{allx}xp\left(x\right)}\\ =\left\{\left(-1\right)\times \frac{1}{50}+0\times \frac{24}{25}+1\times \frac{1}{50}\right\}\\ =0\end{array}$

$\begin{array}{c}E\left(X^2\right)={\displaystyle \sum _{allx}{x}^{2}p\left(x\right)}\\ =\left\{{\left(-1\right)}^2\times \frac{1}{50}+{\left(0\right)}^2\times \frac{24}{25}+{\left(1\right)}^2\times \frac{1}{50}\right\}\\ =\frac{2}{50}=\frac{1}{25}\end{array}$

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2\\ =\frac{1}{25}-0\\ =\frac{1}{25}\end{array}$

Hence,$\sigma =\sqrt{\frac{1}{25}}=\frac{1}{5}$

Hence,

$\begin{array}{c}P\left(\left|X-0\right|\ge 5\times \frac{1}{5}\right)=P\left(\left|X\right|\ge 1\right)\\ =P\left(X=1orX=-1\right)\\ =\frac{1}{50}+\frac{1}{50}\\ =\frac{1}{25}=0.40\end{array}$

Hence, the distribution is

$\underset{\_}{p\left(x\right)=\{\begin{array}{l}\frac{1}{50},\text{If}x=-1\\ \frac{24}{25},\text{If}x=0\\ \frac{1}{50},\text{If}x=1\end{array}}$