Chapter 3.4, Problem 67E

Refer to Chebyshev’s inequality given in Exercise 44. Calculate P(|X - μ| ≥ kσ) for k = 2 and k = 3 when X ∼ Bin(20, .5), and compare to the corresponding upper bound. Repeat for X ∼ Bin(20, .75).

To determine

Find the value of $P\left(\left|X-\mu \right|\ge k\sigma \right)$ for k=2,3 when $X\sim Bin\left(20,0.5\right)$ and $X\sim Bin\left(20,0.75\right)$.

Compare the upper bound for the different k value.

Answer to Problem 67E

For $X\sim Bin\left(20,0.5\right)$

$P\left(\left|X-\mu \right|\ge k\sigma \right)=\{\begin{array}{l}\underset{\_}{0.042},\text{ for }k\text{=2}\\ \underset{\_}{\text{0}\text{.002}}\text{,for }k\text{=3}\end{array}$

for $X\sim Bin\left(20,0.75\right)$

$P\left(\left|X-\mu \right|\ge k\sigma \right)=\{\begin{array}{l}\underset{\_}{0.065},\text{ for }k\text{=2}\\ \underset{\_}{\text{0}\text{.004}}\text{,for }k\text{=3}\end{array}$

In each cases, the values of $P\left(\left|X-\mu \right|\ge k\sigma \right)$ is less than the upper bound of the Chebyshev’s bound.

Explanation of Solution

Given info:

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

$X\sim Bin\left(20,0.5\right)$ and $X\sim Bin\left(20,0.75\right)$.

Calculation:

The value of $P\left(\left|X-\mu \right|\ge k\sigma \right)$ For$X\sim Bin\left(20,0.5\right)$:

The expectation and standard deviation is obtained as given below:

$\begin{array}{c}E\left(X\right)=np\\ =20\times 0.5\\ =10\end{array}$

$\begin{array}{c}SD\left(X\right)=\sqrt{np\left(1-p\right)}\\ =\sqrt{20\times 0.5\times 0.5}\\ =2.236\end{array}$

For k = 2

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=P\left(\left|X-10\right|\ge 2\times 2.236\right)\\ =P\left(\left|X-10\right|\ge 4.472\right)\\ =P\left(X\le 10-4.472\text{or}X\ge 4.472+10\right)\\ =P\left(X\le 5\text{or}X\ge 15\right)\end{array}$

$\begin{array}{c}=P\left(X\le 5\right)+P\left(X\ge 15\right)\\ =P\left(X\le 5\right)+1-P\left(X\le 14\right)\\ =B\left(5:20,0.5\right)+1-B\left(14:20,0.5\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n=20
• Along with n=20, choose x=5,14
• Then, obtain the table value corresponding to p=0.5.

The value of $B\left(5:20,0.5\right)$ is 0.021and $B\left(14:20,0.5\right)$ is 0.979.

Hence,

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=B\left(5:20,0.5\right)+1-B\left(14:20,0.5\right)\\ =0.021+1-0.979\\ =0.042\end{array}$

For k=3

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=P\left(\left|X-10\right|\ge 3\times 2.236\right)\\ =P\left(\left|X-10\right|\ge 6.708\right)\\ =P\left(X\le 10-6.708\text{or}X\ge 6.708+10\right)\\ =P\left(X\le 3\text{or}X\ge 17\right)\end{array}$

$\begin{array}{c}=P\left(X\le 3\right)+P\left(X\ge 17\right)\\ =P\left(X\le 3\right)+1-P\left(X\le 16\right)\\ =B\left(3:20,0.5\right)+1-B\left(16:20,0.5\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n=20
• Along with n=20, choose x=3,16
• Then, obtain the table value corresponding to p=0.5.

The value of $B\left(3:20,0.5\right)$ is 0.001and $B\left(16:20,0.5\right)$ is 0.999.

Hence,

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=B\left(3:20,0.5\right)+1-B\left(16:20,0.5\right)\\ =0.001+1-0.999\\ =0.002\end{array}$

Hence, for $X\sim Bin\left(20,0.5\right)$

$P\left(\left|X-\mu \right|\ge k\sigma \right)=\{\begin{array}{l}0.042,\text{ for }k\text{=2}\\ \text{0}\text{.002,for }k\text{=3}\end{array}$

For$X\sim Bin\left(20,0.75\right)$:

The expectation and standard deviation:

$\begin{array}{c}E\left(X\right)=np\\ =20\times 0.75\\ =15\end{array}$

$\begin{array}{c}SD\left(X\right)=\sqrt{np\left(1-p\right)}\\ =\sqrt{20\times 0.75\times 0.75}\\ =1.937\end{array}$

For k=2

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=P\left(\left|X-15\right|\ge 2\times 1.937\right)\\ =P\left(\left|X-15\right|\ge 3.874\right)\\ =P\left(X\le 15-3.874\text{or}X\ge 3.874+15\right)\\ =P\left(X\le 11\text{or}X\ge 19\right)\end{array}$

$\begin{array}{c}=P\left(X\le 11\right)+P\left(X\ge 19\right)\\ =P\left(X\le 11\right)+1-P\left(X\le 18\right)\\ =B\left(11:20,0.75\right)+1-B\left(18:20,0.75\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n=20
• Along with n=20, choose x=11,18
• Then, obtain the table value corresponding to p=0.75.

The value of $B\left(11:20,0.5\right)$ is 0.041and $B\left(18:20,0.5\right)$ is 0.976.

Hence,

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=B\left(11:20,0.5\right)+1-B\left(18:20,0.5\right)\\ =0.041+1-0.976\\ =0.065\end{array}$

For k=3

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=P\left(\left|X-15\right|\ge 3\times 1.937\right)\\ =P\left(\left|X-15\right|\ge 6.708\right)\\ =P\left(X\le 15-6.708\text{or}X\ge 6.708+15\right)\\ =P\left(X\le 9\text{or}X\ge 22\right)\end{array}$

                            $\begin{array}{c}=P\left(X\le 9\right),\text{as maximum possible value of }x\text{ is 20}\\ =B\left(9:20,0.75\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n=20
• Along with n=20, choose x=9
• Then, obtain the table value corresponding to p=0.75.

The value of $B\left(9:20,0.75\right)$ is 0.004.

Hence,

$\begin{array}{c}P\left(\left|X-\mu \right|\ge k\sigma \right)=B\left(9:20,0.75\right)\\ =0.004\end{array}$

Thus, for $X\sim Bin\left(20,0.75\right)$

$P\left(\left|X-\mu \right|\ge k\sigma \right)=\{\begin{array}{l}0.065,\text{ for }k\text{=2}\\ \text{0}\text{.004,for }k\text{=3}\end{array}$

The Chebyshev’s bound

For k = 2,

$\begin{array}{c}\frac{1}{k^2}=\frac{1}{2^2}\\ =\frac{1}{4}\\ =0.25\end{array}$

For k = 3,

$\begin{array}{c}\frac{1}{k^2}=\frac{1}{3^2}\\ =\frac{1}{9}\\ =0.11\end{array}$

In each cases, the values of $P\left(\left|X-\mu \right|\ge k\sigma \right)$ is less than the upper bound of the Chebyshev’s bound.