Question

Question: Let X be the number appearing on the first die when two fair dice are rolled and let Y be the sum of the numbers appearing on the two dice. Show that\(E(X) E(Y) \ne E(XY)\)

Short answer

Answer:

The inequality \(E(X) E(Y) \ne E(XY)\) has been proved.

Step-by-step solution

Note the given data

Given X be the number appearing on the first die when two fair dice are rolled and let Y be the sum of the numbers appearing on the two dice.

Theorem  

Theorem : If X is a random variable and p(X =r) is the probability that X =r then.\(E\left( X \right) = \sum\limits_{r \in X(s)} {p(X = r)r} \)

Calculation

Let X be the number on the first die. So X = 1,2,3,4,5,6

\(\begin{array}{l}E\left( X \right) = \sum\limits_{r\^I X(s)} {p(X = r)r} \\\begin{array}{*{20}{c}}{}&{}\end{array} = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6}\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{{21}}{6}\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{7}{2}\end{array}\)

Let W be the number on the second die, So W = 1,2,3,4,5,6

\(\begin{array}{l}E\left( W \right) = \sum\limits_{r\^I X(s)} {p(X = r)r} \\\begin{array}{*{20}{c}}{}&{}\end{array} = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6}\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{{21}}{6}\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{7}{2}\end{array}\)

Let Y = X + W

\(\begin{array}{l}E\left( Y \right) = E\left( X \right) + E\left( W \right)\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{7}{2} + \frac{7}{2}\\\begin{array}{*{20}{c}}{}&{}\end{array} = 7\end{array}\)

\(\begin{array}{l}E\left( {{X^2}} \right) = {1^2} \times \frac{1}{6} + {2^2} \times \frac{1}{6} + {3^2} \times \frac{1}{6} + {4^2} \times \frac{1}{6} + {5^2} \times \frac{1}{6} + {6^2} \times \frac{1}{6}\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{{91}}{6}\end{array}\)

\(\begin{array}{l}E\left( {XY} \right) = E(X\left( {X + W} \right))\\\begin{array}{*{20}{c}}{}&{}\end{array} = E\left( {{X^2} + XW} \right)\\\begin{array}{*{20}{c}}{}&{}\end{array} = E\left( {{X^2}} \right) + E\left( {XW} \right)\end{array}\)

Simplification

\(\begin{array}{l}E\left( {XY} \right) = E\left( {{X^2}} \right) + E\left( {XW} \right)\\\begin{array}{*{20}{c}}{}&{}&{}\end{array} = E\left( {{X^2}} \right) + E\left( X \right)E\left( W \right)\\\begin{array}{*{20}{c}}{}&{}&{}\end{array} = \frac{{91}}{6} + \frac{7}{2} \times \frac{7}{2}\\\begin{array}{*{20}{c}}{}&{}&{}\end{array} = \frac{{329}}{{12}}\end{array}\)

Also, \(\begin{array}{l}E\left( X \right)E(Y) = \frac{7}{2} \times 7\\\begin{array}{*{20}{c}}{}&{}\end{array} = \frac{{49}}{2}\end{array}\)

Conclusion  

Thus,\(E(X) E(Y) \ne E(XY)\)