Q. 9.14 - A First Course in Probability

Question

A pair of fair dice is rolled. Let 

$X=\left\{\begin{array}{ll}1& if the sum of the dice is 6\\ 0& otherwise\end{array}\right.$

 and let Y equal the value of the first die. Compute (a) H(Y), (b) HY(X), and (c) H(X, Y).

Step-by-Step Solution

Answer

a) $H\left(Y\right)=2.58$

b) $HY\left(X\right)=0.65$

c) $H(X, Y)=3.12$.

Part (a) Step 1: Given Information

We have given value of $X=\left\{\begin{array}{ll}1& if the sum of the dice is 6\\ 0& otherwise\end{array}\right.$

and $Y$ equal the value of the first die.

We have to compute $H\left(Y\right)$.

Part (a) Step 2: Simplify

We have$Y$ can assume each value $1,......6$  with equal probabilities $\frac{1}{2}$ . So, the entropy of $Y$ is

$$\begin{gathered} H\left(Y\right)-\underset{i}{}{p}_i\text{log}{p}_i \\ =-\text{log}\frac{1}{6} \\ =\text{log}6 \\ =2.58 \end{gathered}$$

Part (b) Step 1: Given Information

We have to compute $H_Y\left(X\right)$.

Part (b) Step 2: Simplify

For $y\in \left\{1,\dots ,5\right\}$, we have that $X=1$given that $Y=y$means that on the first die we have obtained $y$, and in order to obtain 6 in the sum, we need $6-y$ on the second die,

$\frac{X}{Y}=y~\left(\begin{array}{cc}0& 1\\ 5/6& 1/6\end{array}\right)$

So the entropy is given by

$$\begin{gathered} H_{Y=y}\left(X\right)=-\sum _{i=0,1}p\left(i\mid y\right) \text{log }p\left(i\mid y\right) \\ =-\frac{5}{6}\text{log}\frac{5}{6}-\frac{1}{6}\text{log}\frac{1}{6} \\ =0.65 \cup \end{gathered}$$

Part (c) Step 1: Given Information

Need  to compute $H(X,Y)$.

Part (c) Step 2: Explanation

From the proposition in the chapter, we have

$$\begin{gathered} H\left(X,Y\right)=H\left(Y\right)+H_Y\left(X\right) \\ =2.58+0.5417 \\ =3.12 \end{gathered}$$