Question

The final answer is 4.472

2. You roll two fair, six-sided dice. Let X be the number on the first die. Let Y be the number on the second die. Calculate E[max(X,Y)], the expected value of the larger of the two numbers. There are several ways you can do this. You should try to do this by applying 2D LOTUS to the joint distribution of X and Y , which is extremely simple. To check your answer, you can use the p.m.f. of L = max(X, Y) that you derived in Lesson 19.

Answer 1

$\small \text{Since the dices are fair and outcome of two dices are independent of each other, the joint distribution} \\ \text{of } X \text{ and } Y \text{ is given as},\\ P(X = i, Y = j) = P(X = i)*(Y = j) = \frac{1}{6}*\frac{1}{6} = \frac{1}{36}; \ \ \ i,j \in \{1,2,3,4,5,6 \}$$\small \text{Using LOTUS, we get then} \\ \begin{align*} E[\max(X, Y)] &= \sum_{i=1}^6 \sum_{j=1}^6 \max(i,j)*P(X = i, Y = j)\\ &= \sum_{i=1}^6 \sum_{j=1}^6 \max(i,j)*\frac{1}{36} \\ &= \frac{1}{36} \sum_{i=1}^6 \sum_{j=1}^6 \max(i,j) \\ &= \frac{1}{36} \left( \sum_{j=1}^6 \max(1, j) + \sum_{j=1}^6 \max(2, j) + \sum_{j=1}^6 \max(3, j) + \sum_{j=1}^6 \max(4, j) + \sum_{j=1}^6 \max(5, j) + \sum_{j=1}^6 \max(6, j)\right ) \\ &= \frac{1}{36}((1+2+3+4+5+6) + (2+2+3+4+5+6) + (3+3+3+4+5+6) + \\ & \ \ \ \ \ \ \ \ \ \ (4+4+4+4+5+6) + (5+5+5+5+5+6) + (6+6+6+6+6+6)) \\ &= \frac{161}{36} \\ &= 4.472 \end{align*}$