/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Let \(X\) be the value of the fi... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 72

Let \(X\) be the value of the first die and \(Y\) the sum of the values when two dice are rolled. Compute the joint moment generating function of \(X\) and \(Y\).

Short Answer

Expert verified
The joint moment generating function of \(X\) and \(Y\) is given by \(M_{X,Y}(t_1, t_2) = \frac{1}{36}(e^{t_1 + 2t_2} + e^{t_1 + 3t_2} + \dots + e^{t_1 + 7t_2} + e^{2t_1 + 3t_2} + \dots + e^{6t_1 + 12t_2})\).

Step by step solution

01

Finding the joint probability mass function

A die has six faces, each with a probability of \(\frac{1}{6}\) of being rolled. Hence, for any value \(i\) of the first die \(X\) and the sum \(j\) of the two dice, the joint probability mass function can be given as: \(p_{X,Y}(i, j) = P(X = i, Y = j)\) To find the joint probability mass function, we need to consider all possible combinations of the dice rolls that would give us the specified sum \(j\). If the first roll is \(i\), the second roll would be \(j-i\). So, we can define the joint probability mass function as follows: $p_{X,Y}(i, j) =\begin{cases} \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} & \text{if } 1\leq i \leq 6 \text{ and } i+1 \leq j \leq i+6\\ 0 & \text{otherwise} \end{cases}$
02

Computing the joint moment generating function

Now that we have the joint probability mass function, we can compute the joint moment generating function (MGF) using its definition: \(M_{X,Y}(t_1, t_2) = E[e^{t_1X + t_2Y}] = \sum_{i = 1}^6 \sum_{j = i+1}^{i+6} e^{t_1i + t_2j}p_{X,Y}(i, j)\) Using the joint probability mass function defined in step 1, we can write the MGF as follows: \(M_{X,Y}(t_1, t_2) = \sum_{i = 1}^6 \sum_{j = i+1}^{i+6} e^{t_1i + t_2j}\frac{1}{36}\) Now, we'll compute the MGF using this expression:
03

Summation and expansion

Let's compute the sum of the series for the joint MGF. \(M_{X,Y}(t_1, t_2) = \frac{1}{36}\sum_{i = 1}^6 \sum_{j = i+1}^{i+6} e^{t_1i + t_2j}\) Now, expand the summation. \(M_{X,Y}(t_1, t_2) = \frac{1}{36}(e^{t_1(1) + t_2{(1+1)}} + e^{t_1(1) + t_2{(1+2)}} + \dots + e^{t_1(1) + t_2{(1+6)}} + e^{t_1(2) + t_2{(2+1)}} + \dots + e^{t_1(6) + t_2{(6+6)}})\) Finally, the joint moment generating function of \(X\) and \(Y\) is: \(M_{X,Y}(t_1, t_2) = \frac{1}{36}(e^{t_1 + 2t_2} + e^{t_1 + 3t_2} + \dots + e^{t_1 + 7t_2} + e^{2t_1 + 3t_2} + \dots + e^{6t_1 + 12t_2})\)

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Most popular questions from this chapter

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$

The best quadratic predictor of \(Y\) with respect to \(X\) is \(a+b X+c X^{2}\), where \(a, b\), and \(c\) are chosen to minimize \(E\left[\left(Y-\left(a+b X+c X^{2}\right)\right)^{2}\right]\). Determine \(a, b\), and \(c\).

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