A fair die is rolled twice, with outcomes X for the first roll and Y for the second roll. Find the moment generating function MX`Y ptq of X ` Y . Note that your answer should be a function of t and can contain unsimplified finite sums.

Mathematics · College · Thu Feb 04 2021

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 To find the moment generating function (MGF) \(M_{X+Y}(t)\) of the sum \(X + Y\) where \(X\) and \(Y\) are the outcomes of the rolls of a fair six-sided die, we must consider the properties of the MGFs of \(X\) and \(Y\) individually, and then use these properties to find the MGF of their sum. Let's get started.

First, let's find the moment generating function for a single roll of a fair die. The probability of any one outcome, such as rolling a 1, 2, 3, 4, 5, or 6, is \( \frac{1}{6} \). The MGF of a discrete random variable \(Z\) is given by:

\[ M_Z(t) = E[e^{tZ}] = \sum_{\text{all } z} e^{tz} P(Z = z) \]

For a single roll of a die, \(Z\) can take on values 1, 2, 3, 4, 5, or 6 with equal probability. Therefore, the MGF of a single roll is:

\[ M_X(t) = \sum_{x=1}^{6} e^{tx}\frac{1}{6} \]

Since \(X\) and \(Y\) are independent and identically distributed, \(M_Y(t) = M_X(t)\).

Now, to find the MGF of the sum \(X + Y\), we use the property that the MGF of the sum of independent random variables is the product of their individual MGFs:

\[ M_{X+Y}(t) = M_X(t) \cdot M_Y(t) \]

Since both \(X\) and \(Y\) are the outcomes of a fair die roll, their MGFs are identical:

\[ M_{X+Y}(t) = \left( \sum_{x=1}^{6} e^{tx}\frac{1}{6} \right) \left( \sum_{y=1}^{6} e^{ty}\frac{1}{6} \right) \]

Expanding this product, we get:

\[ M_{X+Y}(t) = \left( \frac{e^t}{6} + \frac{e^{2t}}{6} + \frac{e^{3t}}{6} + \frac{e^{4t}}{6} + \frac{e^{5t}}{6} + \frac{e^{6t}}{6} \right)^2 \]

\[ M_{X+Y}(t) = \frac{1}{36} \left( e^{2t} + 2e^{3t} + 3e^{4t} + 4e^{5t} + 5e^{6t} + 6e^{7t} + 5e^{8t} + 4e^{9t} + 3e^{10t} + 2e^{11t} + e^{12t} \right) \]

This expression gives us unsimplified finite sums. It doesn't make further simplification challenging, and it adheres to the requirement that the answer should contain unsimplified finite sums.

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