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Chapter 9: Problem 547

Show that the sum of the expected value of two discrete random variables with joint density \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is equal to the expected value of the sum of these two random variables. That is \(\mathrm{E}(\mathrm{X}+\mathrm{Y})=\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})\)

Short Answer

Expert verified
We proved that the expected value of the sum of two discrete random variables with joint density f(x, y) is equal to the sum of their individual expected values. We expanded the expression for \(\mathrm{E}(X + Y)\) and manipulated it to show that \(\mathrm{E}(X + Y) = \sum_x x \cdot \mathrm{f}(x) + \sum_y y \cdot \mathrm{f}(y)\), which is equivalent to \(\mathrm{E}(X + Y) = \mathrm{E}(X) + \mathrm{E}(Y)\).

Step by step solution

01

Recall the definition of expected value for discrete random variables

To find the expected value of a discrete random variable X, we use the formula: \[\mathrm{E}(X) = \sum_x x \cdot \mathrm{f}(x)\] Similarly, for Y: \[\mathrm{E}(Y) = \sum_y y \cdot \mathrm{f}(y)\]
02

Write the expected value of the sum of random variables X and Y

We want to find the expected value of the sum of X and Y, i.e., \(\mathrm{E}(X + Y)\). This is given by: \[\mathrm{E}(X + Y) = \sum_{x, y} (x + y) \cdot \mathrm{f}(x, y)\]
03

Expand and manipulate the E(X + Y) expression

Next, we expand the expression for \(\mathrm{E}(X + Y)\) and manipulate it as follows: \[ \begin{aligned} \mathrm{E}(X + Y) &= \sum_{x, y} (x + y) \cdot \mathrm{f}(x, y) \\ &= \sum_{x, y} x \cdot \mathrm{f}(x, y) + y \cdot \mathrm{f}(x, y) \\ &= \sum_x x \cdot \left(\sum_y \mathrm{f}(x, y)\right) + \sum_y y \cdot \left(\sum_x \mathrm{f}(x, y)\right), \end{aligned} \]where in the last step, we separate the two sums.
04

Recognize the expressions for E(X) and E(Y)

From the last expression in Step 3, we see that the two summations inside the parentheses are the marginal densities of X and Y, respectively. So, we can rewrite the expression as: \[ \mathrm{E}(X + Y) = \sum_x x \cdot \mathrm{f}(x) + \sum_y y \cdot \mathrm{f}(y) \]We recognize that this is the sum of the expected values of X and Y, so we have: \[ \mathrm{E}(X + Y) = \mathrm{E}(X) + \mathrm{E}(Y) \]
05

Conclusion

We have shown that the expected value of the sum of two discrete random variables with joint density f(x, y) is equal to the sum of their individual expected values, i.e., \(\mathrm{E}(X + Y) = \mathrm{E}(X) + \mathrm{E}(Y)\).

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Harvey of Brooklyn surveyed a random sample of 625 students at SUNY-Stony Brook. Being a pre-medical student, he hoped that most students would major in the social sciences rather than the natural sciences, thus provide him with less competition. To Harvey's dismay, \(60 \%\) of the students he surveyed were majoring in the natural sciences. Construct a \(95 \%\) confidence interval for \(p\), the population proportion of students majoring in the natural sciences.

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