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Chapter 3: Problem 9

Show in the discrete case that if \(X\) and \(Y\) are independent, then $$ E[X \mid Y=y]=E[X] \quad \text { for all } y $$

Short Answer

Expert verified
Using the definition of conditional expectation and the independence of X and Y, we can show the desired result as follows: 1. Recall the definition of conditional expectation: \(E[X | Y = y] = \sum_x x * P(X = x | Y = y)\). 2. Since X and Y are independent, we have \(P(X = x | Y = y) = P(X = x)\) for all x and y. 3. Replace the conditional probability with marginal probability: \(E[X | Y = y] = \sum_x x * P(X = x)\). 4. Recognize that the sum on the right-hand side is the definition of the expected value of X: \(E[X] = \sum_x x * P(X = x)\). 5. Finally, we have \(E[X | Y = y] = E[X]\) for all possible values of y, completing the proof.

Step by step solution

01

Definition of Conditional Expectation

Recall the definition of conditional expectation: \(E[X | Y = y] = \sum_x x * P(X = x | Y = y)\) Where the sum is taken over all possible values of x.
02

Independence of X and Y

Since X and Y are independent, we have: \(P(X = x | Y = y) = P(X = x)\) for all x and y.
03

Replace Conditional Probability with Marginal Probability

Using the independence fact from Step 2, we can replace the conditional probability in the conditional expectation formula from Step 1: \(E[X | Y = y] = \sum_x x * P(X = x)\)
04

Recognize E[X]

Notice that the sum on the right-hand side is the definition of the expected value of X: \(E[X] = \sum_x x * P(X = x)\)
05

Final Result

Combining Steps 3 and 4, we have: \(E[X | Y = y] = E[X]\) for all possible values of y. This completes the proof.

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Most popular questions from this chapter

Suppose \(X\) and \(Y\) are independent continuous random variables. Show that $$ E[X \mid Y=y]=E[X] $$ for all \(y\)

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