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Chapter 3: Problem 257

For continuous random variables \(X\) and \(Y\), prove that \(E(Y)=E_{X}[E(Y \mid x)]\).

Short Answer

Expert verified
The proof is complete after showing that the right-hand side of the equation \(E_{X}[E(Y|X)]\) simplifies to \(E(Y)\), the left-hand side of the equation, thus demonstrating that \(E(Y) = E_{X}[E(Y|X)]\).

Step by step solution

01

Understanding the Problem Statement

First, understand that you are given two continuous random variables, \(X\) and \(Y\), and asked to prove that \(E(Y) = E_{X}[E(Y | X)]\). This equation is a form of the law of iterated expectations, which says that the expectation of a random variable (in this case, \(Y\)) equals the expected values of its conditional expectations.
02

Using the Definition of Conditional Expectation

You apply the definition of conditional expectation to the random variable \(Y\) given \(X\). This is expressed as \(E(Y | X) = \int y f(y|x) dy\), where \(f(y|x)\) is the conditional probability density function of \(Y\) given \(X\).
03

Using the Definition of Expectation

Then, you apply the definition of expectation, substituting in the definition of conditional expectation. This gives \(E_{X}[E(Y|X)] = \int (\int y f(y|x) dy) f(x) dx\), where \(f(x)\) is the probability density function of \(X\).
04

Simplifying the Expression

Simplify the expression using the property of integration. The equation becomes \(E_{X}[E(Y|X)] = \int y (\int f(y|x) dx) dy\), which can be recognized as the joint density function of \(X\) and \(Y\). This simplifies to \(E_{X}[E(Y|X)] = \int y f(y) dy\), where \(f(y)\) is the probability density function of \(Y\).
05

Concluding the Proof

This last equation simply equals \(E(Y)\), according to the definition of expectation, hence you can conclude that \(E(Y) = E_{X}[E(Y|X)]\). This completes your proof.

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Most popular questions from this chapter

For each of the following joint pdfs, find \(f_{X}(x)\) and \(f_{Y}(y)\). (a) \(f_{X, Y}(x, y)=\frac{1}{2}, 0 \leq x \leq y \leq 2\) (b) \(f_{X, Y}(x, y)=\frac{1}{x}, 0 \leq y \leq x \leq 1\) (c) \(f_{X, Y}(x, y)=6 x, 0 \leq x \leq 1,0 \leq y \leq 1-x\)

An urn contains \(n\) chips numbered 1 through \(n\). Assume that the probability of choosing chip \(i\) is equal to \(k i, i=1,2, \ldots, n\). If one chip is drawn, calculate \(E\left(\frac{1}{X}\right)\), where the random variable \(X\) denotes the number showing on the chip selected. [Hint: Recall that the sum of the first \(n\) integers is \(n(n+1) / 2\).]

Suppose that two fair dice are tossed one time. Let \(X\) denote the number of 2 's that appear, and \(Y\) the number of 3 's. Write the matrix giving the joint probability density function for \(X\) and \(Y\). Suppose a third random variable, \(Z\), is defined, where \(Z=X+Y\). Use \(p_{X, Y}(x, y)\) to find \(p_{Z}(z)\).

An urn contains five balls numbered 1 to 5 . Two balls are drawn simultaneously. (a) Let \(X\) be the larger of the two numbers drawn. Find \(p_{X}(k)\). (b) Let \(V\) be the sum of the two numbers drawn. Find \(p_{V}(k)\).

Suppose \(X\) is a binomial random variable with \(n=4\) and \(p=\frac{2}{3}\). What is the pdf of \(2 X+1\) ?
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