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Chapter 2: Problem 6

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=1,-x

Short Answer

Expert verified
The conditional expectation E(Y|X) is 0, which is a horizontal straight line when plotted against X. On the other hand, the conditional expectation E(X|Y) equals (1+y)/2 when y>0 and (1-y)/2 when y<0, thus forming two different lines when plotted against y, which means it is not a straight line pointwise.

Step by step solution

01

Determine E(Y|X)

Remember that the expectation of Y given X = x is, by definition, the integral of y*f(y|x) dy, where f(y|x) is the conditional pdf of Y given X = x. Here though, Y is uniformly distributed over the interval \(-x\) to \(x\). So to find this integral, you'll need to take \(y(-x, x)\), times the conditional pdf \(1/(2x)\), and integrate over the interval \(-x\) to \(x\). Doing this gives E(Y|X) = 0.
02

Graph E(Y|X)

The conditional expectation found in Step 1, E(Y|X) = 0, does not depend on x. Therefore, if you were to graph E(Y|X) as a function of x, it would just be a horizontal line on the y=0 axis. A horizontal line is in fact a straight line, meeting the criteria set out in the problem.
03

Determine E(X|Y)

First, it must be noted there are two possible regions of support, depending if \(y\)>0 or \(y\)is <0. Let's find the expectation in each case. When \(y\)>0, X is uniformly distributed over the interval (\(y\), 1). Thus \(E(X|Y=y)\) is the average of \(y\) and 1. This is a non-constant function of \(y\), namely (1+\(y\))/2. When \(y\)<0, X is uniformly distributed over the interval (-\(y\), 1). Thus \(E(X|Y=y)\) is the average of -\(y\) and 1, which gives (1-\(y\))/2.
04

Graph E(X|Y)

Here, the expression for E(X|Y) changes with y. Therefore, if you were to graph E(X|Y) as a function of y, you would get a different line depending on whether \(y\) is positive or negative. Thus, the graph of E(X|Y) pointwise is not one straight line.

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Most popular questions from this chapter

Let \(X_{1}\) and \(X_{2}\) have the joint pdf \(f\left(x_{1}, x_{2}\right)=x_{1}+x_{2}, 0

Let \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}\) be the common variance of \(X_{1}\) and \(X_{2}\) and let \(\rho\) be the correlation coefficient of \(X_{1}\) and \(X_{2}\). Show that $$P\left[\left|\left(X_{1}-\mu_{1}\right)+\left(X_{2}-\mu_{2}\right)\right| \geq k \sigma\right] \leq \frac{2(1+\rho)}{k^{2}}$$

Suppose \(X_{1}\) and \(X_{2}\) are random variables of the discrete type which have the joint pmf \(p\left(x_{1}, x_{2}\right)=\left(x_{1}+2 x_{2}\right) / 18,\left(x_{1}, x_{2}\right)=(1,1),(1,2),(2,1),(2,2)\), zero elsewhere. Determine the conditional mean and variance of \(X_{2}\), given \(X_{1}=x_{1}\), for \(x_{1}=1\) or 2. Also compute \(E\left(3 X_{1}-2 X_{2}\right)\).

Cast a fair die and let \(X=0\) if 1,2, or 3 spots appear, let \(X=1\) if 4 or 5 spots appear, and let \(X=2\) if 6 spots appear. Do this two independent times, obtaining \(X_{1}\) and \(X_{2} .\) Calculate \(P\left(\left|X_{1}-X_{2}\right|=1\right)\).

Find the probability of the union of the events \(a
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