Question

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

Short answer

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

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.

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.

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.

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.