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.