Expected slope of a random line.

Question

Describe a way to generate a “random line” in the plane; for definiteness, assume the line goes through the origin, and all angles (with respect to the x -axis) are equally likely. Then, under that model, determine what is the expected value of the slope of the line.

So the whole $Y = mX + b$ argument turns into $Y = m X$. Also, I assume since I know (0,0) is a point my slope is just $m = \frac{Y}{X}$. I then have to compute expectations. This portion of the book doesn't really going into Covariance so I'm not sure if this argument has any meaning: $E[Y/X]= E[Y * 1/X] = E[Y]E[1/X]+Cov[X,Y]$

Another issue comes into thinking of it as being a r.v. distributed on something. I know I can't really put the real number line on a distribution, but if I let X and Y be two different integers I feel like I can have them be distributed on something, and then I'm a little unsure if the trick is just that if they're both on say, the discrete uniform distribution, that the covariance is just 0?

Any thoughts?

Answer 1

You have $\theta$ is uniformly distributed on $(-\pi/2; \pi/2)$ and wish to find the expectation for $\tan\theta$.

$\mathsf E(\tan\theta) = \lim\limits_{0>\alpha\to\pi/2} \int_{-\alpha}^{\alpha} \tan\theta~\mathrm d~\theta$.

And you might intuit that it is zero by symmetry.

Answer 2

Another way to see that the expected value for slope doesn't exist, is to work with angles. The possible range of angles is $0\dots\pi$. From this you can easily see the expected angle is $\frac{\pi}2$ - the tangent of which is infinity.