Question

Considering I have a continuous joint distribution of two independent normal random variables (let’s assume the independent vars are on the X and Z axis, and the dependent – the joint probability – is on the Y axis), and I have a line anywhere on the XZ plane, how would I compute the probability of a point falling on one side or the other of that line?

Answer 1

First move everything so that the two normal distributions (the X and the Z) are centered on zero; now the joint ditribution will be a hill centered on the origin.

Now scale one of the axes so that the two distributions have the same variance (or “width”). Now the joint probability should be a rotationally symmetric hill.

Now all that matters is how close the line comes to the origin. Rotate about the origin (this will leave the joint probability unchanged) until the line is parallel to one of the axes, say Z. Now you’re asking for the probability that a random point will have X greater or less than the X-value of the line. That’s determined by one of the scaled ditribution functions (they’re the same), and can be calculated by means of the error function.

I can write out the math if that would be useful.

EDIT: I’ll try to write out the last step. Pardon my crude ascii, I don’t have access to a good math tablet.

Suppose we’ve scaled and centered the distributions so that sigmaX = sigmaZ = 1, and rotated everything:

joint probability: P(x, z) = 1/(2 pi) exp(-(x^2 + z^2)/2)

line: x = c

Now to find the probability that a random point will be on a narrow “vertical” strip between some x and x+dx:

P(x)dx = Int[z=-Inf, z=+Inf]{dz P(x, z)}
       = 1/sqrt(2 pi) exp(-x^2/2) 1/sqrt(2 pi) Int[z=-Inf, z=+Inf]{dz exp(-z^2/2)}
       = 1/sqrt(2 pi) exp(-x^2/2)

But that’s the same as (either) one of the two normal distributions. So the probability that a random point will be, say, to the left of the line is

P(c>x) = Int[-Inf, c]{dx 1/sqrt(2 pi) exp(-x^2/2)}
       = 1/2 (1 - Erf(c/sqrt(2)))