Math Q.

[Ribbo]

Suppose I have two distributions, and for now let's cheat and call them normal. If I know the standard deviations of both, Now suppose I take a random data point from both distributions. What I want to know is how to calculate the chance that the point from the more variant distribution is farther from its mean than the random point from the less variant distribution. The general solution would be quite useful for me right about now.

[Ribbo]

Always nice to see no-one is smart enough to answer it. [img]/images/graemlins/smirk.gif[/img]
Makes me feel less bad about my own lack of math skill.

Easiest way would be set up a Monte Carlo sim and draw 1,000,000 random numbers or so and see how many times your condition holds. This probability will give you a good estimate for that chance.

Normalize the two distributions, and the points you've drawn and then compare their z-scores.

If X is one of the normals with mean X_u and standard deviation X_s, then set Z_X = (X - X_u) / X_s. If you drew x from X, then you've drawn (x - X_u) / X_s from Z_X.

Do the same with Y and your drawn point y.

Now if N(.) is the standard normal distribution, then compare N((x-X_u)/X_s) and N((y-Y_u)/Y_s).

Maybe I can't discern what you want to do from your original post though ?!?

[Cooker]

I think your question isn't so difficult. You can't really nicely display equations on a forum so look at this. I am fairly sure what I did here is correct. Not hard just a tad tedious.