Transformations on monotonic partitions

This one doesn't appear as a named result, though I found it in more than one text and it's pretty crucial for a lot of transformations.

Let X be a random variable with pdf f_X(x). Let Y = g(x). If A₀, ..., A_k are a partition of the sample space of X such that P(A₀) = 0 and f_X(x) is continuous on each A_i then if g(x) can be partitioned into g₁(x), ..., g_k(x) such that:

1. g(x) = g_i(x) for all x in A_i
2. each g_i(x) is monotone on A_i
3. g_i(A_i) = g_j(A_j) for all i, j (that is, the image of each partition is the same under g)
4. each g_i^-1(y) has a continuous derivative on g_i(A_i)

Then



OK, that's a mouthful, but if you can get past the details to see the result for what it is, the details are self-evident (except for the A₀ partition, which doesn't carry any mass and is just a technical device that makes proving the theorem easier).

The transformation goes up and down, but you can chop it into bits where each bit just goes up or just goes down. Further, the projection of each bit is continuous and maps to the same range (this assumption can be generalized away, but it makes the formula more complicated). Finally, the inverse of each bit has a continuous derivative.

If all that is true, then the pdf of y is just the sum of each of the inverses times the absolute value of the inverse derivative.

Another way to think of it: at any point x, f_X(x) is how much weight gets put on that point as X varies along the number line, In the simple monotone result from Wednesday, the transformed pdf, f_Y(y), is just inverse mapped back to that weight, f_X(x), with a "speed adjustment" because Y is "moving" at a different speed than X as indicated by g. All that matters is the speed, not the direction, which is why the absolute value is necessary. In this result, we're saying the same thing except that multiple values of x map to the same y, so we have to add them all up, still adjusting for the speed differences at each point.