Q: Suppose we already know f_

_{X}(x) of rvar X. Now we get an X-derived rvar Y:=y(X), where y() is a "nice" function of X. What's the (unconditional) distribution of Y?
We first find the inverse function of the "nice" function. Call it X=g(Y). Then at any specific value like Y=10, the unconditional density of Y is given by

f_

_{Y}(10) = f__{X}( g(10) ) * g'(10)
, where g'(10) is the curve gradient dx/dy evaluated at the curve point y=10.

Here's a more intuitive interpretation. [[applied stat and prob for engineers]] P161 explains that a density value of 0.31 at x=55 means the "

for a 0.22-narrow strip, Pr( 54.89 < X < 55.11) ~= 0.31 * 0.22 = 6.2%.

for a 0.1-narrow strip, Pr( 54.95 < X < 55.05) ~= 0.31 * 0.1 = 3.1%.

(Note we used X not x because the rvar is X.)

**density**of probability**mass**" is 0.31 in a narrow region around x=55. For eg,for a 0.22-narrow strip, Pr( 54.89 < X < 55.11) ~= 0.31 * 0.22 = 6.2%.

for a 0.1-narrow strip, Pr( 54.95 < X < 55.05) ~= 0.31 * 0.1 = 3.1%.

(Note we used X not x because the rvar is X.)

So what's the density of Y around y=10. Well, y=10 maps to x=55, so we know there's a 3.1% of Y falling into some neighborhood around 10, but Y's density is not 3.1% but "3.1%/width of the neighborhood". If that neighborhood has width = 0.1 for X, but smaller when "projected" onto Y.

The same neighborhood represents an output range. It has a 3.1% total probability mass. 54.95 < X < 55.05, or 9.99 < Y < 10.01, since Y and X has one-to-one mapping.

We use dx/dy at Y=10 to work out the width in Y projected by X's width. For 54.95 < X < 55.05, we get 9.99 < Y < 10.01, so the Y width is 0.02.

Pr( 54.95 < X < 55.05) ~= Pr( 9.99 < Y < 10.01) ~= 3.1%

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