Transformation of Probability Distributions

Given X ~ f(x). What distribution is g(X)?

Solution:

Think in terms of F(x), the CDF, instead of f(x).
Let F_g be the CDF of g(X). Then:


F_g(x) = Prob ( g(X) ≤ x ) = Prob ( X ≤ g^-1(x) )


So the RHS is just: F( g^-1(x) ). Now to get PDF, it's just differentiation of this value.

Examples:

Find the CDF of X², given X ~ f(x), with CDF F(x).

F__x²(x) = Prob (X² ≤ x) = Prob ( -√|x| ≤ X ≤ √|x|) = F(√x) - F(-√x) , noting that domain for X² is [0, ∞).

Taking derivatives, we have:


½ * 1/√x * ( F´(√x) - F´(-√x) ) = ½ * 1/√x * ( f(√x ) - f(-√x ))


Find the CDF of e^X, given X ~ f(x), with CDF F(x).

F__e^X(x) = Prob (e^X ≤ x) = Prob ( X ≤ ln (x) ) = F(ln (x))


Taking derivative of RHS to get PDF:

d/dx (F (ln(x)) = 1/x d/d(ln(x)) F (ln (x)) = 1/x f(ln (x)) (since d/dx F(x) = f(x))