# Thread: Transforming pdf, distribution question

1. ## Transforming pdf, distribution question

A book I am reading has a section on probability density function.

Random var Xi is drawn from a pdf p(x)

Random var Yi is drawn from y(Xi), a one to one function with strictly positive or negative derivative.

Pr( Y <= y(x) ) = Pr( X <= x )

Which I am not understanding how it is derived. Notice It does not make much distinction of X, Xi, and Y, Yi.

Pr is the CDF. The goal of this section is to find the pdf of a transformed random var. The final formula is

p(y) = |dy/dx|^-1 * q(x)

Thanks

2. What you do is differentiate the two CDFs. That produces the densities.

For example if $X\sim U(0,1)$ and $Y=aX+b$

then $F_Y(y)= P(Y\le y)=P(aX+b\le y)=P(X\le (y-b)/a)=F_X((y-b)/a)$

so $f_y(y)= {d\over dy} F_Y(y)= {d\over dy}F_X((y-b)/a)= {1\over a} f_X((y-b)/a)={1\over a}$

where $0<{y-b\over a}<1$ in this case.

3. Thank you for providing the example.

Perhaps it is simple, but what I didn't get is how the claim
Pr( Y<=y(x) ) = Pr( X <=x )
is reasoned from the facts that y(x) is one to one and y(x)'s derivatives are all negative or all positive.

Also, though the absolute value around dy/dx enforces the transformed pdf being non-negative (because dy may be negative), why may the math not work out without it?

Originally Posted by matheagle
What you do is differentiate the two CDFs. That produces the densities.

For example if $X\sim U(0,1)$ and $Y=aX+b$

then $F_Y(y)= P(Y\le y)=P(aX+b\le y)=P(X\le (y-b)/a)=F_X((y-b)/a)$

so $f_y(y)= {d\over dy} F_Y(y)= {d\over dy}F_X((y-b)/a)= {1\over a} f_X((y-b)/a)={1\over a}$

where $0<{y-b\over a}<1$ in this case.