# Thread: transformation of a random variable

1. ## transformation of a random variable

let Y = aX + b, a and b are constants

Fx(x) and fx(x) are the cdf and pdf of X

so fy(y) = 1/|a| fx(y-b/a), why is that?

in other words why is dFx(y-b/a) / dy = 1/|a| fx(y-b/a)?

2. Originally Posted by saoricapital
let Y = aX + b, a and b are constants

Fx(x) and fx(x) are the cdf and pdf of X

so fy(y) = 1/|a| fx(y-b/a), why is that?

in other words why is dFx(y-b/a) / dy = 1/|a| fx(y-b/a)?
Suppose that $\displaystyle a>0$:
Let $\displaystyle F_Y(y)$ be the cdf of $\displaystyle f_Y(y)$. Now $\displaystyle F_Y(y) = P(y\leq Y) = P(y\leq aX+b) = P(X \geq a^{-1}(y-b)) =$$\displaystyle 1 - P(X\leq a^{-1}(y-b)) = 1- F_X(a^{-1}(y-b))$
That means,
$\displaystyle f_Y(y) = F'_Y(y) = 1 - F_X'(a^{-1}(y-b)) = -\frac{1}{a}f_x(a^{-1}(y-b))$.
It seems your formula is wrong.