Finding the pdf of a Transformation

Theorem

Let

• $X$ : Continuous random variable, with
  • pdf $f_X(x)$
  • Support ${\mathcal{S}}_X$
• $g:{\mathcal{S}}_X\to \mathbb{R}$, one-to-one and differentiable
• $Y=g(X)$, with
  • Support ${\mathcal{S}}_Y=\{y=g(x):s\in S_X\}$
• $x=g^{-1}(y)$ : Inverse of $g$
• $\frac{d}{dy}x=\frac{d}{dy}{g}^{-1}(y)$

Then pdf of $Y$ is given by $f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}x\right|,\forall y\in {\mathcal{S}}_Y$

Remark

This theorem can be used in the following simple algorithm:

Assuming $Y=g(X)$ is one-to-one:

1. Find ${\mathcal{S}}_Y$
2. Solve the inverse of the tranformation; i.e., solve for $x$ in terms of $y$ in $y=g(x)$, thereby obtaining $x=g^{-1}(y)$
3. Obtain $\frac{d}{dy}x$
4. The pdf of $Y$ is $f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}x\right|$

See Example 1.7.6 for example