1.9 Direct PDF Method

If $ y=g\left(x\right)\Longrightarrow x=g^{-1}\left(y\right) $ uniquely, then

$ f_{\mathbf{Y}}\left(y\right)=f_{\mathbf{X}}\left(g^{-1}\left(y\right)\right)\left|\frac{dg^{-1}\left(y\right)}{dy}\right|=f_{\mathbf{X}}\left(\mathbf{X}\left(y\right)\right)\left|\frac{d\mathbf{X}\left(y\right)}{dy}\right|\text{ where }\mathbf{X}\left(y\right)=g^{-1}\left(y\right). $

Example. Direct PDF Method

• $ f_{\mathbf{X}}\left(x\right) $ is given and $ g\left(x\right)=ax+b $ where $ a\neq0 $ .

• $ y=ax+b\Longrightarrow x=\frac{y-b}{a}\Longrightarrow\mathbf{X}\left(y\right)=\frac{y-b}{a} $

• $ f_{\mathbf{Y}}\left(y\right)=f_{\mathbf{X}}\left(\mathbf{X}\left(y\right)\right)\left|\frac{d\mathbf{X}\left(y\right)}{dy}\right|=f_{\mathbf{X}}\left(\frac{y-b}{a}\right)\left|\frac{1}{a}\right|=\left|\frac{1}{a}\right|\cdot f_{\mathbf{X}}\left(\frac{y-b}{a}\right) $