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Example. Direct PDF Method | Example. Direct PDF Method | ||
| − | + | *<math>f_{\mathbf{X}}\left(x\right)</math> is given and <math>g\left(x\right)=ax+b</math> where <math>a\neq0</math> . | |
| − | + | *<math>y=ax+b\Longrightarrow x=\frac{y-b}{a}\Longrightarrow\mathbf{X}\left(y\right)=\frac{y-b}{a}</math> | |
| − | + | *<math>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)</math> | |
| + | ---- | ||
| + | [[ECE600|Back to ECE600]] | ||
| + | |||
| + | [[ECE 600 Prerequisites|Back to ECE 600 Prerequisites]] | ||
Revision as of 10:14, 17 November 2010
1.9 Direct PDF Method
From the ECE600 Pre-requisites notes of Sangchun Han, ECE PhD student.
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) $
