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Prove Theorem 3.8.2. (Hint: Either apply Theorem3.8.4 or first compute the cdf. separately for a > 0 and a < 0.)

Short Answer

Expert verified

Pdf of y is \(\begin{array}{l}g\left( y \right) = \left\{ \begin{array}{l}\frac{1}{{\left| a \right|}}f\left( {\frac{{y - b}}{a}} \right)\,for - \infty < y < \infty \\0\,\,otherwise\end{array} \right.\\\end{array}\)

Step by step solution

01

Given information

X be the random variable with the pdf f such that Y=aX+b

02

calculating pdf

If Y=aX+b the inverse function is \(x = \frac{{\left( {y - b} \right)}}{a}\)

Also\(\frac{{dx}}{{dy}} = \frac{1}{a}\)

Therefore,

\(g\left( y \right) = f\left[ {\frac{1}{a}\left( {y - b} \right)} \right]\left[ {\frac{{dx}}{{dy}}} \right]\)

\(g\left( y \right) = \frac{1}{{\left| a \right|}}f\left( {\frac{{y - b}}{a}} \right)\)

Hence,Pdf of y is

\(\begin{array}{l}g\left( y \right) = \left\{ \begin{array}{l}\frac{1}{{\left| a \right|}}f\left( {\frac{{y - b}}{a}} \right)\,for - \infty < y < \infty \\0\,\,otherwise\end{array} \right.\\\end{array}\)

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