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Chapter 3: Q6E (page 174)

Suppose that the p.d.f. of X is as given in Exercise 3.

Determine the p.d.f. of \(Y = 3X + 2\)

Short Answer

Expert verified

Pdf of \(Y = 3X + 2\)is \(g\left( y \right) = \left\{ \begin{array}{l}\frac{1}{{18}}\left( {y - 2} \right)\,for\,2 < y < 8\\0\,\,\,otherwise\end{array} \right.\)

Step by step solution

01

Given information

X be the random variable with pdf \(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}x\,\,for\,0 < x < 2\\0\,otherwise\end{array} \right.\)

02

Calculating pdf

X is lies between 0 and 2 if and only if Y is lies between2 and 8

Therefore, for 2<y<8

\(g\left( y \right) = \frac{1}{3}f\left( {\frac{{y - 2}}{3}} \right)\)

\(g\left( y \right) = \frac{1}{3} \times \frac{1}{2} \times \frac{{y - 2}}{{18}}\)

\(g\left( y \right) = \frac{1}{{18}}\left( {y - 2} \right)\)

Hence pdf of\(Y = 3X + 2\)is:

\(g\left( y \right) = \left\{ \begin{array}{l}\frac{1}{{18}}\left( {y - 2} \right)\,for\,2 < y < 8\\0\,\,\,otherwise\end{array} \right.\)

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