Question

if random variable is uniformly distributed (-2,1) find pdf of y = 2x ^3

Answer 1

Since the random variable $X$ is uniformly distributed with the interval, (-2,1), its pdf is given as,

$f(x)={1\over b-a}={1\over 1--2}={1\over3}$  

Therefore, the probability density function for the random variable $X$ IS,

$f(x)={1\over3},\space -2\leq x\leq 1\\0,\space elsewhere$

To determine the pdf of $y=2x^3$, we shall apply the cumulative density function (CDF) method as described below.

 We determine $G(y),$ the  $cdf$ of $y=2x^3$,

Now,

$G(y)=p(Y\leq y)=p(2x^3\leq y)\implies p(x\leq({y\over2})^{1\over3})$

From definition of probability for continuous distributions,

 $p(x\leq({y\over2})^{1\over3})=\displaystyle\int^{({y\over2})^{1\over3}}_{-2}f(x)dx$

So,

$G(y)=p(x\leq({y\over2})^{1\over3})=\displaystyle\int^{({y\over2})^{1\over3}}_{-2}{1\over3}dx={x\over3}|^ {({y\over2})^{1\over3}} _{-2}= {{({y\over2})^{1\over3}}\over3}+{2\over3}$

To find the $pdf$ of the random variable $y$, we differentiate $G(y)$ with respect to $y$. That is,

$g(y)={dG(y)\over dy}={{1\over3}\times{1\over2}\times({y\over2})^{-{2\over3}}\over3}={1\over18}({y\over2})^{-{2\over3}}$

The limits are,

Lower limit,

$y=2\times(-2)^3=-16$

Upper limit,

$y=2\times(1)^3=2$

Therefore, the probability density function of the random variable $y$  is,

 $g(y)={1\over18}({y\over2})^{-{2\over3}},\space -16\leq y\leq 2\\0,\space elsewhere$