Question

Suppose that the p.d.f. of a random variable X is as follows:

\(f\left( x \right) = \left\{ \begin{array}{l}3{x^2}\,\,for\,0 < x < 1\\0\,\,otherwise\end{array} \right.\)

Also, suppose that\(Y = 1 - {X^2}\) . Determine the p.d.f. of Y

Short answer

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

Step-by-step solution

Given information

X be the random variable with pdf

\(f\left( x \right) = \left\{ \begin{array}{l}3{x^2}\,\,{\rm{for}}\,0 < x < 1\\0\,\,{\rm{otherwise}}\end{array} \right.\)

calculating pdf

Inverse transformation of x:

\(x = {\left( {1 - y} \right)^{\frac{1}{2}}}\)

Derivatives of x:

\( - {\left( {1 - y} \right)^{\frac{{\frac{{ - 1}}{2}}}{2}}}\)

The pdf of Y:

\(g\left( y \right) = f\left( {{{\left[ {1 - y} \right]}^{\frac{1}{2}}}} \right){\left( {1 - y} \right)^{ - \frac{1}{2}}}\)

\(g\left( y \right) = \frac{3}{2}{\left( {1 - y} \right)^{\frac{1}{2}}}\) for 0<y<1

Hence the pdf of Y is ,

\(g\left( y \right) = \left\{ \begin{array}{l}\frac{3}{2}{\left( {1 - y} \right)^{\frac{1}{2}}}\,\,\,for\,0 < y < 1\\0\,\,\,\,otherwise\end{array} \right.\)