Question

If the pdf of \(X\) is \(f(x)=2 x e^{-x^{2}}, 0

Short answer

The pdf of \(Y\), denoted as \(f_{Y}(Y)\), is \(f_{Y}(Y) = e^{-Y}\) for \(Y > 0\), and is zero elsewhere.

Step-by-step solution

Define the Transformation

Define the transformation from \(X\) to \(Y\). Given in the problem that \(Y = X^{2}\).

Find the Jacobian

Find the Jacobian, which is the derivative of the transformation. The Jacobian, denoted by \(J\), for \(Y\) is the derivative of \(Y\) with respect to \(X\), i.e., \(J = dY/dX = 2X\).

Derive the pdf of \(Y\)

Next, substitute the value of \(X = √Y\) and \(J = 2√Y\) into the pdf of \(X\). This gives the equation of the pdf of \(Y\) as \(f_{Y}(Y) = f_{X}(X) |1/J| = 2(√Y)e^{-Y} |1/ 2√Y| = e^{-Y},\) for \(Y > 0\). The pdf is zero elsewhere.