Question

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

Short answer

The pdf of \( Y \) is \( g(y) = e^{-y} \) for \( y > 0 \), and 0 for \( y \leq 0 \).

Step-by-step solution

Compute the Cumulative Distribution Function (CDF) of \( X \)

The CDF of \( X \) can be calculated by integrating the pdf of \( X \) from \( -\infty \) to \( x \). Here, the pdf \( f(x) \) is only defined for \( x > 0 \), so the CDF \( F(x) \) will be: \[ F(x) = \int_0^x f(t) dt = \int_0^x 2 t e^{-t^2} dt \] Using a subsitution of \( u = t^2 \), \( du = 2t dt \), we find: \[ F(x) = \int_0^{x^2} e^{-u} du = 1 - e^{-x^2} \] for \( x > 0 \), and 0 for \( x \leq 0 \).

Find the Cumulative Distribution Function (CDF) of \( Y \)

Now, using the transformation \( Y = X^2 \), we can find the CDF of \( Y \) by substituting \( Y = X^2 \) into \( F(x) \). This gives: \[ G(y) = P(Y \leq y) = P(X^2 \leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y}) \] Since the pdf is only defined for \( x > 0 \), this can be written as: \[ G(y) = P(0 \leq X \leq \sqrt{y}) = F(\sqrt{y}) - F(0) = 1 - e^{-y} - 0 = 1 - e^{-y} \] for \( y > 0 \), and 0 for \( y \leq 0 \).

Compute the Probability Density Function (pdf) of \( Y \)

Finally, the pdf of \( Y \) can be determined by differentiating the CDF of \( Y \). This gives: \[ g(y) = \frac{d}{dy} G(y) = e^{-y} \] for \( y > 0 \), and 0 for \( y \leq 0 \).