If the continuous random variable X has the probability density function (PDF) f(x) = e^(-x) for x >= 0, we can find the PDF of Y = X^3 using the following method:
First, we need to find the cumulative distribution function (CDF) of Y. The CDF of Y is given by F_Y(y) = P(Y <= y) = P(X^3 <= y) = P(X <= y^(1/3)). Since X has the PDF f(x) = e^(-x) for x >= 0, we can find the CDF of X as F_X(x) = integral from 0 to x of f(t) dt = integral from 0 to x of e^(-t) dt = -e^(-t) from 0 to x = 1 - e^(-x) for x >= 0. Therefore, the CDF of Y is given by F_Y(y) = F_X(y^(1/3)) = 1 - e^(-y^(1/3)) for y >= 0.
Next, we can find the PDF of Y by differentiating its CDF with respect to y. The PDF of Y is given by f_Y(y) = dF_Y(y)/dy = d(1 - e^(-y^(1/3)))/dy = (1/3)y^(-2/3)e^(-y^(1/3)) for y >= 0.
Therefore, if the continuous random variable X has the PDF f(x) = e^(-x) for x >= 0, then the continuous random variable Y = X^3 has the PDF f_Y(y) = (1/3)y^(-2/3)e^(-y^(1/3)) for y >= 0.