# PDF Transformation: Finding the PDF of Y = X^3

What is the process to find the probability density function (PDF) of Y = X^3 given that the continuous random variable X has the PDF f(x) = e^(-x) for x >= 0?

The process to find the PDF of Y = X^3 involves calculating the cumulative distribution function (CDF) of Y and then differentiating it to obtain the PDF. Let's break down the steps: First, we need to find the 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 first 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 differentiate the CDF of Y with respect to y to find the PDF of 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.

## Understanding the PDF Transformation Process:

**1. Finding the CDF of Y:**The CDF of Y, denoted as F_Y(y), represents the probability that Y is less than or equal to a specific value y. Since Y = X^3, we need to find the probability that X is less than or equal to y^(1/3) in order to determine the CDF of Y. This involves substituting X <= y^(1/3) into the CDF of X and simplifying to obtain the CDF of Y.

**2. Deriving the PDF of Y:**Once we have the CDF of Y, we differentiate it with respect to y to obtain the PDF of Y, denoted as f_Y(y). This step involves applying the chain rule and simplifying the expression to get the final PDF of the transformed random variable Y = X^3.

**3. Interpreting the Results:**The PDF of the transformed random variable Y provides insights into the distribution of Y based on the transformation applied to the original random variable X. In this case, the PDF of Y = X^3 showcases the relationship between the original PDF f(x) = e^(-x) and the transformed PDF f_Y(y) = (1/3)y^(-2/3)e^(-y^(1/3)). By understanding the process of transforming the PDF of a random variable, we can analyze the characteristics of the resulting distribution and make informed decisions in statistical modeling and analysis.