Final answer:
To find the pdf for Y = X³, knowing that X has a pdf of fX(x) = 3x², we apply a variable transformation to obtain fY(y) = y for 0 ≤ y ≤ 1.
Step-by-step explanation:
The question is asking about finding the probability density function (pdf) for a variable Y, given that it is a function of another variable X with a known pdf. Since Y = X³ and X has the pdf fX(x) = 3x² for 0 ≤ x ≤ 1, we need to apply a transformation to find the pdf for Y. The method to find the pdf of Y involves changing variables in a pdf, which is a common technique in probability theory known as transformation of random variables.
First, to derive the pdf of Y, note that if Y = g(X), then the pdf of Y is given by:
fY(y) = fX(g⁻¹(y)) | d/dy (g⁻¹(y)) |, where g⁻¹ is the inverse function of g.
In this case, since Y = X³, the inverse function g⁻¹(y) is X = y^(1/3). Taking the derivative of the inverse function, we get d/dy (y^(1/3)) = (1/3)y^(-2/3).
Plugging into the formula, we have:
fY(y) = fX(y^(1/3)) | (1/3)y^(-2/3) | = 3(y^(1/3))² * (1/3)y^(-2/3) = y for 0 ≤ y ≤ 1, since 0 ≤ X ≤ 1 implies 0 ≤ X³ ≤ 1.
Hence, the pdf for Y is fY(y) = y for 0 ≤ y ≤ 1.