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Chapter 1: Problem 23

Let \(X\) have the pdf \(f(x)=4 x^{3}, 0

Short Answer

Expert verified
The CDF of \(X\) is \(F(x) = x^4\), for \(0 < x < 1\). The PDF of \(Y\) is \(f_Y(y) = e^{-y}\), for \(0<y<\infty\).

Step by step solution

01

Calculate the CDF of \(X\)

Start with the PDF, \(f(x)= 4x^{3}\), for \(0<x<1\) and find the CDF by integrating the PDF over the range of \(X\): \[F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 4t^{3} dt = [t^4]_{0}^{x} = x^4\]
02

Define the transformation

Here, we need to calculate the PDF of \(Y=- \ln(X^4) = -4 \ln(X)\). Using the formula for transformation of random variables, we know that the PDF of \(Y\) is \(f_Y(y) = f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|\) where \(g(.)\) is the function that relates \(X\) and \(Y\) (in this case \(g(x)=-4 \ln(x)\)) and \(g^{-1}(.)\) is its inverse.
03

Compute Inverse and its Derivative

Specifically, we first need to find the inverse of \(g(x) = y\), which is \(g^{-1}(y) = e^{-y/4}\). Then, find its derivative with respect to \(y\), \(\frac{d}{dy}g^{-1}(y) = -\frac{1}{4}e^{-y/4}\).
04

Substitute into the Formula

Now substitute \(g^{-1}(y)\) and its derivative into the formula, we get: \[f_Y(y) = f_X(e^{-y/4}) \times |\frac{d}{dy}e^{-y/4}| = 4(e^{-y/4})^3 \times \frac{1}{4}e^{-y/4} = e^{-y}, 0<y<\infty\]

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