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Chapter 1: Problem 23

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

Short Answer

Expert verified
The Cumulative Distribution Function (CDF) of \(Y\) is \(F_Y(y) = e^{-y}\) for \(y > 0\), zero elsewhere. The Probability Density Function (PDF) of \(Y\) is \(f_Y(y) = -e^{-y}\) for \(y > 0\), zero elsewhere.

Step by step solution

01

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

The cumulative distribution function \(F_X(x)\) of \(X\) is the integral of its PDF from -infinity to \(x\), and considering that \(f(x) = 0\) when \(x < 0\) and \(x > 1\), we integrate from \(0\) to \(x\): \[ F_X(x) = \int_0^x 4u^{3} du = u^{4} \bigg|_0^x = x^{4}, for \(0 < x < 1\)\]
02

Change of Variables

We have \(Y = -\ln X^{4}\) or equivalently \(Y = -4 \ln X\). Solving this for \(X\), we get \(X = e^{(-Y/4)}\). Now we obtain the CDF \(F_Y(y)\) of \(Y\) by substituting \(X\) in \(F_X(x)\). The resulting function is monotonically increasing, \(F_Y(y) = (e^{(-y/4)})^{4}\), so it can directly represent the cumulative distribution of \(Y\). Thus, \(F_Y(y) = e^{-y}, for \(y > 0\) and zero elsewhere.
03

Calculation of the Probability Density Function (PDF) of \(Y\)

The PDF \(f_Y(y)\) of \(Y\) is the derivative of its CDF with respect to \(y\). So, \(f_Y(y) = -e^{-y}\), for \(y > 0\) and zero elsewhere.

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