/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Let \(X\) have the uniform distr... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 16

Let \(X\) have the uniform distribution with pdf \(f(x)=1,0

Short Answer

Expert verified
The Cumulative Distribution Function (CDF) of \(Y\) is: \(F_Y(y) = 1 - e^{-y/2}\), for \(0 < y < \infty\), and the Probability Density Function (PDF) of \(Y\) is: \(f_Y(y) = e^{-y/2}/2\), for \(0 < y < \infty\).

Step by step solution

01

Specify the range of \(Y\)

First, we need to deal with the range of the transformed variable \(Y\). Since we are given \(0<X<1\), we have \(-\infty < \log X < 0\), and consequently \(0 < -2\log X = Y < \infty\). In the first inequality, the order has been reversed as we are taking the log of a quantity between 0 and 1 and in the second inequality, we multiply by -2 to get the range of \(Y\).
02

Calculate the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of \(Y\), is defined as: \(F_Y(y) = P(Y \leq y)\). Insert \(Y = -2 \log X\) we get: \(F_Y(y) = P(-2 \log X \leq y)\). As \(Y\) is increasing, we can transform this to: \(F_Y(y) = P(X \geq e^{-y/2})\) for \(0 < y < \infty\). As \(X\) has a uniform distribution between 0 and 1 the cumulative distribution becomes: \(F_Y(y) = 1 - e^{-y/2}\), \(0 < y < \infty\) .
03

Calculate the Probability Density Function (PDF)

The probability density function (PDF) of \(Y\) would be the derivative of its cumulative distribution function (CDF). So we calculate the derivative of \(F_Y(y)\) that we found in step 2: \(f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}(1 - e^{-y/2}) = e^{-y/2}/2\), defined for \(0 < y < \infty\) . The PDF of \(Y\) is a one sided exponential distribution with rate \(2\).

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