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Chapter 1: Problem 13

Consider the cdf \(F(x)=1-e^{-x}-x e^{-x}, 0 \leq x<\infty\), zero elsewhere. Find the pdf, the mode, and the median (by numerical methods) of this distribution.

Short Answer

Expert verified
The PDF is \(f(x)=xe^{-x}\), the mode is 1, and the median needs to be approximated numerically.

Step by step solution

01

Find the PDF

We have the CDF as \(F(x)=1-e^{-x}-x e^{-x}, 0 \leq x<\infty \).The PDF \(f(x)\) is the derivative of the CDF. So, using the chain rule for differentiation, we get: \(f(x)= \frac{d}{dx}(1 - e^{-x}) + \frac{d}{dx}(-xe^{-x}) = e^{-x} + xe^{-x} - e^{-x} = xe^{-x}\).
02

Find the Mode

To find the mode, we take the derivative of the PDF and set it to zero: \(\frac{d}{dx}(xe^{-x}) = 0\). The derivative of \(xe^{-x}\) is \(-xe^{-x} + e^{-x}\). Setting this to zero gives us \(x=1\). Therefore, the mode of this distribution is 1.
03

Find the Median

To find the median, we need to solve for \(x\) in the equation \(F(x) = 0.5\), where \(F(x)\) is the given CDF. However, this equation is not analytically solvable, so numerical methods must be utilized. A possible approach is to employ the bisection method, or Newton's method to solve this equation.

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