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Chapter 1: Problem 8

Let \(f(x)=2 x, 0

Short Answer

Expert verified
The solution to the exercise is as follows: (a) \(E(1/X) = 2\). (b) The cdf of \(Y = 1/X\) is \(F_Y(y) = 1-1/y\) when \(0<y<1\), zero elsewhere. And the pdf of \(Y\) is \(f_Y(y) = 1/y^2\) when \(1<y<\infty\), zero elsewhere. (c) \(E(Y)= \infty\) and it is not equal to the visualization in part (a).

Step by step solution

01

Compute E(1/X)

The expectation of \(1/X\), denoted by \(E(1/X)\), is given as the integral from 0 to 1 of \(1/x\) times the pdf of \(X\). So, to calculate this expectation, we perform this integration: \[E(1/X) = \int_0^1 1/x \cdot f(x) dx = \int_0^1 1/x \cdot 2x \cdot dx = 2 \int_0^1 dx = 2 \cdot [x]_0^1 = 2.\]
02

Find the cdf and pdf of Y = 1/X

We first find the cdf \(F_Y(y)\) of \(Y\) by transforming the variable \(x\) to \(y\ =\ 1/x\). So when \(0<y<1\), the cdf \(F_Y(y)\) is given as 1-\(1/y\), and 0 elsewhere. Then, the pdf \(f_Y(y)\) of \(Y\) is obtained by differentiating the cdf \(F_Y(y)\) with respect to \(y\), which is \(f_Y(y)\ =\ 1/y^2\) when \(1<y<\infty\), zero elsewhere.
03

Compute E(Y) and compare the result with part (a)

The expectation of \(Y\), denoted by \(E(Y)\), is given as the integral from 1 to infinity of \(y\) times the pdf of \(Y\). So, to calculate this expectation, we perform this integration: \[E(Y) = \int_1^\infty y \cdot f_Y(y) dy = \int_1^\infty y/y^2 \cdot dy = \int_1^\infty \frac{1}{y} \cdot dy = \infty.\] If we compare this result with the part (a) result, \(2\neq\infty\).

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