(adsbygoogle = window.adsbygoogle || []).push({}); REVISED: Expectation of a function of a continuous RV

Given:

[tex]f_{X}(x)=1[/tex]

[tex]0 \leq x \leq 1[/tex]

and 0 everywhere else.

We are asked to findE[e^{X}]

The way my book does it is as follows:

page 191 in A First Course in Probability 8th ed by Sheldon Ross said:[tex]Y = e^{X}[/tex]

[tex]F_{Y}(x) = P(X\leq Ln (x)) = \int_{0}^{ln x} f_{Y}(y) dy = ln (x)[/tex]

[tex]f_{Y}(x) = \frac{d}{dx} \left[ln (x)\right]=\frac{1}{x}[/tex]

[tex]1 \leq x \leq e[/tex]

[tex] E[e^{X}] = \int_{-\infty}^{\infty} x f_{Y}(x) dx = \int_{1}^{e} dx = e - 1[/tex]I understand how to do it as follows. I don't understand the author's way of doing it.

[tex]Y = e^{X}[/tex]

[tex]0 \leq x \leq 1[/tex]

[tex]1 \leq y \leq e[/tex]

[tex]F_{Y}(y) = P(X\leq Ln (y)) = \int_{0}^{ln y} f_{X}(x) dx = \int_{0}^{ln y} dx = ln (y)[/tex]

[tex]f_{Y}(y) = \frac{d}{dy} \left[ln (y)\right]=\frac{1}{y}[/tex]

[tex] E[e^{X}] = \int_{-\infty}^{\infty} y f_{Y}(y) dy = \int_{1}^{e} dy = e - 1[/tex]

Can someone please explain to me why the author does it his way instead of mine?

I know I can make life easier by just taking:

[tex]\int_{-\infty}^{\infty}g(x)f(x) dx [/tex]

but I want to understand what's going on here.

Any help is much appreciated!

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# CDP of a function of a continuous RV

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