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CDP of a function of a continuous RV

  1. Apr 17, 2010 #1
    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 find E[eX]

    The way my book does it is as follows:
    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!
     
    Last edited: Apr 18, 2010
  2. jcsd
  3. Apr 18, 2010 #2
    Any ideas? Sorry to bump, just redid the whole post, so it's essentially a new question :).
     
  4. Apr 18, 2010 #3
    A general rule if you're given [tex]f_{X}(x)[/tex] to calculate [tex]E[g(X)][/tex] is:


    [tex]
    E[g(X)] = \int{g(x) \, f_{X}(x) \, dx}
    [/tex]
     
  5. Apr 18, 2010 #4
    Yeah, I noted that I'm aware of that at the bottom of my post. :smile:

    I just wanna understand why the book does it this way.
     
  6. Apr 18, 2010 #5
    I think the book did it the same way as you, but used a different dummy variable in the last integral.
     
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