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Math Help - [SOLVED] question on expectations and cumulative distribution functions

  1. #1
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    [SOLVED] question on expectations and cumulative distribution functions

    I want to prove that E(X) = \int_{0}^{\infty} (1-F_X(X))\, dx is true with the only information given being that X is a continuous RV such that P(X > 0) = 1.

    The hint that I've received from my teacher is to replace  F_X(X) with the integral of the pdf, to get E(X) = \int_{0}^{\infty} (1- \int_{0}^{\infty} f_T(t)\, dt )\, dx (using dummy variable t) and then switch the order I do the integrals.

    Also, with  E(X) = \int_{-\infty}^{\infty} xf(x)\, dx (the definition of E(X)) I'm not sure how I would integrate this without knowing exactly what f(x) is.

    Thanks in advance for any help
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  2. #2
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    Ok, I realized that I should be working with
    E(X) = \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx

    but I'm still trying to figure out reversing the order of integration.
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  3. #3
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    Quote Originally Posted by joop View Post
    Ok, I realized that I should be working with
    E(X) = \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx

    but I'm still trying to figure out reversing the order of integration.
    I was going to point that out but had no time.

    \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx

    = \int_{0}^{\infty} \int_{x}^{+\infty} f_T(t)\, dt \, dx

    It should be smooth from here.
    Last edited by mr fantastic; October 27th 2008 at 02:35 AM. Reason: fixed the latex integral
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  4. #4
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    Would you please explain to me how you got
    \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx
    = \int_{0}^{\infty} \int_{x}^{\infty} f_T(t)\, dt \, dx

    To the best of my understanding,
    \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx
    = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{0}^{x} f_T(t)\,dt \,dx

    and reversing the order,
    = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{t}^{\infty} f_T(t)\,dx \,dt

    Thanks in advance.
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  5. #5
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    Quote Originally Posted by joop View Post
    Would you please explain to me how you got
    \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx
    = \int_{0}^{\infty} \int_{x}^{\infty} f_T(t)\, dt \, dx

    To the best of my understanding,
    \int_{0}^{\infty} (1- \int_{0}^{x} f_T(t)\, dt )\, dx
    = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{0}^{x} f_T(t)\,dt \,dx

    and reversing the order,
    = \int_{0}^{\infty} \, dx - \int_{0}^{\infty} \int_{t}^{\infty} f_T(t)\,dx \,dt Mr F says: I'd be interested to see how you do the first integral ......

    Thanks in advance.
    You should have seen that the second line follows from:

    1 = \int_{0}^{x} f_T(t)\, dt + \int_{x}^{+\infty} f_T(t)\, dt

    \Rightarrow 1 - \int_{0}^{x} f_T(t)\, dt = \int_{x}^{+\infty} f_T(t)\, dt.
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  6. #6
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    Thanks. Here is what I got (which I'm sure is correct):

    Starting from Mr. F's help, and reversing the order of integration:
    \int_{0}^{\infty} \int_{0}^{t} f(t) \, dt
    = \int_{0}^{\infty} t f(t) \, dt
    rename t to x
    <br />
\int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x f(x) \, dx<br />
    (Defn of E(X) = what I just got)

    Thanks again.
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  7. #7
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    Quote Originally Posted by joop View Post
    Thanks. Here is what I got (which I'm sure is correct):

    Starting from Mr. F's help, and reversing the order of integration:
    \int_{0}^{\infty} \int_{0}^{t} f(t) \, {\color{red}dx} \, dt
    = \int_{0}^{\infty} t f(t) \, dt
    rename t to x
    <br />
\int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x f(x) \, dx<br />
    (Defn of E(X) = what I just got)

    Thanks again.
    You're welcome. I've corrected (in red) a small omission.
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