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Math Help - Mean and Variance

  1. #1
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    Mean and Variance

    Given the pdf and cdf of a continuous random variable, say X, describe how you would find the mean and variance of X without using integration.
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  2. #2
    Super Member Anonymous1's Avatar
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    Well you could use the MGF but I guess that is defined by integration...
    Last edited by Anonymous1; March 31st 2010 at 02:24 PM.
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  3. #3
    Super Member Anonymous1's Avatar
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    Oh wait! I remember how to do this:

    Here is an example:
    X~ exp(\lambda)

    Differentiating the density identity (ofcourse I'm assuming we can swap the order of differentiation and integration).

    \frac{d}{d\lambda} \int_0^{\infty}\lambda e^{-\lambda x}dx = 1

    \Rightarrow \int_0^{\infty} (-\lambda x +1) e^{-\lambda x}dx = 0

    Which simplifies to

    \int_0^{\infty} \lambda x e^{-\lambda x} = \int_0^{\infty} xe^{-\lambda x} = \frac{1}{\lambda} = E[X]

    Where the last equality holds by an identity
    Last edited by Anonymous1; March 31st 2010 at 11:37 AM.
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  4. #4
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    Thank you for the response, but I'm not quite following. Can you use the following as an example and find the mean using the method you described using no integration:

    pdf: fx(X) = ((x-5)^3)/4 for 5 <= x <= 7 and 0 elsewhere
    cdf: Fx(X) = ((x-5)^4)/16 FOR 5 <= x <= 7

    Thanks
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  5. #5
    Super Member Anonymous1's Avatar
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    Start like this

    \frac{d}{dx}\int_5^7 \frac{x^3 - 15x^2 + 75x - 125}{4} dx = \frac{d}{dx}1

    \Rightarrow \int_5^7 \frac{3x^2-30x+75}{4} = 0

    Um... gimme a sec.
    Last edited by Anonymous1; March 31st 2010 at 12:43 PM.
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