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Math Help - desperate...moment generating function

  1. #1
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    desperate...moment generating function

    f(x)=3/64x^2(4-x)

    what is the moment generating function?
    how do i find the mean and variance?

    any help would really be appreciated since i dont understand anything concerning this

    thankx
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  2. #2
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by sasasa View Post
    f(x)=3/64x^2(4-x)

    what is the moment generating function?
    how do i find the mean and variance?

    any help would really be appreciated since i dont understand anything concerning this

    thankx
    Do you mean f(x)=\frac{3}{64}x^{2(4-x)} ?
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  3. #3
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    Var(x) = E(x^2) - E(x)^2

    E(x) = M'(0)

    E(x^2) = M''(0)

    MGF of f(X) = E(e^{f(x)}) so you'll have to integrate e^{\frac{3}{64}x^{2(4-x)}} over a certain region.
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  4. #4
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    Quote Originally Posted by statmajor View Post
    MGF of f(X) = E(e^{f(x)}) so you'll have to integrate e^{\frac{3}{64}x^{2(4-x)}} over a certain region.
    This is not right. The moment generating function is defined as

    M_X (t) = \mathbb{E}e^{Xt} = \int_{\mathbb{R}} e^{xt}f(x) dx

    (for continuous distributions). I would help with the computation but I can't figure out what the density is supposed to be from the OP.
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  5. #5
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    Quote Originally Posted by sasasa View Post
    f(x)=3/64x^2(4-x)

    what is the moment generating function?
    how do i find the mean and variance?

    any help would really be appreciated since i dont understand anything concerning this

    thankx
    I assume you mean f(x) = \frac{3}{64} x^2 (4 - x). This pdf is incomplete as you do not include the interval over which this expression is defined (the support).

    You should know that you need to calculate E\left(e^{tX}\right) = \int e^{tx} \frac{3}{64} x^2 (4 - x) \, dx (and I have not included the integral terminals since you did not completely define the pdf). So please show all your working and clearly state where you are stuck.
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  6. #6
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    Darn, can't believe I wrote that. Thanks for correcting me.

    But the other parts of my post should be right.

    Thanks again.
    Last edited by mr fantastic; November 4th 2009 at 04:29 PM. Reason: m --> r
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  7. #7
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    Sorry (so many functions)… so overwhelmed that i even posted the wrong function. The right one is:
    f(x)=0.15e^(-0.15(x-0.5)), when x is bigger or equal to 0.5
    Thanks again for all your precious help
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  8. #8
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    Quote Originally Posted by sasasa View Post
    Sorry (so many functions)… so overwhelmed that i even posted the wrong function. The right one is:
    f(x)=0.15e^(-0.15(x-0.5)), when x is bigger or equal to 0.5
    Thanks again for all your precious help
    Can you set up the required integral using the definition of the moment generating function? Can you do the integration? Please show all that you can do and say where you're stuck.

    (And please use normal size font so that we don't have to use microscopes to read it).
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  9. #9
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    this is what i got:
    Mx(t)=(-0.15e^(0.5t))/(t-0.15) but i am not sure if it is right

    Can you please tell me if this is right:

    Mean=E(X)=Mx´(0)=7.166

    thanks
    Last edited by sasasa; November 5th 2009 at 03:07 PM.
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  10. #10
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    Quote Originally Posted by sasasa View Post
    this is what i got:
    Mx(t)=(-0.15e^(0.5t))/(t-0.15) but i am not sure if it is right

    Can you please tell me if this is right:

    Mean=E(X)=Mx´(0)=7.166

    thanks
    Your mgf is OK: - Wolfram|Alpha

    I assume at this level you can correctly differentiate and substitute t = 0. You can check your own answer using the definition of E(X): Calculate 0.15 \int_{1/2}^{+\infty} x e^{-0.15(x - 0.5)} \, dx.
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  11. #11
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    i think the mean is right...

    did you get 0,25 as the variance?

    thanks for all your help
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  12. #12
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    Quote Originally Posted by sasasa View Post
    i think the mean is right...

    did you get 0,25 as the variance?

    thanks for all your help
    I don't plan to do the calculation since you know how to answer the question. If you have found the correct values of E(X) and E(X^2) then your answer for Var(X) = E(X^2) - (E(X))^2 will probably be correct.
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  13. #13
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    The random variable looks like a member of the location family of an exponential, i.e. if Z ~ Exponential(1/.15), X = Z + .5. That should simplify checking your answer.
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