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Math Help - k-th moment of X

  1. #1
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    k-th moment of X

    With the probability density function:

    f(x)=\left\{\begin{array}{cc}\lambda e^{-\lambda x},&\mbox{ if } x>0;\lambda>0\\0, & \mbox otherwise \end{array}\right.

    i am asked to find the k-th moment of X, i.e. E(X^k).

    How do i do this?
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  2. #2
    MHF Contributor matheagle's Avatar
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    You can do this by parts, but there's a nice trick to doing this in general for all gamma densities.

    Let  X\sim \Gamma(\alpha, \beta)

    Then  f(x)={1\over \Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}I(x>0)

    And E(X^k)={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}x^{k+\  alpha-1}e^{-x/\beta}dx

    = {\Gamma(k+\alpha)\beta^{k+\alpha}\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}{1\ove  r \Gamma(k+\alpha)\beta^{k+\alpha}}x^{k+\alpha-1}e^{-x/\beta}dx

    = {\Gamma(k+\alpha)\beta^k\over \Gamma(\alpha)}

    Since that is a valid density in the integrand.
    NOW you should know how the gamma function reduces (via parts)
    and note that this is true even if k is not an integer.

    Finally, let \alpha=1 and \beta=1/\lambda and you have your answer.
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  3. #3
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    As a follow-on it asks for the coefficient of skewness of X but doesn't give any \alpha , \beta or k.

    Is it possible to find an actual coefficient for this or does it just come out as a general solution?
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  4. #4
    Moo
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    I don't understand your question. In the end, you won't have anymore \alpha,\beta

    baldeagle calculated the k-th moment of a gamma distribution with parameters \alpha and \beta

    Then just notice that if you substitute \alpha=1, \beta=\frac 1\lambda in the pdf of a gamma distribution, you'll get the pdf you have.

    ----------------------------------------
    Another method... :

    Let's calculate the moment generating function of your distribution.

    \begin{aligned}M_X(t)&=\mathbb{E}(e^{tX}) \\<br />
&=\lambda \int_0^\infty e^{tx}e^{-\lambda x} ~dx \\<br />
&=\frac{\lambda}{\lambda-t} \end{aligned}
    (defined for t<\lambda)

    Then we know that the k-th moment of X will be the k-th derivative of its mgf, taken at t=0.

    Find the Taylor series of \frac{\lambda}{\lambda-t} when t is near 0 and its coefficients will be the derivatives.

    \frac{\lambda}{\lambda-t}=\frac{1}{1-\frac t\lambda} which is like a geometric series... :

    M_X(t)=\sum_{k=0}^\infty \left(\frac t\lambda\right)^k=\sum_{k=0}^\infty t^k \cdot \frac{k!}{\lambda^k}\cdot \frac{1}{k!}

    But the Taylor series of M_X(t) is \sum_{k=0}^\infty t^k \cdot \frac{M^{(k)}(0)}{k!}
    where M^{(k)}(0) denotes the k-th derivative of M taken at t=0.

    By identification, it follows that \mathbb{E}(X^k)=M^{(k)}(0)=\frac{k!}{\lambda^k}
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  5. #5
    MHF Contributor matheagle's Avatar
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    But, be aware I calculated any moment as long as k>-\alpha
    That include noninteger and negative moments.
    Using the MGF you are only getting the positive integer moments.
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  6. #6
    Moo
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    And since \alpha=1, it's like calculating for k\geq 0... So it's not about negative moments.
    And how can one define a negative moment ? oO
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Moo View Post
    And since \alpha=1, it's like calculating for k\geq 0... So it's not about negative moments.
    And how can one define a negative moment ? oO

    No, for k>-1, in that case. Of course one can find  E(X^{-1/2})
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  8. #8
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    Ah, sorry, the first time i saw the page the bottom part about \alpha=1, \beta=\frac{1}{\lambda} didn't appear.
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