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Thread: [SOLVED] Density function

  1. #1
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    [SOLVED] Density function

    I've got this problem:
    f(y) = $\displaystyle (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0
    0, elsewhere

    Now I'm supposed to find $\displaystyle E(Y^k)$ for any positive integer k.

    I start out like this:

    $\displaystyle E(Y^k)$ = $\displaystyle \int_0^\infty y^{k}f(y) dy$ = $\displaystyle \int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$

    I then move out $\displaystyle (1/\alpha)$ and m outside the integral =

    $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$

    and from here I don't know how to go on. The correct answer is $\displaystyle \Gamma (k/m+1)\alpha ^{k/m}$

    and I know that $\displaystyle \Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$

    but I don't see how $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.
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  2. #2
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    Quote Originally Posted by approx View Post
    I've got this problem:
    f(y) = $\displaystyle (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0
    0, elsewhere

    Now I'm supposed to find $\displaystyle E(Y^k)$ for any positive integer k.

    I start out like this:

    $\displaystyle E(Y^k)$ = $\displaystyle \int_0^\infty y^{k}f(y) dy$ = $\displaystyle \int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$

    I then move out $\displaystyle (1/\alpha)$ and m outside the integral =

    $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$

    and from here I don't know how to go on. The correct answer is $\displaystyle \Gamma (k/m+1)\alpha ^{k/m}$

    and I know that $\displaystyle \Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$

    but I don't see how $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.
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  3. #3
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    mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
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  4. #4
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    Quote Originally Posted by approx View Post
    mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
    The first three lines of the reference I've given you are crystal clear I would have thought. What exactly don't you understand?
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  5. #5
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    Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:

    $\displaystyle u = y^m/\alpha$ ?

    which gives $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-u}} du$ ?
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  6. #6
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    Quote Originally Posted by approx View Post
    Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:

    $\displaystyle u = y^m/\alpha$ ?

    which gives $\displaystyle (1/\alpha)m$$\displaystyle \int_0^\infty y^{k+m-1}e^{{-u}} du$ ?
    Your substitution is incorrect for a number of reasons:

    1. You still have y in the integral, there should only be u's.
    2. You have not substituted the correct exprssion for dy. $\displaystyle dy \neq du$ !

    Note that $\displaystyle u = \frac{y^m}{\alpha} \Rightarrow dy = \frac{\alpha}{m \, y^{m-1}}\, du$ and $\displaystyle y = (\alpha \, u)^{1/m}$.

    It's expected that at this level you can correctly substitute a change of variable in an integral.
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  7. #7
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    Ok. So I've done some thinking and came up with this substitution:

    $\displaystyle u = y^m$, which gives $\displaystyle y = u^{1/m}$ and $\displaystyle dy = (1/m) u^{{(1/m)}-1} du$. Right?

    And then I go on:

    $\displaystyle (m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

    I move $\displaystyle 1/m$ outside which gives $\displaystyle (1/m)(m/\alpha) = (1/\alpha)$ outside the integral.

    $\displaystyle (1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

    =

    $\displaystyle (1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$

    I put together the u:s and receives:

    $\displaystyle (1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$

    Am I right so far?
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  8. #8
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    Quote Originally Posted by approx View Post
    Ok. So I've done some thinking and came up with this substitution:

    $\displaystyle u = y^m$, which gives $\displaystyle y = u^{1/m}$ and $\displaystyle dy = (1/m) u^{{(1/m)}-1} du$. Right?

    And then I go on:

    $\displaystyle (m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

    I move $\displaystyle 1/m$ outside which gives $\displaystyle (1/m)(m/\alpha) = (1/\alpha)$ outside the integral.

    $\displaystyle (1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$

    =

    $\displaystyle (1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$

    I put together the u:s and receives:

    $\displaystyle (1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$

    Am I right so far?
    Yes. Now substitute $\displaystyle w = \frac{u}{\alpha} \Rightarrow u = \alpha w$.
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  9. #9
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    Thank you! I got the right answer after the last substitution.
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