Results 1 to 9 of 9

Math Help - [SOLVED] Density function

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    23

    [SOLVED] Density function

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

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

    I start out like this:

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

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

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

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

    but I don't see how (1/\alpha)m \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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by approx View Post
    I've got this problem:
    f(y) = (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}, y>0
    0, elsewhere

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

    I start out like this:

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

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

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

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

    but I don't see how (1/\alpha)m \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.
    Read post #5 of this thread: http://www.mathhelpforum.com/math-he...-function.html
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    23
    mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    23
    Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:

    u = y^m/\alpha ?

    which gives (1/\alpha)m \int_0^\infty y^{k+m-1}e^{{-u}} du ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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:

    u = y^m/\alpha ?

    which gives (1/\alpha)m \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. dy \neq du !

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

    It's expected that at this level you can correctly substitute a change of variable in an integral.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2008
    Posts
    23
    Ok. So I've done some thinking and came up with this substitution:

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

    And then I go on:

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

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

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

    =

     (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:

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

    Am I right so far?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by approx View Post
    Ok. So I've done some thinking and came up with this substitution:

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

    And then I go on:

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

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

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

    =

     (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:

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

    Am I right so far?
    Yes. Now substitute w = \frac{u}{\alpha} \Rightarrow u = \alpha w.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2008
    Posts
    23
    Thank you! I got the right answer after the last substitution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a Cumulative Distribution Function and Density Function
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: November 17th 2011, 10:41 AM
  2. Replies: 4
    Last Post: October 27th 2010, 06:41 AM
  3. [SOLVED] joint probability density function
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 24th 2010, 03:36 AM
  4. Replies: 5
    Last Post: December 6th 2009, 12:30 AM
  5. [SOLVED] Probability density function problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 4th 2009, 09:54 AM

Search Tags


/mathhelpforum @mathhelpforum