Results 1 to 4 of 4

Math Help - Finding a moment-generating function of a continuous r.v. given its pdf

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    27

    Finding a moment-generating function of a continuous r.v. given its pdf

    Hi,

    I'm trying to make sense of a problem we saw in my probability class. mr. fantastic helped me dust off my calculus skills for the integration part but now I'm having trouble understanding the very last part. Here's the problem and its solution:

    The probability density of a continuous random variable Y is given by
    f(y) = \begin{cases}\alpha{e^{-\alpha y}}&\text{ for 0} < y < \infty \text{ for some integer k},\\0&\text{otherwise},\end{cases}
    where \alpha >0

    Prove that the moment generating function, m(t), of Y is given by
    <br />
m(t) = \frac{\alpha}{\alpha - t} for t<\alpha

    Solution:

    m(t) = E[e^{ty}] = \int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy

     = \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy<br />


    <br />
= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)

    <br />
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}<br />


    Can someone please explain
    <br />
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}<br />
to me? I'm confused by the very end. If k is the one being taken to \infty then why should the outcome depend on t? Please explain how we get \frac{\alpha}{\alpha -t} and \infty.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Canadian0469 View Post
    Hi,

    I'm trying to make sense of a problem we saw in my probability class. mr. fantastic helped me dust off my calculus skills for the integration part but now I'm having trouble understanding the very last part. Here's the problem and its solution:

    The probability density of a continuous random variable Y is given by
    f(y) = \begin{cases}\alpha{e^{-\alpha y}}&\text{ for 0} < y < \infty \text{ for some integer k},\\0&\text{otherwise},\end{cases}
    where \alpha >0

    Prove that the moment generating function, m(t), of Y is given by
    <br />
m(t) = \frac{\alpha}{\alpha - t} for t<\alpha

    Solution:

    m(t) = E[e^{ty}] = \int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy

     = \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy<br />


    <br />
= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)

    <br />
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}<br />


    Can someone please explain
    <br />
= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}<br />
to me? I'm confused by the very end. If k is the one being taken to \infty then why should the outcome depend on t? Please explain how we get \frac{\alpha}{\alpha -t} and \infty.

    Thanks
    In my opinion, it should be [and it makes more sense this way] <br />
  \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < {\color{red}\alpha} ,\\\infty &\text{if t} \geq {\color{red}\alpha},\end{cases}<br />

    If t\geq\alpha, then t-\alpha\geq0 and the limit would diverge for t>\alpha, making M(t)=\infty and M(t)=0 if t=\alpha. [which really doesn't make sense].

    If t<\alpha, then t-\alpha<0 and the limit would converge to -1, causing M(t)=-\frac{\alpha}{t-\alpha}=\frac{\alpha}{\alpha-t}.

    Does this help?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    27
    If t\geq\alpha, then t-\alpha\geq0 and the limit would diverge for t>\alpha, making M(t)=\infty and M(t)=0 if t=\alpha. [which really doesn't make sense].
    When you say "which really doesn't make sense", what are you referring to? Are you suggesting that there ought to be a third case for t=\alpha?

    BTW, I should have noticed the correction you made myself... The question actually states "for t < \alpha"

    Thanks Chris!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Canadian0469 View Post
    When you say "which really doesn't make sense", what are you referring to?
    I was referring to the moment generating functions...

    It doesn't make sense to have M(t)=\infty or M(t)=0....

    --Chris
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding E[XY] in termes of the moment generating function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 24th 2011, 05:34 AM
  2. Finding Derivative, Moment Generating Function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 23rd 2011, 03:19 PM
  3. Finding probability function of moment generating function
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: July 4th 2011, 04:03 PM
  4. Replies: 5
    Last Post: August 7th 2009, 11:22 PM
  5. moment-generating function
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: May 3rd 2008, 07:34 PM

Search Tags


/mathhelpforum @mathhelpforum