Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By HallsofIvy

Thread: Question mostly on sum transformation than on actual marginal dsitribution

  1. #1
    Senior Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    277
    Thanks
    21

    Question mostly on sum transformation than on actual marginal dsitribution

    Hi,

    So I have a problem given as example in class, but with a whole bunch of steps skipped (don't know if intentionally...):

    For two RVs, $X_1, X_2$, the joint pmf is given by:

    $p(m, n) = {{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m}$

    Now, marginal distrib. of $X_1$ is defined as:

    $p_{X_1}(m) = \sum_{\text{all } n} p(m,n) = \sum_{n = m}^{\infty}{{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m} = $

    $ p_1p_2^{m}(1-p_1)^{m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n-m}(1-p_2)^{n-m} = $
    .
    . steps missing which I can't figure out!!!
    .
    $ = p_1p_2^{m}(1-p_1)^{m} [ 1- (1-p_{1})(1-p_2)]^{-(m+1)}$

    I would really appreciate if you could help we understand the transforms,
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,029
    Thanks
    2765

    Re: Question mostly on sum transformation than on actual marginal dsitribution

    Quote Originally Posted by dokrbb View Post
    Hi,

    So I have a problem given as example in class, but with a whole bunch of steps skipped (don't know if intentionally...):

    For two RVs, $X_1, X_2$, the joint pmf is given by:

    $p(m, n) = {{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m}$

    Now, marginal distrib. of $X_1$ is defined as:

    $p_{X_1}(m) = \sum_{\text{all } n} p(m,n) = \sum_{n = m}^{\infty}{{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m} = $

    $ p_1p_2^{m}(1-p_1)^{m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n-m}(1-p_2)^{n-m} = $
    You know, I am sure, that $(1- p_1)^{n-m}= (1- p_1)^n(1- p_1)^{-m}$ and similarly for $(1- p_2)^{n-m}$. So the first thing I would do is factor $(1- p_1)^{-m}$ and $(1- p_2)^{-m}$ to give
    $ p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n}(1-p_2)^{n} $
    Now try using the "binomial theorem"- $(a- b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$

    steps missing which I can't figure out!!!
    .
    $ = p_1p_2^{m}(1-p_1)^{m} [ 1- (1-p_{1})(1-p_2)]^{-(m+1)}$

    I would really appreciate if you could help we understand the transforms,
    Thanks
    Thanks from dokrbb
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    277
    Thanks
    21

    Re: Question mostly on sum transformation than on actual marginal dsitribution

    OMG, Thank you, very much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member dokrbb's Avatar
    Joined
    Feb 2013
    From
    ...from out there ;)
    Posts
    277
    Thanks
    21

    Re: Question mostly on sum transformation than on actual marginal dsitribution

    Quote Originally Posted by HallsofIvy View Post
    You know, I am sure, that $(1- p_1)^{n-m}= (1- p_1)^n(1- p_1)^{-m}$ and similarly for $(1- p_2)^{n-m}$. So the first thing I would do is factor $(1- p_1)^{-m}$ and $(1- p_2)^{-m}$ to give
    $ p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n}(1-p_2)^{n} $
    Now try using the "binomial theorem"- $(a- b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$
    So, I thought I got it, but...

    $ p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}[(1-p_1)(1-p_2)]^{n}$

    now, by the binomial we have $(a + b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$

    so, how he is getting the $(1 - (term^n)^{-m-1})$...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Marginal distributions (and marginal density), three variables
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Feb 24th 2016, 01:55 PM
  2. Question about problem statement (marginal distribution)
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Dec 10th 2013, 06:27 PM
  3. Marginal Productivity Question?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Nov 15th 2011, 12:54 PM
  4. Marginal Profit Question
    Posted in the Business Math Forum
    Replies: 3
    Last Post: Jun 25th 2010, 11:31 AM
  5. Marginal revenue and the marginal cost
    Posted in the Business Math Forum
    Replies: 1
    Last Post: Nov 28th 2008, 05:22 PM

/mathhelpforum @mathhelpforum