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Negative binomial transformation and mgf

  1. Jul 27, 2014 #1
    Imo, this problem is crazy hard.

    1. The problem statement, all variables and given/known data

    Let X have the negative binomial distribution with pmf:

    [itex] f_X(x) = \binom{r+x-1}{x}p^{r}(1-p)^{x}[/itex], x=0.1.2...,

    where [itex]0<p<1[/itex] and r is a positive integer.
    (a) Calculate the mgf (moment generating function) of X.

    (b) Define a new random variable by Y=2pX. Show that as [itex]p \downarrow 0[/itex], the mgf of Y converges to that of a chi squared random variable with 2r degrees of freedom by showing that

    [itex] lim_{p \to 0} M_Y(t) = (\frac 1{1-2t})^{r}, |t|< \frac 12[/itex].


    2. Relevant equations

    the moment generating function for a discrete random variable, denoted as [itex]M_X(t)[/itex] is equal to

    [itex] \sum_x e^{tx}f_X(x) [/itex]

    3. The attempt at a solution
    So I think I've figured out part (a), but I'm stuck on part (b).

    For part (a) the mgf is

    [itex]\sum_{x=0}^{\infty} e^{tx} \binom{r+x-1}{x}p^{r}(1-p)^{x} = \sum_{x=0}^{\infty} \binom{r+x-1}{x}p^{r}(e^{t}(1-p))^{x} [/itex].

    I would think that...

    [itex]\sum_{x=0}^{\infty} \binom{r+x-1}{x}(1-e^{t}(1-p))^{r}(e^{t}(1-p))^{x} = 1[/itex],

    since this is another negative binomial pmf (probability mass function), whose sum must be 1.

    So now I do my sneaky trick....

    [itex] \frac {(1-(1-p)e^{t})^{r}}{(1-(1-p)e^{t})^{r}} \frac {p^{r}}{p^{r}} \sum_{x=0}^{\infty} \binom{r+x-1}{x}p^{r}(e^{t}(1-p))^{x} = \frac {p^{r}}{(1-(1-p)e^{t})^{r}} \cdot \sum_{x=0}^{\infty} \binom{r+x-1}{x}(1-e^{t}(1-p))^{r}(e^{t}(1-p))^{x} =
    \frac {p^{r}}{(1-(1-p)e^{t})^{r}}[/itex].

    So if this is correct, that takes care of part (a).

    For part (b) I have [itex] Y = 2pX [/itex], so [itex] y = 2px \iff x = \frac {y}{2p} [/itex]. I think this is a bijection, so I have

    [itex] f_Y(y) = \sum_{x \in g^{-1}(y)} f_X(x) = f_X(\frac {y}{2p}) [/itex].

    I'm not too sure about this transformation, but continuing on, I have mgf of Y is

    [itex] \sum_y e^{ty}f_Y(y) = \sum_y e^{yt} \binom {r + \frac {y}{2p} - 1}{\frac {y}{2p}}p^{r}(1-p)^{\frac {y}{2p}} [/itex]...

    I could continue on, but the trick I used earlier doesn't seem to work on this summation. Please help.
     
  2. jcsd
  3. Jul 27, 2014 #2

    Ray Vickson

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    The mgf of ##Y = 2pX## is ##M_Y(t) = E e^{tY} = E e^{2pt X} = M_X(2pt)##.
     
  4. Jul 27, 2014 #3
    That seems simple.

    So [itex]M_X(2pt) = (\frac {p}{1-(1-p)e^{2pt}})^{r}[/itex], right?
     
  5. Jul 28, 2014 #4
    Ok, then taking the limit as p approaches 0, and ignoring the exponent r, for now, I have....

    [itex] lim_{p \to 0} \frac {p}{(1-(1-p)e^{2pt})} = lim_{p \to 0} \frac {p}{(1-e^{2pt}+pe^{2pt})} = \frac 00 [/itex].

    Applying L'Hospital's rule, I have...

    [itex] lim_{p \to 0} \frac {\frac {d}{dp}}{\frac {d}{dp}}\frac {p}{(1-e^{2pt}+pe^{2pt})} = lim_{p \to 0} \frac 1{(-2te^{2pt}+e^{2pt}+pe^{2pt})} = \frac 1{(1-2t)}[/itex].

    Putting everything together I have...

    [itex] lim_{p \to 0} M_Y(t) = (\frac 1{(1-2t)})^{r} [/itex].

    Is it legit to ignore the exponent while taking the limit?
     
  6. Jul 29, 2014 #5

    Ray Vickson

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    Yes, why not? The function f(w) = w^r is continuous in w for fixed integer r > 0.
     
    Last edited: Jul 29, 2014
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