1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Yes, why not? The function f(w) = w^r is continuous in w for fixed integer r > 0.
    Last edited: Jul 29, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted