Negative binomial transformation and mgf

[Mogarrr]

Imo, this problem is crazy hard.

Homework Statement

Let X have the negative binomial distribution with pmf:

$f_X(x) = \binom{r+x-1}{x}p^{r}(1-p)^{x}$, x=0.1.2...,

where $0<p<1$ 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 $p \downarrow 0$, the mgf of Y converges to that of a chi squared random variable with 2r degrees of freedom by showing that

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

Homework Equations

the moment generating function for a discrete random variable, denoted as $M_X(t)$ is equal to

$\sum_x e^{tx}f_X(x)$

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

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

I would think that...

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

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

So now I do my sneaky trick...

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

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

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

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

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

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

I could continue on, but the trick I used earlier doesn't seem to work on this summation. Please help.

 

[Ray Vickson]

Imo, this problem is crazy hard.

Homework Statement

Let X have the negative binomial distribution with pmf:

$f_X(x) = \binom{r+x-1}{x}p^{r}(1-p)^{x}$, x=0.1.2...,

where $0<p<1$ 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 $p \downarrow 0$, the mgf of Y converges to that of a chi squared random variable with 2r degrees of freedom by showing that

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

Homework Equations

the moment generating function for a discrete random variable, denoted as $M_X(t)$ is equal to

$\sum_x e^{tx}f_X(x)$

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

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

I would think that...

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

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

So now I do my sneaky trick...

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

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

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

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

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

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

I could continue on, but the trick I used earlier doesn't seem to work on this summation. Please help.

The mgf of $Y = 2pX$ is $M_Y(t) = E e^{tY} = E e^{2pt X} = M_X(2pt)$.

 

[Mogarrr]

The mgf of $Y = 2pX$ is $M_Y(t) = E e^{tY} = E e^{2pt X} = M_X(2pt)$.

That seems simple.

So $M_X(2pt) = (\frac {p}{1-(1-p)e^{2pt}})^{r}$, right?

 

[Mogarrr]

Ok, then taking the limit as p approaches 0, and ignoring the exponent r, for now, I have...

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

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

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

Putting everything together I have...

$lim_{p \to 0} M_Y(t) = (\frac 1{(1-2t)})^{r}$.

Is it legit to ignore the exponent while taking the limit?

 

[Ray Vickson]

Yes, why not? The function f(w) = w^r is continuous in w for fixed integer r > 0.