- #1
Mogarrr
- 120
- 6
Imo, this problem is crazy hard.
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.
the moment generating function for a discrete random variable, denoted as M_X(t) is equal to
\sum_x e^{tx}f_X(x)
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} = <br /> \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.
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} = <br /> \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.