Let $X_1, X_2, · · · , X_n$ be a random sample of size n from a geometric distribution withpmf $f(x) = 0.75 · 0.25^{ x-1} , x = 1, 2, 3, ··· .$

(a) Find the mgf $M_{Y_n} (t)$ of $Y_n = X_1 + X_2 + ··· + X_n$. Then name the distribution of $Y_n$.

I found this to be

$$M_{Y_n} (t) = \frac{(0.75e^t)^n}{(1-0.25e^t)^n}$$ Hence $Y_n$ is a negative binomial distribution

(b) Find the mgf $M_{\bar{Y_{n}}} (t)$ of the sample mean $\bar{Y{_{n}}} = \frac{Y{_{n}}}{n}$

Found this to be $$M_{Y_n} (t) = \frac{(0.75e^\frac{t}{n})^n}{(1-0.25e^\frac{t}{n})^n}$$

(c) Find the limit ${\lim_{ n \to \infty }} M_{\bar{Y_{n}}} (t)$ using the result of (c). What distribution does the limiting mgf correspond to?

Found this to be

$${\lim_{ n \to \infty }} M_{\bar{Y_{n}}} (t) = e^\frac{4t}{3}$$

Now here is where my problem begins. What distribution does the limiting mgf correspond to? I've went through the formula sheet and it doesn't seem to correspond to any of the basic distributions. I'm pretty sure everything else above is correct.

(d) Let $Z_{n} = \frac{3}{2}\sqrt{n}\bar{Y_{n}} - 2{\sqrt{n}}$ . Find $M_{Z{_{n}}} (t)$ , the mgf of $Z_n$. Then find the limiting mgf ${\lim_{n \to \infty }} M_{Z_{n}} (t)$. (We were given a hint that $Z_n$ would be a normal distribution).

I have no idea how to do part (d) and would be really grateful if someone can help with it.


(a) Yes, the sum of $n$ iid Geometric Distributed Random Variables has a Negative Binomial distribution, and that is the right moment generating function for the given one.

(b) is okay, and see also (d) below.

(c) Well, $\mathsf e^{4t/3}$ is the moment generating function for a Degenerate Distribution.

In this case $\lim\limits_{n\to\infty}\overline{Y_n}\sim \delta({4/3})$, or simple $\lim\limits_{n\to\infty}\overline{Y_n}=4/3\text{ a.s.}$

Does that make sense?

For (d) use the definition of Moment Generation Function: $$\begin{align}\mathsf M_{aX+b}(t) =&~ \mathsf E(\mathsf e^{t(aX+b)})\\ =&~ \mathsf e^{tb}\mathsf E(\mathsf e^{(ta)X)})\\ =&~ \mathsf e^{tb}\mathsf M_X(at)\end{align}$$

PS: Is this not what you did in part (b)?

  • $\begingroup$ Ohhh thanks heaps. Never heard of that distribution, thank you. Also, do you know how to do part (d). I tried from the definition, ie $M_{z_n} (t) = E(e^{t(Z_n)})$ but I can't seem to progress at all $\endgroup$ – Jim Wilson May 26 '16 at 8:50
  • $\begingroup$ So then $M_{z_n} (t) = e^{-2\sqrt{n}t} M_{\bar{Y_{n}}} (\frac{3}{2} \sqrt{n}t)$ right? $\endgroup$ – Jim Wilson May 26 '16 at 9:44

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