# Math Help - Moment generating function to deduce limiting distribution

1. ## Moment generating function to deduce limiting distribution

"Let Y_n be the number of trials up to and including the first success in a sequence of independent trials, where each trial has a probability of success of k/n.

Find the moment generating function of (1/n)Y_n. Hence show that, as n -> infinity, the limiting distribution of (1/n)Y_n is exponential with parameter k."

I should be able to manage the "deduction" part of the question providing that I can find the moment generating function of (1/n)Y_n in the first place. This is proving to be quite tricky for me though. How do we go about doing this? And then for the second part, presumably we can let n -> infinity in the moment generating function and spot that this gives the same generating function as an exponential with parameter k, and then appeal to uniqueness of generating functions? Thanks.

2. Originally Posted by Amanda1990
"Let Y_n be the number of trials up to and including the first success in a sequence of independent trials, where each trial has a probability of success of k/n.

Find the moment generating function of (1/n)Y_n. Hence show that, as n -> infinity, the limiting distribution of (1/n)Y_n is exponential with parameter k."

I should be able to manage the "deduction" part of the question providing that I can find the moment generating function of (1/n)Y_n in the first place. This is proving to be quite tricky for me though. How do we go about doing this? And then for the second part, presumably we can let n -> infinity in the moment generating function and spot that this gives the same generating function as an exponential with parameter k, and then appeal to uniqueness of generating functions? Thanks.
Hi,

what is the probability that the first success is strictly after time $m$? This means that there have been at least $n$ failures so, using independence, you get $P(Y_n>m)=\left(\frac{k}{n}\right)^m$. From there, you can deduce $P(Y_n=m)$ using $P(Y_n=m)=P(m-1m-1)-P(Y_n>m+1)$ and recognize a geometric distribution. (Or you can recognize it directly from the distribution function $P(Y_n\geq m)=P(Y_n>m-1)$)

For the second part, you can either use generating functions indeed (so you have to compute the generating function of $\frac{1}{n}Y_n$), or you can use the distribution function (the distribution function of $\frac{1}{n}Y_n$ is obtained by $P(\frac{1}{n}Y_n\geq t)=P(Y_n\geq nt)=P(Y_n\geq \left\lceil nt \right\rceil)$ where $\lceil\cdot \rceil$ is the upper integer part)

3. Sorry, but what I'm really struggling over is finding the moment generating function of 1/n Y_n in the first place. This seems quite tricky, so any help would be very useful. I've found the distribution of 1/n Y_n, namely P(1/n Y_n = m) = (1- k/n)^m-1 . (k/n), but I can't find the expectation of exp(1/n Y_n t) to find the moment generating function.

4. Originally Posted by Amanda1990
Sorry, but what I'm really struggling over is finding the moment generating function of 1/n Y_n in the first place. This seems quite tricky, so any help would be very useful. I've found the distribution of 1/n Y_n, namely P(1/n Y_n = m) = (1- k/n)^m-1 . (k/n), but I can't find the expectation of exp(1/n Y_n t) to find the moment generating function.
The moment generating function of $Y_n$ is $M_{Y_n}(t) = \frac{p e^t}{1 - (1-p)e^t} = \frac{\frac{k}{n} e^t}{1 - \left( 1 - \frac{k}{n}\right)e^t } = \frac{ke^t}{n - (n-k) e^t}$.

Therefore the moment generating function of $X_n = \frac{1}{n} Y_n$ is $M_{X_n} (t) = M_{Y_n} \left( \frac{t}{n}\right) = \frac{ke^{t/n}}{n - (n-k) e^{t/n}}$.

Substitute the Maclaurin series for $e^{t/n}$:

$M_{X_n} (t) = \frac{k \left( 1 + \frac{t}{n} + \, .... \right) }{n - (n-k) \left( 1 + \frac{t}{n} + \, .... \right)}$ $= \frac{k \left( 1 + \frac{t}{n} + \, .... \right) }{n - \left( n + t - k - \frac{kt}{n} + \, .... \right)}$ $= \frac{k \left( 1 + \frac{t}{n} + \, .... \right) }{- t + k + \frac{kt}{n} - \, .... }$.

Now take the limit $n \rightarrow + \infty$:

$M_{X_n} (t) = \frac{k}{k - t} = \frac{1}{1 - \frac{t}{k}} = \left( 1 - \frac{t}{k} \right)^{-1}$

which is readily recognised as the moment generating function of the exponential distribution with parameter $k$.