Geometric Distribution

Discrete Random Variable

$\bigstar$ X~Geom(p) : the # of Bernoulli trials needed to get ONE success, each with probability P.

$\bigstar$ $P(X=k)=(1-p)^{k-1}\cdot p$,  k=1,2,3...
$\bigstar$ E(X)= $\frac {1}{p}$, Var(x)= $\frac {1-p}{p^2}= \frac {q}{p^2}$ 


Method of Moments(MoM) in Geometric Distribution
Proof
$M(t)=E(e^{tx})=\sum_{x=1}^{\infty}e^{tx}\cdot P(X=x)= P \sum_{x=1}^{\infty}e^{tx} \cdot (1-p)^{x-1}\cdot$ $ \frac {e^t}{e^t}$
          =  $p\cdot e^t \sum_{x=1}^{\infty}e^{t(x-1)}\cdot (1-p)^{x-1}=p \cdot e^t \sum_{x=1}^{\infty}(e^t \cdot (1-p))^{x-1}$  $\leftarrow$ Geometric Series!
        (we multiply by   $ \frac {e^t}{e^t}$ to make a geometric series)
        = $p \cdot e^t \sum_{x=0}^{\infty} (e^t(x-p))^x = \frac {p\cdot e^t}{1-(e^t(1-p))}$, $|e^t(1-p)| < 1$ 

$M^{(1)}(t)= \frac {d}{dt} \frac {p\cdot e^t}{1-(e^t(1-p))}= \frac {p \cdot e^t[1-(e^t(1-p))]-p \cdot e^t \cdot e^t(1-p)}{[1-(e^t(1-p))]^2}$
$M^{(1)}(0)= \frac {p \cdot[1-(1-p)]+(1-p)p}{[1-(1-p)]^2}= \frac {p^2+p-p^2}{(-p)^2} = \frac {p}{p^2}= \frac {1}{p} =\mu_{1}$  

$p= \frac {1}{\mu_{1}}$, $\therefore \hat{p}= \frac {1}{\hat{\mu_{1}}}$ 


Maximum Likelihood Estimate (MLE) in Geometric Distribution 
Proof
$L(p)=\prod P(Y_{i}=y_{i}|p)=p^n \cdot (1-p)^{\sum{(y_{i}-1)}}$
$l(p)=n \log p + \sum (y_{i}-1) \cdot \log(1-p)$ 
$l'(p)= \frac {n}{p}- \frac{\sum (y_{i}-1)}{1-p}=(set)0$ $\rightarrow n(1-p)=p \cdot \sum (y_i-1) \rightarrow \hat{p}= \frac {1}{\bar{y}}$ 

Example) Mathematical Statistics and Data Analysis, 3ED, Chapter8. Q8

In an ecological study of the feeding behavior of birds, the number of hos between flights was counted for several birds. For the following data, (a) fit a geometric distribution, (b) find an approximate 95% confidence interval for p. 

(# Hops, Frequency)= (1, 48) (2, 31) (3, 20) (4, 9) (5, 6) (6,5) (7,4) (8,2) (9,1) (10,1) (11,2) (12,1)

Solution??!!