1. ## Geometric Distribution MLE

Consider carrying out a sequence of Bernoulli trials where the probability of success is pi in each trial. We stop when the first success occurs. Let Y denote the number of trials carried out. It is known that Y follows the so-called geometric distribution defined by:

P(Y = k) = pi*(1-pi)^k-1, k = 1,2,...,

a) Apply this distribution to an example of a "game" where a person tosses a balanced six-sided die repeatedly and "wins" the game when they obtain an ace (one). Calculate the probability that they obtain an ace on the fourth toss. Identify pi and k first.

b) Assume pi is unknown. Derive the MLE of pi if a single observation Y = y is observed.

c) Calculate the MLE when Y = 3 is observed using your formula from part b)

d) Determine and plot (using SAS) the relative likelihood function for the observed data Y=3. Does the likelihood function attain its maximum at the MLE obtained in part c) ? explain why or why not.

A) pi = 1/6, k = 4. Then, P (Y = 4) = 1/6 * 5/6^3 = 0.0964506173

B) for a single observation P(Y=y) the likelihood function is:

L(pi) = pi * (1-pi)^y-1

The log-likelihood function l(pi) = log(pi) + (y-1) log (1-pi)

Differentiating this with respect to pi and setting it equal to 0 gives:

1/pi - (y-1)/(1-pi) = 0

1/pi = (y-1)/(1-pi)

1 - pi = pi(y-1)

(1 - pi)/pi = y-1

1/pi - 1 = y-1

y = 1/pi

Thus the MLE is 1 / pi

C) Y = 3. Thus the MLE = 1/3

D) I haven't got SAS on hand at the moment, but I guess any hints might be helpful

2. anyone have any ideas?

3. Should just be a matter of plotting

l( pi ; y = 3 ) = ln( pi ) + ln(1 - pi)*2

as a function of pi.

It should have max at pi = 1/3.

4. Thanks for the reply Fed. I did the plotting at school yesterday using SAS, and I indeed got a maximum occurring at 1/3. Does everything else look okay though?

Thanks again,

James

Edit:

Sorry, I plotted the relative likelihood function and got the MLE occurring at 1/3

R(p) = l(p) / l(3) (the likelihood function, not the log likelihood)

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