Maximum Likelihood Estimation Exercise (short)

Miau Piau

New Member
Hi

I have found the following exercise while trying to understand the Maximum Likelihood Estimation in Practice:

"In a random sample of 1000 Swiss employees, 78% stated that
they received a pay increase for the current year.
What is the MLE for the share of employees with a pay raise?"

I have no idea how to solve this question. My only idea was that, to be able to make a Maximum Likelihood Estimation at all, I would have to chose a distribution first (and I would chose a normal distribution here) - but else? I am grateful for any help!
Thanks

BGM

TS Contributor
The distribution you are looking for is the Bernoulli distribution - Two outcomes (yes/no for pay increase) which is mutually exclusive and collectively exhaustive (partition), for each independent trial (employee). So you will be interested in the proportion parameter/probability of success (share of employees with a pay increase).

Miau Piau

New Member
Hi

Thanks for the reply!

Then, would that mean, that our estimation for the parameter of the Bernoullidistribution would be that p = 0.78?
Where does Maximum Likelihood comes into play? Is it just that in this case p = 0.78 is the parameter best suiting for this particular observation?

Thanks

JesperHP

TS Contributor
Find a definition of the maximulikelyhood function in whatever book your reading in class. Plugin the Bernoullidistribution. Take logs of likelyhood function and differentiate.

Miau Piau

New Member
Ok, so the Maximum Likelihood Estimator for the Bernoulli distribution's parameter is:

T = 1/n sum_(i=1)^n X_i

So in this case that would be nothing else then 780/1000.

Is that correct?

Thanks for your help people!