# 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

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?