I like Serena said:
In statistics we typically try to find a probability distribution that describes a population.
It helps us to understand the population, and moreover to make predictions.
Often enough that's a normal distribution with a mean value combined and a standard deviation.
It's usually fairly straight forward to find and approximation for those particular parameters.
More generally a population is described the assumption of a distribution combined with the parameters defining it, like a normal distribution characterized by $\mu$ and $\sigma$, but we can also have different distributions, parameters, and it may not be so straight forward to find the parameters.
Maximizing the likelihood function means we find the most likely approximation of parameters given an assumed distribution. (Thinking)
Let's consider a specific example.
An employee starts work around 8:00 am The general beginning of duty varies by up to $2$ minutes up or down. We have the following:
[table="width: 700"]
[tr]
[td]$X$ = Beginning of duty (Difference to 8 o'clock in minutes) [/td]
[td]$-2$ [/td]
[td]$-1$ [/td]
[td]$1$ [/td]
[td]$2$[/td]
[/tr]
[tr]
[td]$\mathbb{P}(X=x)$ [/td]
[td]$0.2\, \theta$[/td]
[td]$0.3\, \theta$[/td]
[td]$0.5\, \theta$[/td]
[td]$1-\theta$[/td]
[/tr]
[/table]
For $10$ consecutive working days, the following values have occurred:
\begin{equation*}-1, \ \ \ 2, \ -2, \ -2, \ \ \ 1, \ \ \ 1, \ \ \ 2, \ -1, \ \ \ 1, \ -1\end{equation*}
From these infotmation we get the Likelihood function:
\begin{equation*}L(-1, 2, -2, -2, 1, 1, 2, -1, 1, -1 \mid \theta ) = 0.000135\cdot \theta^8\cdot (1-\theta)^2 \end{equation*}
The maximum Likelihood estimator is $\hat{\theta}=\frac{4}{5}$.
Do we calculate each probabilities $\mathbb{P}(X=x_i)$ with this value of $\theta$ and then we have the predictions what time the duty begins? (Wondering)
