[1] 0.05399097
Statistics: Interested in estimating population-level characteristics; i.e., the parameters
\[ \begin{align*} y \rightarrow& f(y|\boldsymbol{\theta}) \\ \end{align*} \]
All the evidence/information in a sample (\(\textbf{y}\), i.e., data) relevant to making inference on model parameters (\(\theta\)) is contained in the likelihood function.
The sample data, \(\textbf{y}\)
A probability function for \(\textbf{y}\):
\(f(\textbf{y};\theta)\) or \([\textbf{y}|\theta]\) or \(P(\textbf{y}|\theta)\)
the unknown parameter(s) (\(\theta\)) of the probability function
\[ \begin{align*} \mathcal{L}(\boldsymbol{\theta}|y) = P(y|\boldsymbol{\theta}) = f(y|\boldsymbol{\theta}) \end{align*} \]
The likelihood (\(\mathcal{L}\)) of the unknown parameters, given our data, can be calculated using our probability function.
For example, for \(y_{1} \sim \text{Normal}(\mu,\sigma)\)
CODE:
[1] 0.05399097
If we knew the mean is truly 8, it would also be the probability density of the observation y = 10. But, we don’t know what the mean truly is.
The key is to understand that the likelihood values are relative, which means we need many guesses.
Central Tenet: evidence is relative.
Parameters are not RVs. They are not defined by a PDF/PMF.
MLEs are consistent. As sample size increases, they will converge to the true parameter value.
MLEs are asymptotically unbiased. The \(E[\hat{\theta}]\) converges to \(\theta\) as the sample size gets larger.
No guarantee that MLE is unbiased at small sample size. Can be tested!
MLEs will have the minimum variance among all estimators, as the sample size gets larger.
What is the mean height of King Penguins?
We go and collect data,
\(\boldsymbol{y} = \begin{matrix} [4.34 & 3.53 & 3.75] \end{matrix}\)
Let’s decide to use the Normal Distribution as our PDF.
\[ \begin{align*} f(y_1 = 4.34|\mu,\sigma) &= \frac{1}{\sigma\sqrt(2\pi)}e^{-\frac{1}{2}(\frac{y_{1}-\mu}{\sigma})^2} \\ \end{align*} \]
AND
\[ \begin{align*} f(y_2 = 3.53|\mu,\sigma) &= \frac{1}{\sigma\sqrt(2\pi)}e^{-\frac{1}{2}(\frac{y_{2}-\mu}{\sigma})^2} \\ \end{align*} \]
AND
\[ \begin{align*} f(y_3 = 3.75|\mu,\sigma) &= \frac{1}{\sigma\sqrt(2\pi)}e^{-\frac{1}{2}(\frac{y_{3}-\mu}{\sigma})^2} \\ \end{align*} \]
Or simply,
\[ \textbf{y} \stackrel{iid}{\sim} \text{Normal}(\mu, \sigma) \]
\(iid\) = independent and identically distributed
The joint probability of our data with shared parameters \(\mu\) and \(\sigma\),
\[ \begin{align*} & P(Y_{1} = y_1,Y_{2} = y_2, Y_{3} = y_3 | \mu, \sigma) \\ \end{align*} \]
\[ \begin{align*} & P(Y_{1} = 4.34,Y_{2} = 3.53, Y_{3} = 3.75 | \mu, \sigma) \\ &= \mathcal{L}(\mu, \sigma|\textbf{y}) \end{align*} \]
IF each \(y_{i}\) is independent, the joint probability of our data are simply the multiplication of all three probability densities,
\[ \begin{align*} =& f(4.34|\mu, \sigma)\times f(3.53|\mu, \sigma)\times f(3.75|\mu, \sigma) \end{align*} \] \[ \begin{align*} =& \prod_{i=1}^{3} f(y_{i}|\mu, \sigma) \\ =& \mathcal{L}(\mu, \sigma|y_{1},y_{2},y_{3}) \end{align*} \]
We can do this because we are assuming knowing one observation does not tell us any new information about another observation. Such that,
\(P(y_{2}|y_{1}) = P(y_{2})\)
Translate the math to code…
Calculate likelihood of many guesses of \(\mu\) and \(\sigma\) simultaneously,
What happens to the likelihood if we increase the sample size to N=100?