# Maximum Likelihood Estimation (MLE) example: Bernouilli Distribution

Link to other examples: Exponential and geometric distributions

Observations: k successes in n Bernoulli trials.

$f(x)=\left(\frac{n!}{x!\left(n-x \right)!} \right){p}^{x}{\left(1-p \right)}^{n-x}$

$L(p)=\prod_{i=1}^{n}f({x}_{i})=\prod_{i=1}^{n}\left(\frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right){p}^{{x}_{i}}{\left(1-p \right)}^{n-{x}_{i}}$

$L(p)=\left( \prod_{i=1}^{n}\left(\frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right)\right){p}^{\sum_{i=1}^{n}{x}_{i}}{\left(1-p \right)}^{n-\sum_{i=1}^{n}{x}_{i}}$

$lnL(p)=\sum_{i=1}^{n}{x}_{i}ln(p)+\left(n-\sum_{i=1}^{n}{x}_{i} \right)ln\left(1-p \right)$

$\frac{dlnL(p)}{dp}=\frac{1}{p}\sum_{i=1}^{n}{x}_{i}+\frac{1}{1-p}\left(n-\sum_{i=1}^{n}{x}_{i} \right)=0$

$\left(1-\hat{p}\right)\sum_{i=1}^{n}{x}_{i}+p\left(n-\sum_{i=1}^{n}{x}_{i} \right)=0$

$\hat{p}=\frac{\sum_{i=1}^{n}{x}_{i}}{n}=\frac{k}{n}$

### Poisson Distribution

Observations: ${X}_{1}, {X}_{2}, {X}_{3}.....{X}_{n}$

$f(x)=\frac{{\lambda}^{x}{e}^{-\lambda}}{x!} x=0, 1, 2,$...

$L(\lambda)=\prod_{i=1}^{n}\frac{{\lambda}^{{x}_{i}}{e}^{-\lambda}}{{x}_{i}!} = {e}^{-n\lambda} \frac{{\lambda}^{\sum_{1}^{n}{x}_{i}}}{\prod_{i=1}^{n}{x}_{i}}$

$lnL(\lambda)=-n\lambda+\sum_{1}^{n}{x}_{i}ln(\lambda)-ln\left(\prod_{i=1}^{n}{x}_{i}\right)$

$\frac{dlnL(\lambda)}{dp}=-n+\sum_{1}^{n}{x}_{i}\frac{1}{\lambda}$

$\hat{\lambda}=\frac{\sum_{i=1}^{n}{x}_{i}}{n}$

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett