Poisson distribution
As we can see in previous slide, we need to use bernouilli / binomial distribution when we deal with specific event which make only two result (success / failure). In the case of dealing with another event like how many cholcolate chips distributed in each cookies, we need to use poisson distribution.
Poisson distribution has one parameter lambda. We can also think of conjugate distribution of poisson distribution. That is gamma distribution. $$Y_i \sim \text{Pois}(\lambda)$$ $$f(\tilde y|\lambda) = \frac{\lambda^{\Sigma y_i}e^{-n\lambda}}{\Pi^n_{i=1}y_i!} \text{for} \lambda \gt 0$$ $$\text{Gamma prior}\;\lambda \sim \Gamma(\alpha, \beta)$$ $$f(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta \lambda}$$ $$f(\lambda|\tilde y) \varpropto f(\tilde y|\lambda)f(\lambda) \varpropto \lambda^{\Sigma y_i}e^{-n\lambda}\lambda^{\alpha-1}e^{-\beta \lambda}=\lambda^{(\alpha+\Sigma y_i)-1}e^{-(n+\beta)\lambda}$$ $$\text{Posterior}\;\Gamma(\alpha+\Sigma y_i,\beta + n)$$
Mean of prior gamma distribution and posterior is like this $$\text{Prior mean}\;\frac{\alpha}{\beta}$$ $$\text{Posterior mean}\;\frac{\alpha+\Sigma y_i}{\beta + n}=\frac{\beta}{\beta+n}\frac{\alpha}{\beta}+\frac{n}{\beta+n}\frac{\Sigma y_i}{n}$$ $$\text{Effective sample size : } \beta$$
Exponential distribution
When we wait for bus which is coming every 10 minute averagely, we can ues exponential distribtuion when bus will come. The parameter of distribution is lambda which is showing how oftern the event will happer. The important thing is gamma distribution is also conjugate of exponential distribution.
For example, if the bus come every 10 minutes averagely, the parameter of exponential distribution is 1 over 10. $$Y \sim \text{Exp}(\lambda)$$ $$\text{Every 10 minutes : } \lambda = \text{Prior mean}=\frac{1}{10}$$ $$\text{Prior mean is same with }\Gamma(100,1000)$$ $$\text{Prior std.dev }= \frac{1}{100}$$ $$\text{mean} \pm \text{two std.dev }= 0.1 \pm 0.02$$
When there are one bus coming after 12 minutes, we can update posterior with our new data. $$f(\lambda|y) \varpropto f(y|\lambda)f(\lambda) \varpropto \lambda e^{-\lambda y}\lambda^{\alpha -1}e^{-\beta\lambda}=\lambda^{(\alpha+1)-1}e^{-(\beta+y)\lambda}$$ $$\text{Posterior } \Gamma(\alpha +1, \beta + y)$$ $$\text{So, } f(\lambda|y) \sim \Gamma(100,1012)$$ $$\text{Posterior mean }= \frac{101}{1012} = \frac{1}{10.02}$$ We can see waiting time is larger than 10 minutes which is prior value. However, there are no big affect from one data.