Repeated observations discrete distributions

Binomial

Binomial distribution

If we repeat a Bernoulli trials with the same parameter and sum the results, we have the binomial distribution.

We therefore have two parameters, $$p$$ and $$n$$.

$$P(X=x)={n\choose x }p^x(1-p)^{n-x}$$

Moments of the binomial distribution

The mean is $$np$$, which can be seen as the trials are independent.

Similarly, the variances can be addeded together giving $$np(1-p)$$.

Multinomial distribution

The mass function for the binomial case is:

$$f(x)=\dfrac{n!}{x!(n-x)!}p^k(1-p)^{n-k}$$

The multinomial distribution

This generalises the binomial distribution where there are more than $$2$$ outcomes.

$$f(x_1,...,x_n)=\dfrac{n!}{\prod_i x_i!}\prod_i p_i^{x_i}$$

Poisson

Definition

We can use the Poisson distribution to model the number of indepedent events that occur in an a time period.

For a very short time period the chance of us observing an event is a Bernoulli trial.

$$P(1)=p$$

$$P(0)=1-p$$

Chance of no observations

Let’s consider the chance of repeatedly getting $$0$$: $$P(0;t)$$.

We can see that: $$P(0;t+\delta t)=P(0;t)(1-p)$$.

And therefore:

$$P(0;t+\delta t)-P(0;t)=-pP(0;t))$$

By setting $$p=\lambda \delta t$$:

$$\dfrac{P(0;t+\delta t)-P(0;t)}{\delta t}=-\lambda P(0;t))$$

$$\dfrac{\delta P(0;t)}{\delta t}=-\lambda P(0;t)$$

$$P(0;t)=Ce^{-\lambda t}$$

If $$t=0$$ then $$P(0;t)=0$$ and so $$C=1$$.

$$P(0;t)=e^{-\lambda t}$$