Repeated observations discrete distributions


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:


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 distribution


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.



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 \(C=1\).

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

Deriving the Poisson distribution