Suppose X is a random variable that is binomially distributed with probability of success \( \lambda/n \) and number of trials \( n \), that is \(X \sim B(n, \lambda/n) \). The probability that X takes a special value \( x \) is then given by:\[ P(X=x)=\binom{n}{x} \left( \frac{ \lambda}{n} \right)^x \left( 1-\frac{\lambda}{n} \right)^{n-x} \]This can be written as:\[ P(X=x)=\frac{n(n-1)(n-2) ... (n-x+1)}{x!} \times \frac{\lambda^x}{n^x}\left(1-\frac{\lambda}{n} \right)^{n-x} \]Again, this can be rearranged to:\[ P(X=x)=\frac{\lambda^x}{x!} \times \frac{n-1}{n} \times \frac{n-2}{n} \times ... \times \frac{n-x+1}{n} \times \left(1-\frac{\lambda}{n} \right)^{n-x} \]An \( n \to \infty \), the expression becomes: \[P(X=x)=\frac{\lambda^x}{x!}e^{-\lambda} \]This formula then represents the Poisson distribution, a suitable model for events which:
- occur randomly in space or time
- occur singly, that is events cannot occur simultaneously
- occur independently
- occur at a constant rate, that is a mean number of events in a given time interval is proportional to the size of the interval
A discrete random variable that follows a Poisson distribution with parameter \( \lambda \) is written as \(X \sim Po(\lambda) \) and the mean of this distribution is \( \lambda \) and the variance is also \( \lambda \). Thus the mean and variance of a Poisson distribution are equal.
In Google Sheets, the syntax for the Poisson distribution function is shown below:
No comments:
Post a Comment