Proof of the Poission Distribution

[h2]Introduction[/h2] The Binomial Distribution is used when there are a certain number of trials, out of which there are successes and failures, give p. The Poission Distribution is instead used when you want to calculate the number of events in some period, given an average rate. E.g) Say we are told that 5 people buy something in a shop every hour on average, calculate the probability of 9 people buying. [ul][li]Split the hour instead into 5 minute (20) intervals, then the probability of someone buying something per interval is 0.25[/li] [li]So now we have 20 intervals, a probability of 0.25 per interval, and 9 successes, we can model this as a Binomial probability[/li] [li]And we get about 0.02706[/li] [li]If we increase the intervals to 40, and then divide the average properly (5/40, so 0.125), and plug it in, we get 0.03246[/li] [/ul] And when we plug it into the Poisson Distribution (avg events = 5, looking for P(X=9)), we get 0.03627. As you can see, by increasing the intervals of the Binomial Distribution, we get closer to the Poission Distribution. This is because the Binomial Distribution is discrete, so there are finite number of discrete steps, unlike for Poission. These steps can only be successes or failures, we can't deal with multiple events in one interval, so more steps are needed. But if there are infinitessimal intervals which can be successes or failures, it works, and that is the Poission Distribution. And we can also check for as high a number of events in that timer period we want (so as there is no n, it can be anything > 0). [h3]Preliminary to Proof[/h3] $$\binom{7}{5}=\frac{7!}{5!*2!}=\frac{7...*2*1}{(5...*2*1)*(2*1)}=\frac{7*6...*3}{5!}=\frac{7(7-1)(7-2)...(7-5+1)}{5!}$$ And so you can probably see then, that it's fairly obvious that the following holds: $$\binom{n}{x}=\frac{n(n-1)(n-2)...(n-x+1)}{x!}$$ [h2]Proof[/h2] The format of the Binomial approaching infinity (where the RHS is the formula for the Poission Distribution): $$\lim_{n\to \infty}\binom{n}{x}p^{x}(1-p)^{n-x} = \frac{e^{-\lambda}\lambda^{x}}{x!}$$ Where [b]p = [lambda]/n[/b], where [lambda] is the average number of events per interval, and n is the number of intervals. $$\lim_{n\to \infty}\binom{n}{x}p^{x}(1-p)^{n-x} = \lim_{n\to \infty}\frac{n(n-1)(n-2)...(n-x+1)}{x!}(\frac{\lambda}{n})^{x}(1-p)^{n-x}$$ $$= \lim_{n\to \infty}\frac{n(n-1)(n-2)...(n-x+1)}{x!}(\frac{\lambda^{x}}{n^{x}})(1-p)^{n-x}$$ $$= \lim_{n\to \infty}\frac{\frac{n}{n}\frac{(n-1)}{n}\frac{(n-2)}{n}...\frac{(n-x+1)}{n}}{x!}(\lambda^{x})(1-p)^{n-x}$$ As there are necessarily [i]x[/i] elements in the numerator's expansion, dividing by n is equivalent to multiplying by 1/(x^n). (Because there are x elements, and multiplying 1/n by itself - every term of the numerator expansion - is x^n, hence the above). $$= \lim_{n\to \infty}\frac{1(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{x-1}{n})}{x!}(\lambda^{x})(1-p)^{n-x}$$ And then it becomes clear that the entire fraction's numerator would evaluate to 1. (Because all elements greater than 1 are divided by n, which approaches infinity, so they evaluate to 0). So the only bit of the fraction left is [i]1/x![/i], and we have the term which doesn't evaluate any further [i][lambda]^x[/i]. Then we are left only with one term left to evaluate in the limit (the other 2 are constant so we can take them out): $$\lim_{n\to \infty}(1-\frac{\lambda}{n})^{n-x}$$ We can then use a standard result to evaluate that: $$\lim_{n\to \infty}(1-\frac{\lambda}{n})^{n-x}=e^{-\lambda}$$ [link href='https://erewhon.xyz/notes/categories.php?category_id=827&subject_id=25? subject_name=Maths&subjectName=Maths&categoryGroup=limit-proofs']Standard Result Proof[/link] And then we have 3 terms from the limit left over: [i]1/x![/i], [i][lambda]^x[/i] and [i]e^-[lambda][/i], which we can combine, to get: $$\frac{e^{-\lambda}\lambda^{x}}{x!}$$ So, instead of doing X~B(n, [lambda]/n) with high values of n to get an approximation, we can do X~P([lambda]): $$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$$