Question

How to get \[\frac{\lambda ^k}{k!}e^{-\lambda}\] from \[\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}\]when \(n \rightarrow \infty\)

Answer 1

taking a limit i believe

Answer 2

i am surprised this is not in your text, can look it up and give you a clue if you like

Answer 3

first off we need to state exactly what the assumption is, namely that as \(n\to \infty\) and \(p\to 0\) you have \(np\to \lambda>0\)

Answer 4

then you write \(np =\lambda+\epsilon_n\) and now the assumption is \(\epsilon_n\to 0\) ans \(n\to \infty\)

Answer 5

I was trying to work it out using Stirling formula, but looks complicated (I haven't finished it yet...)
Attached is what I've got in the notes, but don't quite understand it.

Answer 6

yeah it is a bunch of algebra, i don't think sterling is involved

Answer 7

Yes, Stirling's doesn't work..

Answer 8

you can think of it this way
\[\binom{n}{k}p^k(1-p)^{n-k}\] now stick the multiply and divide by the \(n^k\) and get
\[\frac{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...(1-\frac{(m-1)}{n})}{k!}(mp)^k\left(1-\frac{\lambda+\epsilon_n}{n}\right)^{n-k}\]

Answer 9

wow i screwed that all up

Answer 10

i meant pull out an \(n^k\) and stick it in the second part
\[\frac{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...(1-\frac{(k-1)}{n})}{k!}(np)^k\left(1-\frac{\lambda+\epsilon_n}{n}\right)^{n-k}\]

Answer 11

now let \(n\to \infty\) and you get \[\frac{\lambda^k}{k!}\lim_{n\to \infty}\left(\frac{1-\lambda +\epsilon_n}{n}\right)^{n-m}\]

Answer 12

i meant k, but in any case it is fixed, so the second term gives \(e^{-\lambda}\)
whichever version, it is really algebraic manipulation that gets it

Answer 13

Poisson is a limiting case of binomial considering infinite number of trials and constant expectation of success no matter how many trials.
\[\begin{array}{l}\mu=np=cons\tan t\rightarrow p=\frac\mu n\\\lim_{n\rightarrow\infty}\left(\begin{array}{c}n\\k\end{array}\right)p^k\left(1-p\right)^{n-k}=\lim_{n\rightarrow\infty}\left(\begin{array}{c}n\\k\end{array}\right)\left(\frac\mu n\right)^k\left(1-\frac\mu n\right)^{n-k}=\\=\lim_{n\rightarrow\infty}\frac{n\left(n-1\right)...\left(n-k+1\right)}{n^k}\left(\frac{\mu^k}{k!}\right)\left(1-\frac\mu n\right)^{-k}\left[\left(1-\frac1{\displaystyle\frac n\mu}\right)^{-\frac n\mu}\right]^{-\mu}\end{array}\]where\[\begin{array}{l}\lim_{n\rightarrow\infty}\frac{n\left(n-1\right)...\left(n-k+1\right)}{n^k}=1\\\lim_{n\rightarrow\infty}\left(1-\frac\mu n\right)^{-k}=1\\\lim_{n\rightarrow\infty}\left[\left(1-\frac1{\displaystyle\frac n\mu}\right)^{-\frac n\mu}\right]^{-\mu}=e^{-\mu}\end{array}\]then,\[\lim_{n\rightarrow\infty}\left(\begin{array}{c}n\\k\end{array}\right)p^k\left(1-p\right)^{n-k}=\frac{\mu^k}{k!}e^{-\mu}\]