Question

let X be a binomial variable with parameters n and p .can someone help with this question

Answer 1

man this must be a hell of a question

Answer 2

let X be a binpmial variable with parameters n and p , such that np=\[\lambda \.show that for fixed k ,as \[ n \rightarrow \infty ,with \lambda fixed ,\] \[ p(X=k)\rightarrow (1/k !)\lambda^{k}e^{-\lambda}\]

Answer 3

this is the poisson distribution. proof will be in any text

Answer 4

but i dont get the answer

Answer 5

i can grab a text and start copying, but really it is in any book under the heading something like "the poisson as an approximation to the binomial"
are you reading from a book? as i recall it is not hard

Answer 6

if so let me know we can go through it step by step

Answer 7

can u solve it to me i'm stuck

Answer 8

i have in front of me three proofs. all are basically the same and require one somewhat tricky piece of algebra.
are you looking at a proof?

Answer 9

yes

Answer 10

but it does not have step they just says is a poisson then they give that answer i've showed you

Answer 11

ok so we know binomial is
\[P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\] and now \(\lambda =np \implies p=\frac{\lambda}{n}\) so step one is to put
\[P(X=k)=\dbinom{n}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k}\]

Answer 12

actually maybe that is the last step. here is a more straightforward solution
we have
\[\dbinom{n}{k}p^k(1-p)^{n-k}=\frac{n(n-1)...(n-k+1)}{k!}p^k(1-p)^{n-k}\] and keeping in mind that as \(n\to \infty\) we have \(np -\to \lambda\) we can put
\[\lambda^k=(np)^k=n^kp^k\] so replace and divide each term in the numerator by \(n^k\) to get
\[\frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{(k-1)}{n})}{k!}(np)^k(1-p)^{n-k}\]

also in the limit \(p\to \frac{\lambda}{n}\) so we can write
\[\lim_{n\to \infty}\frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{(k-1)}{n})}{k!}(np)^k(1-p)^{n-k}\]
\[=\lim_{n\to \infty}\frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{(k-1)}{n})}{k!}(\lambda)^k(1-\frac{\lambda}{n})^{n-k}\]
\[=\lim_{n\to \infty}\frac{\lambda^k}{k!}(1-\frac{\lambda}{n})^{n-k}\] because the limit in that ugly numerator is 1

Answer 13

last step is to compute
\[\lim_{n\to \infty}(1-\frac{\lambda}{n})^{n-k}\] but k is fixed and only \(n\to \infty\) so this is the same as
\[\lim_{n\to \infty}(1-\frac{\lambda}{n})^n\] which is certainly
\[e^{-\lambda}\]

Answer 14

i will be proficient \(\LaTeX\) soon

Answer 15

thanks

Answer 16

yw