# Thread: Converging distributions - Binomial to Poisson

1. ## Converging distributions - Binomial to Poisson

Let Yn be a binomial random variable with n trials and with success probability p. Consider that n tends to infinity and p tends to zero in a way that np remains fixed at np=ƛ. Prove that the distribution of Yn converges to a Poisson distribtution with mean ƛ.

I am supposed to use mgfs but am unsure of how to do so.

2. For the binomial distribution, $\displaystyle M_{x}(t) = [pe^{t}+(1-p)]^{n}$

Now let $\displaystyle p = \frac{\lambda}{n}$

then $\displaystyle M_{x}(t) = [\frac{\lambda}{n}e^{t}+(1-\frac{\lambda}{n})]^{n} = [1+ \frac{(e^{t}-1)\lambda}{n}]^{n}$

now let n go to infinity

$\displaystyle \lim_{n \to \infty} M_{x}(t) = \lim_{n \to \infty} [1+ \frac{(e^{t}-1)\lambda}{n}]^{n}= e^{(e^{t}-1)\lambda}$ which is the MGF of the Poisson distribution

3. I usually just use the distribution...........

$\displaystyle {n\choose x}p^x(1-p)^{n-x}$

$\displaystyle ={n!\over (n-x)!x!} \left({\lambda\over n}\right)^x \left(1-{\lambda\over n}\right)^{n-x}$

$\displaystyle ={n(n-1)\cdots (n-x+1)\over n^x} \left({\lambda^x\over x!}\right) \left(1-{\lambda\over n}\right)^n\left(1-{\lambda\over n}\right)^{-x}$

$\displaystyle \to\left(1\right) \left({\lambda^x\over x!}\right) e^{-\lambda}\left(1\right)$

$\displaystyle ={\lambda^xe^{-\lambda} \over x!}$