Binomial Distribution at Infinity

Probability Level 1

Consider a binomial distribution of X ∼ B ( n , p ) X\sim B(n,p) .

It can be easily shown that P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X=k)={n\choose k}p^k{(1-p)}^{n-k} for k = 0 , 1 , 2 , 3 , … , n k=0,1,2,3,\ldots,n .

Now, let's take the limit of the above using n → ∞ n \to \infty . Instead of having an infinitesimal p p , let's assume that it is given that n p np , the mean of the probability distribution function, is some finite value m m .

Find P ( X = k ) P(X=k) in terms of m m and k k for this new distribution, where k = 0 , 1 , 2 , 3 , … k=0,1,2,3,\ldots , without looking anything up or reciting any formulas from memory.

0 0 e − m k ! m \frac{e^{-m}}{k!\ m} e − k m ! k \frac{e^{-k}}{m!\ k} m k e − m k ! \frac{m^ke^{-m}}{k!} m k ! ln ⁡ m \frac{m}{k! \ln{m}} ln ⁡ k k ! ln ⁡ m \frac{\ln{k}}{k! \ln{m}} e − k ln ⁡ m m ! k \frac{e^{-k}\ln{m}}{m!\ k} Limit does not exist

Nick Turtle by Nick Turtle , 21, USA

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1 solution

Nick Turtle
May 2, 2018

You probably recognize that the new probability distribution function is the Poisson distribution.

Without using the formula, consider the binomial distribution: P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X=k)={n\choose k}p^k{\left(1-p\right)}^{n-k}

= n ! k ! ( n − k ) ! p k ( 1 − p ) n − k =\frac{n!}{k!(n-k)!}p^k{\left(1-p\right)}^{n-k}

Substitute m = n p m=np , or p = m n p=\frac{m}{n} : = n ! k ! ( n − k ) ! ( m n ) k ( 1 − m n ) n − k =\frac{n!}{k!(n-k)!}{\left(\frac{m}{n}\right)}^k{\left(1-\frac{m}{n}\right)}^{n-k}

= n ! k ! ( n − k ) ! m k n k ( 1 − m n ) n ( 1 − m n ) − k =\frac{n!}{k!(n-k)!}\frac{m^k}{n^k}{\left(1-\frac{m}{n}\right)}^n{\left(1-\frac{m}{n}\right)}^{-k}

Slightly rearrange = n ! n k ( n − k ) ! ( 1 − m n ) − k m k k ! ( 1 − m n ) n =\frac{n!}{n^k(n-k)!}{\left(1-\frac{m}{n}\right)}^{-k}\frac{m^k}{k!}{\left(1-\frac{m}{n}\right)}^n

Note that lim ⁡ n → ∞ n ! n k ( n − k ) ! = 1 \lim_{n\to\infty}\frac{n!}{n^k(n-k)!}=1 and lim ⁡ n → ∞ ( 1 − m n ) − k = 1 \lim_{n\to\infty}{\left(1-\frac{m}{n}\right)}^{-k}=1 and lim ⁡ n → ∞ ( 1 − m n ) n = e − m \lim_{n\to\infty}{\left(1-\frac{m}{n}\right)}^n=e^{-m}

Thus, we have the final result of = m k e − m k ! =\frac{m^k e^{-m}}{k!}

which is equal to the formula for the Poisson distribution.

102 Helpful 32 Interesting 29 Brilliant 39 Confused

Could you elaborate on how you calculated the first limit (the one with all the factorials)?

Piotr Trochim - 1 year, 11 months ago

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=n(n-1)(n-2)...(n-k+1)(n-k)!/n^k (n-k)! =n(n-1)(n-2)...(n-k+1)/n^k =(n/n) ((n-1)/n) ((n-2)/n) ...(n-k+1)/n =1

Rajesh basak - 1 year, 9 months ago

Can you elaborate on calculating the third limit

Chaithanya Laxminarayan - 1 year, 10 months ago

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lim(1-m/n)^n = x

log(x) = lim[ n*log(1-m/n) ] = (using L'Hospital) lim[ mn/(m-n) ] = -m

log(x) = -m

x = e^-m

Matt Zhang - 10 months, 4 weeks ago

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Could you elaborate on the second step? How to apply the L'Hospital rule? Thx!

KE WEI - 1 month, 3 weeks ago

could you explain why we substitute p=m/n?

Ethel Chew - 7 months, 1 week ago

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because the question asks us to find P(X=k) in terms of m and k

He Lin Chooi - 5 months, 1 week ago

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