Binomial Approximation to Poisson Distribution

Poisson distribution as a limiting case of binomial probability distribution under the following conditions

1)n= number of trials indefinitely large. 2)p= constant probability of success for each trial is indefinitely small 3)np=m is finite

In binomial distribution, the probability of x successes p(x)=ncx(p^x) q^(n-x),

Put now p=m/n and q=1-p=1-(m/n) we get

p(x)=n(n-1) (n-2)…(n-x+1)/x! (m/n)^x (1-(m/n)^(n-x)

=1 [1-(1/n)][1-(2/n)]…[1-{(x-1)/n}] m^x. [(1-(m/n))^n]

—————————————-—-—— ——————

x! [(1-(m/n))^x]

Now,for fixed x,as n -->∞

[1-(1/n)][1-(2/n)]………[1-(x-1)/n] all tend to 1 and [1-(m/n)]^(-n) to. e^(-m)

Can anyone tell me how [1-(m/n)]^(-x) is = e^(-m)

Re: Binomial Approximation to Poisson Distribution

Quote:

Originally Posted by

**Vinod** Poisson distribution as a limiting case of binomial probability distribution under the following conditions

1)n= number of trials indefinitely large. 2)p= constant probability of success for each trial is indefinitely small 3)np=m is finite

In binomial distribution, the probability of x successes p(x)=n_C_x (p^x) q^(n-x),

Put now p=m/n and q=1-p=1-(m/n) we get

p(x)=n(n-1) (n-2)…(n-x+1)/x! (m/n)^x (1-(m/n)^(n-x)

=1 [1-(1/n)][1-(2/n)]…[1-{(x-1)/n}] m^x. [(1-(m/n))^n]

—————————————-—-—— ——————

x! [(1-(m/n))^x]

Now,for fixed x,as n -->∞

[1-(1/n)][1-(2/n)]………[1-(x-1)/n] all tend to 1 and [1-(m/n)]^(-n) to. e^(-m)

Can anyone tell me how [1-(m/n)]^(-n) is = e^(-m)[/tex]

See the overview >>here<<, it is one of the definitions of the exponential function (which you should have net in Calculus I(?)).

CB