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de Moivre-Laplace Theorem

where M ⊂ {0, 1, 2, . . .}. In particular, if a set M contains only one number k, then

k
λ
2
e −λ < np , (4.7)
P n (k) −
k!

We can consider an expression λ k e −λ as a function in two variables k and λ. There is a table of
k!
evaluated values for this expression.
We denote P(k) = λ k e −λ . A set
k!
{P(k) : k = 0, 1, 2, . . .}

is called Poisson distribution with a parameter λ > 0.

Example 4.5. A company has 500 workers. Find probability that two workers will
celebrate their birthday in one same day, if probability of birth in given day is
1/365. ,

Solution. We have n = 500, p = 1/365, λ = np = 500 = 100 , k = 2,
365 73
( ) 2
100
73 − 100
P 500 (2) ≈ e 73 = 0, 2384517.
2
Using Bernoulli formula we obtain P 500 (2) = 0, 2388347. Hence
2
|P 500 (2) − 0, 2384517| = 0, 000383 < 0, 003753 = np .

Thus an estimate (4.7) is valid for this case.

de Moivre-Laplace Theorem

Theorem 4.3. (de Moivre-Laplace Theorem). Let a probability of success in every n
independent experiments of Bernoulli trials equals p, 0 < p < 1. Then for large n a
probability P n (k) of k successes in n experiments evaluates by approximate formula

f(x 0 )
P n (k) ≈ √ , (4.8)
npq
⋆

k−np 1 − x 2
where x 0 = √ , f(x) = √ e 2 is a Gauss function.
npq 2π
It is proved that a relative error of (4.8) tends to zero when n → ∞.
A Gauss functions is tabulated. There are values f(x) for 0 ≤ x ≤ 3, 99 in most textbooks. If
−3, 99 ≤ x ≤ 0 we can use that f(x) is an even function and for |x| > 3, 99 we put f(x) = 0.

Theorem 4.4. (de Moivre-Laplace Integral Theorem). Let a probability of success in every
n independent experiments of Bernoulli trials equals p, 0 < p < 1. And µ n be a number of
successes in n experiments. Then

b
{ } ∫
µ n − np 1 x 2
lim P a ≤ √ ≤ b = √ e − 2 dx
n→∞ npq 2π
a
for arbitrary a, b (−∞ ≤ a ≤ b ≤ +∞). ⋆

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