Texts: (De Moivre-Laplace Theorem) Sterling's formula gives quite an accurate approximation of the factorial: n! ~ β(2Οn)(n/e), where the symbol ~ means that the ratio between both sides tends to 1 as n tends to infinity, i.e., n!/[β(2Οn)(n/e)] = 1 + o(1) as n β β. Using Sterling's formula, we want to show that a β€ (x - np)/β(npq) β€ b as n β β, where q = 1 - p and 0 < p < 1.
a. Show that log(pq^n-r) = (r/n)*(1 - x/n) - e. Show that exp(1) = (x - np)^2/(npq) - log(pq^n) + o(1), where z = (x - np)/β(npq) β [a, b].
d. Show that P(9 β€ log(1 + β2) β€ -mp) = (1/2β2)log(11 + 2β2) - (1 - β(1 - β2)).ng(npq)^n(x/n)(1 - x/n) = β(npq)(x/n)(1 - x/n) where z = (x - np)/β(npq) β [a, b].
c. Using the Taylor expansion (1 + x)^2 log(1 + x) = x + o(x^3), show that 1 - np + β(npq)z^2 - nq + npq + o(1), - (2) p^2 q^2 ~ β(2Ο) exp(-1(x - np)^2) pβ(npq) npq P
