I need help when I try to prove this theorem, DE MOIVRE-LAPLACE from textbook Probability Theory I page 22-23.
The theorem says: In the Bernoulli case with p > 0, q = 1- p > - , as n -> \infty,
de Moivre (1732):
Pn(x) = P[Sn = j] ~ $\frac{1}{\sqrt{2\pi npq}} e^{-\frac{x^{2}}{2}}$, x = \frac{j-np}{\sqrt{npq}}
uniformly on every finite interval [a,b] of values of x;
Laplace (1801):
p[a <= \frac{Sn - np}{\sqrt{npq}} <= b] -> \frac{1}{\sqrt{2\pi}}\int_{a}^{b} e^{-\frac{x^{2}}{2}} dx
The relation an ~ bn means that an/bn -> 1.
Now, I almost went over this proof from the textbook, and I have one question at the last step
when we need to show..
How do people prove ..
On account of the first assertion, uniformly in j, Pn(x_{nj}) ~ \frac{1}{\sqrt{2 \pi npq}}e^{-x^{2}_{nj}/2} and p[a <= \frac{Sn - np}{\sqrt{npq}} <= b] = \sum_{j} P_n{x_{nj}} ~ \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{npq}} \sum_{j} e^{-x^{2}_{nj}/2}
From the textbook hint, we can use a Riemann sum approximating the integral \frac{1}{\sqrt{2\pi}\int_{a}^{b} e^{-x^{2} / 2}
Thus, the question becomes to show that 1/sqrt{2\pi} 1/sqrt{npq} \sum_{j} e^{-\frac{x^{2}_{nj}}{2}} = \frac{1}{\sqrt{2\pi}} \int_{a}^{b} e^{-\frac{x^{2}}{2}} dx
Any thoughts? suggestions? Thanks in advance!
The theorem says: In the Bernoulli case with p > 0, q = 1- p > - , as n -> \infty,
de Moivre (1732):
Pn(x) = P[Sn = j] ~ $\frac{1}{\sqrt{2\pi npq}} e^{-\frac{x^{2}}{2}}$, x = \frac{j-np}{\sqrt{npq}}
uniformly on every finite interval [a,b] of values of x;
Laplace (1801):
p[a <= \frac{Sn - np}{\sqrt{npq}} <= b] -> \frac{1}{\sqrt{2\pi}}\int_{a}^{b} e^{-\frac{x^{2}}{2}} dx
The relation an ~ bn means that an/bn -> 1.
Now, I almost went over this proof from the textbook, and I have one question at the last step
when we need to show..
How do people prove ..
On account of the first assertion, uniformly in j, Pn(x_{nj}) ~ \frac{1}{\sqrt{2 \pi npq}}e^{-x^{2}_{nj}/2} and p[a <= \frac{Sn - np}{\sqrt{npq}} <= b] = \sum_{j} P_n{x_{nj}} ~ \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{npq}} \sum_{j} e^{-x^{2}_{nj}/2}
From the textbook hint, we can use a Riemann sum approximating the integral \frac{1}{\sqrt{2\pi}\int_{a}^{b} e^{-x^{2} / 2}
Thus, the question becomes to show that 1/sqrt{2\pi} 1/sqrt{npq} \sum_{j} e^{-\frac{x^{2}_{nj}}{2}} = \frac{1}{\sqrt{2\pi}} \int_{a}^{b} e^{-\frac{x^{2}}{2}} dx
Any thoughts? suggestions? Thanks in advance!
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