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Gaussian distribution
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In the limit of similar probabilities the binomial distribution is reduced in the continuous limit to the Gaussean distribution.

>Model

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Distribución binomial
Equation

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Con la probabilidad de que se de un numero definido de pasos a la derecha e izquierda esta dada por

W_N(n_1,n_2)=\displaystyle\frac{N!}{n_1!n_2!}p^{n_1}q^{n_2}



con el número total de pasos es

N=n_1+n_2



y solo existe la probabilidad de ir a la derecha o a la izquierda, con se tiene para las probabilidades que

p+q=1



por lo que con se tiene la distribución binomial

W_N(n) =\displaystyle\frac{ N !}{ n !( N - n )!} p ^ n (1- p )^{ N - n }

ID:(8961, 0)


Approach for N!
Equation

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With the Stirling approximation

equation=8966

and the change of variables

equation=8996

you get that

equation

ID:(8998, 0)


Approximation for n!
Equation

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With the Stirling approximation

equation=8966

and the change of variables

equation=11431

you get that

equation

ID:(9003, 0)


Approximation for (N-n)!
Equation

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With the Stirling approximation

equation=8966

and the change of variables

equation=8997

the expression is

equation

ID:(8999, 0)


Factor N!/N!(N-n)! For N\gg 1, n\gg 1 and N>n
Equation

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In the case of medium probabilities (p \sim q \sim 1/2) and large numbers N it can be shown with

N!\sim\sqrt{2\pi N}\left(\displaystyle\frac{N}{e}\right)^N



n!\sim\sqrt{2\pi n}\left(\displaystyle\frac{n}{e}\right)^n



and

(N-n)!\sim\sqrt{2\pi(N-n)}\left(\displaystyle\frac{N-n}{e}\right)^{N-n}



is obtained

\displaystyle\frac{N!}{n!(N-n)!}\sim\displaystyle\frac{1}{\sqrt{2\pi N}}\left(\displaystyle\frac{n}{N}\right)^{-n-1/2}\left(\displaystyle\frac{N-n}{N}\right)^{-N+n-1/2}

ID:(507, 0)


Limit of large Numbers and middle Probabilities
Equation

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The expression

W_N(n) =\displaystyle\frac{ N !}{ n !( N - n )!} p ^ n (1- p )^{ N - n }



is reduced by

\displaystyle\frac{N!}{n!(N-n)!}\sim\displaystyle\frac{1}{\sqrt{2\pi N}}\left(\displaystyle\frac{n}{N}\right)^{-n-1/2}\left(\displaystyle\frac{N-n}{N}\right)^{-N+n-1/2}



to representation

W_N(n)\sim\displaystyle\frac{1}{\sqrt{2\pi N}}\left(\displaystyle\frac{n}{N}\right)^{-n-1/2}\left(\displaystyle\frac{N-n}{N}\right)^{-N+n-1/2}p^n(1-p)^{N-n}

ID:(506, 0)


Average position
Equation

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If total N steps are taken with a probability p in the right direction and these have a length a the expected final position will be

\mu=aNp

ID:(9008, 0)


Change variables by offset x=(n-Np)a
Equation

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To obtain the Gaussian distribution it is necessary to develop the distribution around its deviation from its mean position that can be given by

x=(n-Np)a

ID:(8973, 0)


Factor n/N depending on the path x
Equation

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As the way is

x=(n-Np)a



factor n/N can be written as

\displaystyle\frac{n}{N}=p\left(1+\displaystyle\frac{x}{aNp}\right)

ID:(9004, 0)


Factor N-n/N depending on the path x
Equation

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As the way is

x=(n-Np)a



factor n/N can be written as

\displaystyle\frac{N-n}{N}=(1-p)\left(1-\displaystyle\frac{x}{aN(1-p)}\right)

ID:(9005, 0)


Binomial distribution as a function of deviation
Equation

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If large numbers and probabilities around 1/2 are entered in the binomial distribution for the case

W_N(n)\sim\displaystyle\frac{1}{\sqrt{2\pi N}}\left(\displaystyle\frac{n}{N}\right)^{-n-1/2}\left(\displaystyle\frac{N-n}{N}\right)^{-N+n-1/2}p^n(1-p)^{N-n}



the expressions

\displaystyle\frac{n}{N}=p\left(1+\displaystyle\frac{x}{aNp}\right)



and

\displaystyle\frac{N-n}{N}=(1-p)\left(1-\displaystyle\frac{x}{aN(1-p)}\right)




a distribution of the form is obtained

W_N(n)\sim\displaystyle\frac{1}{\sqrt{2\pi N}p(1-p)}\left(1+\displaystyle\frac{x}{aNp}\right)^{-n-1/2}\left(1-\displaystyle\frac{x}{aN(1-p)}\right)^{-N+n-1/2}

ID:(8974, 0)


Variable change u=x/aNp
Equation

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To develop the 1+x/aNp factor you can work with the variable change

u=\displaystyle\frac{x}{aNp}

ID:(9021, 0)


Factor 1+x/aNp for N\gg 1 and p\sim 1/2
Equation

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With the approximation

1+u\sim e^{u-\frac{1}{2}u^2}



it has to

\left(1+\displaystyle\frac{x}{aNp}\right)\sim e^{x/aNp-x^2/2a^2N^2p^2}

ID:(9006, 0)


Variable change u=x/aN(1-p)
Equation

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To develop the factor 1+x/aN(1-p) you can work with the variable change

u=\displaystyle\frac{x}{aN(1-p)}

ID:(9022, 0)


Factor 1-x /aN(1-p) for N\gg 1 and p\sim 1/2
Equation

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With the approximation

1+u\sim e^{u-\frac{1}{2}u^2}



it has to

\left(1-\displaystyle\frac{x}{aN(1-p)}\right)\sim e^{-x/aN(1-p)-x^2/2a^2N^2(1-p)^2}

ID:(9007, 0)


Probability for large N and middle p
Equation

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It can be shown that for a large number N and probability p neither too small nor too close to 1, the binomial distribution is reduced to a gausseana for the position x=na:

P(x)=\displaystyle\frac{1}{\sqrt{2\pi N^2p(1-p)}}e^{-(x-aNp)^2/2N^2p(1-p)}

In this case, the probability q was replaced by 1-p.

ID:(3367, 0)


Generalization Limit Big Numbers
Equation

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$\begin{matrix}

P(x) & = & \displaystyle\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}\\

\sigma^2 & = & Np(1-p)\\

\end{matrix}

$

ID:(3368, 0)


Gaussian distribution standard deviation
Equation

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The standard deviation of the binomial distribution at the limit N large and p medium is

\sigma^2 = N ^2 p (1- p )

ID:(8963, 0)


Example comparison with Gaussian distribution
Image

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If we study the binomial distribution for large numbers N and probabilities around 1/2

P(x)=\displaystyle\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}



which is represented below:

ID:(7793, 0)


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Video

Video: Gaussian distribution