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# System prepended metadata

title: 各種 Distribution 的證明
tags: [probability]

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# 各種 Distribution 的證明
## Binomial:

$f(x) = C^n_x \cdot p^x \cdot (1-p)^{n-x}$ <br/>

$M(t) = E(e^{xt}) = \Sigma^n_{x = 0} e^{xt} \cdot C^n_x \cdot p^x \cdot (1-p)^{n-x}$ <br/> $= ((1-p) + pe^t)^n$

## Negative Binomial

$f(x) = C^{x-1}_{r-1} \cdot p^r \cdot (1-p)^{x-r}$ <br/>

$M(t) = E(e^{xt}) = \Sigma^{\infty}_{x = r} e^{xt} \cdot C^{x-1}_{r-1} \cdot p^r \cdot (1-p)^{x-r}$ 

$= (\cfrac{p}{1-p})^r \cdot \Sigma^{\infty}_{x = r} \cdot C^{x-1}_{r-1} \cdot (e^t(1-p))^x$

$令 y = x-r, x = y+r \Longrightarrow \because 在上述 \Sigma中 x = r$

$\Longrightarrow = (\cfrac{p}{1-p})^r \cdot \Sigma^{\infty}_{y = 0} \cdot C^{y+r-1}_{r-1} \cdot (e^t(1-p))^{(y+r)}$

$by 泰勒展開式 : (1-x)^{-r} = \Sigma^{\infty}_0C^{r+k-1}_{k-1} x^k$

$\Longrightarrow = (pe^t)^r \cdot (1-e^t(1-p))^{-r} = (\cfrac{pe^t}{1-e^t(1-p)})^r$

## Poisson

$f(x) = \cfrac{\lambda^xe^{-\lambda}}{x!}$

$M(t) = E(e^{xt}) = \Sigma^{\infty}_{x=0}e^{xt}(\cfrac{\lambda^xe^{-\lambda}}{x!}) = e^{-\lambda} \cdot \Sigma^{\infty}_{x=0}(\cfrac{(\lambda e^t)^x}{x!})$

$\because e^x = 1 + \Sigma^{\infty}_{x=1} \cfrac{(x)^n}{n!}$

$\Longrightarrow = e^{-\lambda} \cdot e^{\lambda e^t} = e^{\lambda (e^t-1)}$



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## Exponential

$f(x) = \cfrac{1}{\theta} e^{({-x}/{\theta})}$

$M(t) = E(e^{xt}) = \int^{\infty}_0e^{xt}(\cfrac{1}{\theta})e^{({-x}/{\theta})}dx = \int^{\infty}_0 \cfrac{1}{\theta}e^{(-x(1- \theta t)/ \theta )}dx$

$= [- \cfrac{e^{-x(1-\theta t)/ \theta}}{1- \theta t}]^{\infty}_{0} = \cfrac{1}{1- \theta t}$

## Gamma

$f(x) = \cfrac{x^{(\alpha -1)}e^{-x/ \theta}}{(\alpha -1)! \theta^{ \alpha}}$

$M(t) = E(e^{xt}) = \int^{\infty}_0e^{xt} \cdot \cfrac{x^{(\alpha -1)}e^{-x/ \theta}}{(\alpha -1)! \theta^{ \alpha}}dx$

$= \cfrac{1}{(\alpha -1)! \theta^{ \alpha}} \int^{\infty}_0x^{(\alpha -1)}e^{x(t-(1/ \theta))}dx = \cfrac{1}{(1- \theta t)^{\alpha}}$

## Normal

$f(x) = \cfrac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^2/(2 \sigma ^2)}$

$M(t) = E(e^{xt}) = \int^{\infty}_0e^{xt} \cdot \cfrac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^2/(2 \sigma ^2)}$

$= e^{(\mu t + \sigma ^2 t^2/2)}$


###### tags: `probability`