各種 Distribution 的證明

Binomial:

f (x) = C_{x}^{n} \cdot p^{x} \cdot (1 - p)^{n - x}

M (t) = E (e^{x t}) = Σ_{x = 0}^{n} e^{x t} \cdot C_{x}^{n} \cdot p^{x} \cdot (1 - p)^{n - x}

= ((1 - p) + p e^{t})^{n}

f (x) = C_{r - 1}^{x - 1} \cdot p^{r} \cdot (1 - p)^{x - r}

M (t) = E (e^{x t}) = Σ_{x = r}^{\infty} e^{x t} \cdot C_{r - 1}^{x - 1} \cdot p^{r} \cdot (1 - p)^{x - r}

= (\frac{p}{1 - p})^{r} \cdot Σ_{x = r}^{\infty} \cdot C_{r - 1}^{x - 1} \cdot (e^{t} (1 - p))^{x}

令 y = x - r, x = y + r ⟹∵ 在 上 述 Σ 中 x = r

⟹= (\frac{p}{1 - p})^{r} \cdot Σ_{y = 0}^{\infty} \cdot C_{r - 1}^{y + r - 1} \cdot (e^{t} (1 - p))^{(y + r)}

b y 泰 勒 展 開 式 : (1 - x)^{- r} = Σ_{0}^{\infty} C_{k - 1}^{r + k - 1} x^{k}

⟹= (p e^{t})^{r} \cdot (1 - e^{t} (1 - p))^{- r} = (\frac{p e^{t}}{1 - e^{t} (1 - p)})^{r}

f (x) = \frac{λ^{x} e^{- λ}}{x!}

M (t) = E (e^{x t}) = Σ_{x = 0}^{\infty} e^{x t} (\frac{λ^{x} e^{- λ}}{x!}) = e^{- λ} \cdot Σ_{x = 0}^{\infty} (\frac{(λ e^{t})^{x}}{x!})

∵ e^{x} = 1 + Σ_{x = 1}^{\infty} \frac{(x)^{n}}{n!}

⟹= e^{- λ} \cdot e^{λ e^{t}} = e^{λ (e^{t} - 1)}

f (x) = \frac{1}{θ} e^{(- x / θ)}

M (t) = E (e^{x t}) = \int_{0}^{\infty} e^{x t} (\frac{1}{θ}) e^{(- x / θ)} d x = \int_{0}^{\infty} \frac{1}{θ} e^{(- x (1 - θ t) / θ)} d x

= [- \frac{e^{- x (1 - θ t) / θ}}{1 - θ t}]_{0}^{\infty} = \frac{1}{1 - θ t}

f (x) = \frac{x^{(α - 1)} e^{- x / θ}}{(α - 1)! θ^{α}}

M (t) = E (e^{x t}) = \int_{0}^{\infty} e^{x t} \cdot \frac{x^{(α - 1)} e^{- x / θ}}{(α - 1)! θ^{α}} d x

= \frac{1}{(α - 1)! θ^{α}} \int_{0}^{\infty} x^{(α - 1)} e^{x (t - (1 / θ))} d x = \frac{1}{(1 - θ t)^{α}}

f (x) = \frac{1}{σ \sqrt{2 π}} e^{- (x - μ)^{2} / (2 σ^{2})}

M (t) = E (e^{x t}) = \int_{0}^{\infty} e^{x t} \cdot \frac{1}{σ \sqrt{2 π}} e^{- (x - μ)^{2} / (2 σ^{2})}

= e^{(μ t + σ^{2} t^{2} / 2)}