Expectation of X^n; X~N(u,sigma^2)

출처: http://en.wikipedia.org/wiki/Moment-generating_function , http://en.wikipedia.org/wiki/Normal_distribution

When X~N(u,sigma^2)

then Expectation of X^n can be solved easily using MGF(Moment Generating Function)

Order	Raw moment	Central moment	Cumulant
1	μ	0	μ
2	μ² + σ²	σ²	σ²
3	μ³ + 3μσ²	0	0
4	μ⁴ + 6μ²σ² + 3σ⁴	3σ⁴	0
5	μ⁵ + 10μ³σ² + 15μσ⁴	0	0
6	μ⁶ + 15μ⁴σ² + 45μ²σ⁴ + 15σ⁶	15σ⁶	0
7	μ⁷ + 21μ⁵σ² + 105μ³σ⁴ + 105μσ⁶	0	0
8	μ⁸ + 28μ⁶σ² + 210μ⁴σ⁴ + 420μ²σ⁶ + 105σ⁸	105σ⁸	0

일반적인 경우의 MGF는 다음과 같다.

If X has a continuous probability density function ƒ(x), then M_X(−t) is the two-sided Laplace transform of ƒ(x).

$\begin{align} M_X(t) & = \int_{-\infty}^\infty e^{tx} f(x)\,dx \\ & = \int_{-\infty}^\infty \left( 1+ tx + \frac{t^2x^2}{2!} + \cdots\right) f(x)\,dx \\ & = 1 + tm_1 + \frac{t^2m_2}{2!} +\cdots, \end{align}$

where m_i is the ith moment.

여러 분포 함수들의 MGF 는 다음과 같다.

Distribution	Moment-generating function M_X(t)	Characteristic function φ(t)
Bernoulli $\, P(X=1)=p$	$\, 1-p+pe^t$	$\, 1-p+pe^{it}$
Geometric $(1 - p)^{k-1}\,p\!$	$\frac{pe^t}{1-(1-p) e^t}\!$ , for $t<-\ln(1-p)\!$	$\frac{pe^{it}}{1-(1-p)\,e^{it}}\!$
Binomial B(n, p)	$\, (1-p+pe^t)^n$	$\, (1-p+pe^{it})^n$
Poisson Pois(λ)	$\, e^{\lambda(e^t-1)}$	$\, e^{\lambda(e^{it}-1)}$
Uniform U(a, b)	$\, \frac{e^{tb} - e^{ta}}{t(b-a)}$	$\, \frac{e^{itb} - e^{ita}}{it(b-a)}$
Normal N(μ, σ²)	$\, e^{t\mu + \frac{1}{2}\sigma^2t^2}$	$\, e^{it\mu - \frac{1}{2}\sigma^2t^2}$
Chi-square χ²_k	$\, (1 - 2t)^{-k/2}$	$\, (1 - 2it)^{-k/2}$
Gamma Γ(k, θ)	$\, (1 - t\theta)^{-k}$	$\, (1 - it\theta)^{-k}$
Exponential Exp(λ)	$\, (1 - t\lambda^{1})^{-1}$	$\, (1 - it\lambda^{1})^{-1}$
Multivariate normal N(μ, Σ)	$\, e^{t^\mathrm{T} \mu + \frac{1}{2} t^\mathrm{T} \Sigma t}$	$\, e^{i t^\mathrm{T} \mu - \frac{1}{2} t^\mathrm{T} \Sigma t}$
Degenerate δ_a	$\, e^{ta}$	$\, e^{ita}$
Laplace L(μ, b)	$\, \frac{e^{t\mu}}{1 - b^2t^2}$	$\, \frac{e^{it\mu}}{1 + b^2t^2}$
Cauchy Cauchy(μ, θ)	not defined	$\, e^{it\mu -\theta\|t\|}$
Negative Binomial NB(r, p)	$\, \frac{(1-p)^r}{(1-pe^t)^r}$	$\, \frac{(1-p)^r}{(1-pe^{it})^r}$

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Expectation of X^n; X~N(u,sigma^2)

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