Skewness / kurtosis bergerak tertimbang eksponensial

Ada rumus on-line yang terkenal untuk menghitung rata-rata bergerak tertimbang secara eksponensial dan standar deviasi dari suatu proses $(x_n)_{n=0,1,2,\dots}$ . Untuk rata-rata,

$\mu_n = (1-\alpha) \mu_{n-1} + \alpha x_n$

dan untuk varians

$\sigma_n^2 = (1-\alpha) \sigma_{n-1}^2 + \alpha(x_n - \mu_{n-1})(x_n - \mu_n)$

dari mana Anda dapat menghitung simpangan baku.

Apakah ada rumus yang sama untuk perhitungan on-line dari momen ketiga dan keempat terpusat eksponensial? Intuisi saya adalah mereka harus mengambil formulir

$M_{3,n} = (1-\alpha) M_{3,n-1} + \alpha f(x_n,\mu_n,\mu_{n-1},S_n,S_{n-1})$

dan

$M_{4,n} = (1-\alpha) M_{4,n-1} + \alpha f(x_n,\mu_n,\mu_{n-1},S_n,S_{n-1},M_{3,n},M_{3,n-1})$

dari mana Anda dapat menghitung kemiringan dan kurtosis k n = M 4 , n / σ 4 n tetapi saya tidak dapat menemukan ekspresi bentuk-tertutup yang sederhana untuk fungsi-fungsi tersebut f dan g .$\gamma_n = M_{3,n} / \sigma_n^3$$k_n = M_{4,n}/\sigma_n^4$$f$$g$

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Sunting: Beberapa informasi lebih lanjut. Rumus pembaruan untuk varian bergerak adalah kasus khusus dari rumus untuk kovarians bergerak tertimbang eksponensial, yang dapat dihitung melalui

$C_n(x,y) = (1-\alpha) C_{n-1}(x,y) + \alpha (x_n - \bar{x}_n) (y_n - \bar{y}_{n-1})$

di mana dan ˉ y n adalah cara bergerak eksponensial dari x dan y . Asimetri antara x dan y adalah ilusi, dan menghilang ketika Anda memperhatikan bahwa y - ˉ y n = ( 1 - α ) ( y - ˉ y n - 1 ) .$\bar{x}_n$$\bar{y}_n$$x$$y$$x$$y$$y-\bar{y}_n = (1-\alpha) (y-\bar{y}_{n-1})$

Formulas like this can be computed by writing the central moment as an expectation $E_n(\cdot)$, where weights in the expectation are understood to be exponential, and using the fact that for any function $f(x)$ we have

$E_n(f(x)) = \alpha f(x_n) + (1-\alpha) E_{n-1}(f(x))$

It's easy to derive the updating formulas for the mean and variance using this relation, but it's proving to be more tricky for the third and fourth central moments.

The formulas are straightforward but they are not as simple as intimated in the question.

$Y$$X = x_n$$Y$$Z = \alpha X + (1 - \alpha)Y$ for a constant value $\alpha$. For notational convenience, set $\beta = 1-\alpha$. Let $F$ denote the CDF of a random variable and $\phi$ denote its moment generating function, so that

$$\phi_X(t) = \mathbb{E}_F[\exp(t X)] = \int_\mathbb{R}{\exp(t x) dF_X(x)}.$$

With Kendall and Stuart, let $\mu_k^{'}(Z)$ denote the non-central moment of order $k$ for the random variable $Z$; that is, $\mu_k^{'}(Z) = \mathbb{E}[Z^k]$. The skewness and kurtosis are expressible in terms of the $\mu_k^{'}$ for $k = 1,2,3,4$; for example, the skewness is defined as $\mu_3 / \mu_2^{3/2}$ where

$$\mu_3 = \mu_3^{'} - 3 \mu_2^{'}\mu_1^{'} + 2{\mu_1^{'}}^3 \text{ and }\mu_2 = \mu_2^{'} - {\mu_1^{'}}^2$$

are the third and second central moments, respectively.

By standard elementary results,

$$\eqalign{ &1 + \mu_1^{'}(Z) t + \frac{1}{2!} \mu_2^{'}(Z) t^2 + \frac{1}{3!} \mu_3^{'}(Z) t^3 + \frac{1}{4!} \mu_4^{'}(Z) t^4 +O(t^5) \cr &= \phi_Z(t) \cr &= \phi_{\alpha X}(t) \phi_{\beta Y}(t) \cr &= \phi_X(\alpha t) \phi_Y(\beta t) \cr &= (1 + \mu_1^{'}(X) \alpha t + \frac{1}{2!} \mu_2^{'}(X) \alpha^2 t^2 + \cdots) (1 + \mu_1^{'}(Y) \beta t + \frac{1}{2!} \mu_2^{'}(Y) \beta^2 t^2 + \cdots). }$$

To obtain the desired non-central moments, multiply the latter power series through fourth order in $t$ and equate the result term-by-term with the terms in $\phi_Z(t)$.

I think that the following updating formula works for the third moment, although I'd be glad to have someone check it:

$M_{3,n} = (1-\alpha)M_{3,n-1} + \alpha \Big[ x_n(x_n-\mu_n)(x_n-2\mu_n) - x_n\mu_{n-1}(\mu_{n-1}-2\mu_n) - \dots$ $\dots - \mu_{n-1}(\mu_n-\mu_{n-1})^2 - 3(x_n-\mu_n) \sigma_{n-1}^2 \Big]$

Updating formula for the kurtosis still open...