k個の相関ランダム変数の積の分散

$k$ 相関ランダム変数の積の分散はどのくらいですか？

variance random-variable

回答:

このトピックに関する必要以上の情報は、Goodman（1962）：「K Random Variablesの積の分散」で見つけることができます。以前の論文（Goodman、1960）で、正確に2つのランダム変数の積の式が導出されました。これはやや単純です（まだかなり厄介ですが）。。

ただし、完全を期すために、このようにします。

2つの変数

以下を想定します。

$x$ と $y$ は2つのランダム変数です
$X$ および $Y$ は（ゼロ以外の）期待値です
$V(x)$ および $V(y)$ はそれらの分散です
$\delta_x = (x-X)/X$ （及び同様のため $\delta_y$ ）
$D_{i,j} = E \left[ (\delta_x)^i (\delta_y)^j\right]$
$\Delta_x = x-X$ （及び同様のため $\Delta_y$ ）
$E_{i,j} = E\left[(\Delta_x)^i (\Delta_y)^j\right]$
$G(x)$ は2乗変動係数： $V(x)/X^2$ （も同様 $G(Y)$ ）

その後：またはそれと同等：

V (x y) = (X Y)^{2} [G (y) + G (x) + 2 D_{1, 1} + 2 D_{1, 2} + 2 D_{2, 1} + D_{2, 2} - D_{1, 1}^{2}]

$V(xy) = (XY)^2[G(y) + G(x) + 2D_{1,1} + 2D_{1,2} + 2D_{2,1} + D_{2,2} - D_{1,1}^2]$

V (x y) = X^{2} V (y) + Y^{2} V (x) + 2 X Y E_{1, 1} + 2 X E_{1, 2} + 2 Y E_{2, 1} + E_{2, 2} - E_{1, 1}^{2}

$V(xy) = X^2V(y) + Y^2V(x) + 2XYE_{1,1} + 2XE_{1,2} + 2YE_{2,1} + E_{2,2} - E_{1,1}^2$

3つ以上の変数

1960年の論文は、これが読者にとっての演習であることを示唆しています（1962年の論文の動機付けになったようです！）。

表記は似ていますが、いくつかの拡張子があります。

および代わりにランダム変数 $(x_1, x_2, \ldots x_n)$ $x$ $y$
$M = E\left( \prod_{i=1}^k x_i \right)$
$A = \left(M / \prod_{i=1}^k X_i\right) - 1$
0、1、または2 = $s_i$ $i = 1, 2, \ldots k$
=の1の数 $u$ $(s_1, s_2, \ldots s_k)$
= $m$ $(s_1, s_2, \ldots s_k)$
のためのおよびのために、 $D(u,m) = 2^u - 2$ $m=0$ $2^u$ $m>1$
$C(s_1, s_2, \ldots, s_k) = D(u,m) \cdot E \left( \prod_{i=1}^k \delta_{x_i}^{s_i} \right)$
は、セットの合計を示しここで、 $\sum_{s_1 \cdots s_k}$ $3^k - k -1$ $(s_1, s_2, \ldots s_k)$ $2m + u > 1$

そして、ついに：

V (\prod_{i = 1}^{k} x_{i}) = \prod X_{i}^{2} (\sum_{s_{1} \dots s_{k}} C (s_{1}, s_{2} \dots s_{k}) - A^{2})

$V\left(\prod_{i=1}^k x_i\right) = \prod X_i^2 \left( \sum_{s_1 \cdots s_k} C(s_1, s_2 \ldots s_k) - A^2\right)$

See the papers for details and slightly more tractable approximations!

— Matt Krause
ソース

please note, that the above answer from Matt Krause contains a mistake as well as the paper itself. In the definition of the function C(s1,...,sk) it should be a product instead of a sum.

— Nicolas Gisler

Could you elaborate a little bit more..? "Because I - an anonymous person from the Internet - say so" is not really an answer...

— Tim

独立したランダム変数の分散var（x * y）を取得しようとすると、任意のkの式を介して、合計ではなく積のみが正しい答えを与えることがわかります。さらに、紙を見ると、それも見ることができます。紙の59ページ（少なくとも私のバージョンでは）で、彼は合計ではなく製品を使用していました。

— ニコラスギスラー

V (x y) = X^{2} V (y) + Y^{2} V (x) + 2 X Y E_{1, 1} + 2 X E_{1, 2} + 2 Y E_{2, 1} + E_{2, 2} - E_{1, 1}^{2},

$V(xy) = X^2V(y) + Y^2V(x) + 2XYE_{1,1} + 2XE_{1,2} + 2YE_{2,1} + E_{2,2} - E_{1,1}^2,$ viz., the thicket of notation conceals the essential fact that there are terms in it whose value cannot be determined unless we know cov

(x^{2}, y^{2})

$(x^2,y^2)$ , or enough about the joint density of the two random variables to determine this quantity.

— Dilip Sarwate

An edit suggestion, that should really have been a comment, suggested that the original paper contained a typo where a sum and product were mixed up and this answer should be amended. See stats.stackexchange.com/review/suggested-edits/83662

— Silverfish

Just to add to the awesome answer of Matt Krause (in fact easily derivable from there). If x, y are independent then,

\begin{aligned} E_{1, 1} & = E [(x - E [x]) (y - E [y])] = C o v (x, y) = 0 \\ E_{1, 2} & = E [(x - E [x]) (y - E [y])^{2}] \\ = E [x - E (x)] E [(y - E [y])^{2}] \\ = (E [x] - E [x]) E [(y - E [y])^{2}] = 0 \\ E_{2, 1} & = 0 \\ E_{2, 2} & = E [(x - E [x])^{2} (y - E [y])^{2}] \\ = E [(x - E [x])^{2}] E [(y - E [y])^{2} \\ = V [x] V [y] \\ V [x y] & = E [x]^{2} V [y] + E [y]^{2} V [x] + V [x] V [y] \end{aligned}

$\begin{equation*} \begin{split} E_{1,1} &= E[(x-E[x])(y-E[y])] = Cov(x,y) = 0\\ E_{1,2} &= E[(x-E[x])(y-E[y])^2] \\ &= E[x-E(x)]E[(y-E[y])^2] \\ &= (E[x]-E[x])E[(y-E[y])^2]=0\\ E_{2,1} &= 0\\ E_{2,2} &= E[(x-E[x])^2(y-E[y])^2]\\ &= E[(x-E[x])^2]E[(y-E[y])^2\\ &= V[x]V[y]\\ V[xy] &= E[x]^2 V[y] + E[y]^2 V[x] + V[x]V[y] \end{split} \end{equation*}$

— Ananda
ソース

The result for the case of

n

$n$ independent random variables has been discussed here.

— Dilip Sarwate

In addition to the general formula given by Matt it may be worth noting that there is a somewhat more explicit formula for zero mean Gaussian random variables. It follows from Isserlis' theorem, see also Higher moments for the centered multivariate normal distribution.

Suppose that $(x_1, \ldots, x_k)$ follows a multivariate normal distribution with mean 0 and covariance matrix $\Sigma$ . If the number of variables $k$ is odd, $E\left(\prod_i x_i\right) = 0$ and

V (\prod_{i} x_{i}) = E (\prod_{i} x_{i}^{2}) = \sum \prod {\tilde{Σ}}_{i, j}

$V\left(\prod_i x_i\right) = E\left( \prod_i x_i^2\right) = \sum \prod \tilde{\Sigma}_{i,j}$ where

Σ \prod

$\Sigma \prod$ means sum over all partitions of

{1, \dots, 2 k}

$\{1, \ldots, 2k\}$ into

k

$k$ disjoint pairs

{i, j}

$\{i, j\}$ with each term being a product of the corresponding

k

$k$

{\tilde{Σ}}_{i, j}

$\tilde{\Sigma}_{i,j}$ 's, and where

\tilde{Σ} = (\begin{array}{cc} Σ & Σ \\ Σ & Σ \end{array})

$\tilde{\Sigma} = \left( \begin{array}{cc} \Sigma & \Sigma \\ \Sigma & \Sigma \end{array} \right)$ is the covariance matrix for

(x_{1}, \dots, x_{k}, x_{1}, \dots, x_{k})

$(x_1, \ldots, x_k, x_1, \ldots, x_k)$ . If

k

$k$ is even,

V (\prod_{i} x_{i}) = \sum \prod {\tilde{Σ}}_{i, j} - {(\sum \prod Σ_{i, j})}^{2} .

$V\left(\prod_i x_i\right) = \sum \prod \tilde{\Sigma}_{i,j} - \left(\sum \prod \Sigma_{i,j}\right)^2.$ In the case

k = 2

$k = 2$ we get

V (x_{1} x_{2}) = Σ_{1, 1} Σ_{2, 2} + 2 (Σ_{1, 2})^{2} - Σ_{1, 2}^{2} = Σ_{1, 1} Σ_{2, 2} + (Σ_{1, 2})^{2} .

$V(x_1x_2) = \Sigma_{1,1} \Sigma_{2,2} + 2 (\Sigma_{1,2})^2 - \Sigma_{1,2}^2 = \Sigma_{1,1} \Sigma_{2,2} + (\Sigma_{1,2})^2.$ If

k = 3

$k = 3$ we get

V (x_{1} x_{2} x_{3}) = \sum Σ_{i, j} Σ_{k, l} Σ_{r, t},

$V(x_1x_2x_3) = \sum \Sigma_{i,j}\Sigma_{k,l}\Sigma_{r,t},$ where there are 15 terms in the sum.

It is, in fact, possible to implement the general formula. The most difficult part appears to be the computation of the required partitions. In R, this can be done with the function setparts from the package partitions. Using this package it was no problem to generate the 2,027,025 partitions for $k = 8$ , the 34,459,425 partitions for $k = 9$ could also be generated, but not the 654,729,075 partitions for $k = 10$ (on my 16 GB laptop).

A couple of other things are worth noting. First, for Gaussian variables with non-zero mean it should be possible to derive an expression as well from Isserlis' theorem. Second, it is unclear (to me) if the above formula is robust against deviations from normality, that is, if it can be used as an approximation even if the variables are not multivariate normally distributed. Third, though the formulas above are correct, it is questionable how much the variance tells about the distribution of the products. Even for $k = 2$ the distribution of the product is quite leptokurtic, and for larger $k$ it quickly becomes extremely leptokurtic.

— NRH
ソース

Neat approach! For what it's worth, the formula in my answer also has a combinatorial blow-up: the summation over C involves summing

O (3^{k})

$O(3^k)$ terms.

— Matt Krause