Varianz des Produkts von k korrelierten Zufallsvariablen

Was ist die Varianz des Produkts von korrelierten Zufallsvariablen? $k$

variance random-variable

— Jafar Mansouri
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Antworten:

Weitere Informationen zu diesem Thema, als Sie wahrscheinlich benötigen, finden Sie in Goodman (1962): "Die Varianz des Produkts von K Zufallsvariablen" , das Formeln sowohl für unabhängige Zufallsvariablen als auch für potenziell korrelierte Zufallsvariablen zusammen mit einigen Annäherungen herleitet. In einer früheren Veröffentlichung ( Goodman, 1960 ) wurde die Formel für das Produkt aus genau zwei Zufallsvariablen abgeleitet, die etwas einfacher (wenn auch immer noch ziemlich knorrig) ist. Wenn Sie die Ableitung verstehen möchten, ist dies möglicherweise ein besserer Ausgangspunkt .

Der Vollständigkeit halber geht es aber so.

Zwei Variablen

Es sei folgendes angenommen:

$x$ und $y$ sind zwei Zufallsvariablen
$X$ und $Y$ sind ihre (von Null verschiedenen) Erwartungen
$V(x)$ und $V(y)$ sind ihre Varianzen
$\delta_x = (x-X)/X$ (und ebenfalls für $\delta_y$ )
$D_{i,j} = E \left[ (\delta_x)^i (\delta_y)^j\right]$
$\Delta_x = x-X$ (und ebenfalls für $\Delta_y$ )
$E_{i,j} = E\left[(\Delta_x)^i (\Delta_y)^j\right]$
$G(x)$ ist der quadratische Variationskoeffizient: $V(x)/X^2$ (ebenfalls für $G(Y)$ )

Dann gilt: oder gleichwertig:

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}

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

Mehr als zwei Variablen

Das Papier von 1960 schlägt vor, dass dies eine Übung für den Leser ist (was anscheinend das Papier von 1962 motiviert hat!).

Die Notation ist ähnlich, mit einigen Erweiterungen:

sind die Zufallsvariablen anstelle von und $(x_1, x_2, \ldots x_n)$ $x$ $y$
$M = E\left( \prod_{i=1}^k x_i \right)$
$A = \left(M / \prod_{i=1}^k X_i\right) - 1$
= 0, 1 oder 2 für $s_i$ $i = 1, 2, \ldots k$
= Anzahl der Einsen in $u$ $(s_1, s_2, \ldots s_k)$
= Anzahl der 2er in $m$ $(s_1, s_2, \ldots s_k)$
für und für , $D(u,m) = 2^u - 2$ $m=0$ $2^u$ $m>1$
$C(s_1, s_2, \ldots, s_k) = D(u,m) \cdot E \left( \prod_{i=1}^k \delta_{x_i}^{s_i} \right)$
gibt die Summe der Mengen von mit $\sum_{s_1 \cdots s_k}$ $3^k - k -1$ $(s_1, s_2, \ldots s_k)$ $2m + u > 1$

Dann endlich:

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
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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

If you try to get the variance var(x*y) for independent random variables, via the formula for arbitrary k you can see that only a product and not a sum gives you the correct answer. In addition, if you look at the paper you can see it as well, on page 59 of the paper (at least in my version) he used a product instead of a sum.

— Nicolas Gisler

For the case of two random variables, an easier-to-read formula for the variance of the product of two correlated random variables can be found in this answer by @macro. This answer also points out the essential problem in

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
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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
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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