Quais são as distribuições no quadrante k-dimensional positivo com matriz de covariância parametrizável?

Após a pergunta de zzk sobre seu problema com simulações negativas, estou me perguntando quais são as famílias parametrizadas de distribuições no quadrante k-dimensional positivo, para o qual a matriz de covariância pode ser definida. $\mathbb{R}_+^k$ $\Sigma$

Como discutido com ZZK , partindo de uma distribuição em $\mathbb{R}_+^k$ e aplicando a transformar linear $X \longrightarrow\Sigma^{1/2} (X-\mu) + \mu$ não funciona.

distributions multivariate-analysis covariance

— Xi'an
fonte

Respostas:

Suponha que tenhamos um vetor aleatório normal multivariado com e matriz definida positiva simétrica de posição completa .

(\log X_{1}, \dots, \log X_{k}) \sim N (μ, Σ),

$(\log X_1,\dots,\log X_k) \sim N(\mu,\Sigma) \, ,$

μ \in R^{k}

$\mu\in\mathbb{R}^k$

k \times k

$k\times k$

Σ = (σ_{i j})

$\Sigma=(\sigma_{ij})$

Para o lognormal , não é difícil provar que $(X_1,\dots,X_k)$

m_{i} := E [X_{i}] = e^{μ_{i} + σ_{i i} / 2}, i = 1, \dots, k,

$m_i := \textrm{E}[X_i] = e^{\mu_i + \sigma_{ii}/2} \, , \quad i=1,\dots,k\, ,$

c_{i j} := Cov [X_{i}, X_{j}] = m_{i} m_{j} (e^{σ_{i j}} - 1), i, j = 1, \dots, k,

$c_{ij} := \textrm{Cov}[X_i,X_j] = m_i \,m_j \,(e^{\sigma_{ij}} - 1) \, , \quad i,j=1,\dots,k\, ,$

e segue-se que . $c_{ij}>-m_im_j$

Portanto, podemos fazer a pergunta inversa: dado e matriz definida positiva simétrica , satisfazendo , se deixarmos $m=(m_1,\dots,m_k)\in\mathbb{R}^k_+$ $k\times k$ $C=(c_{ij})$ $c_{ij}>-m_im_j$

μ_{i} = \log m_{i} - \frac{1}{2} \log (\frac{c_{i i}}{m_{i}^{2}} + 1), i = 1, \dots, k,

$\mu_i = \log m_i - \frac{1}{2} \log\left(\frac{c_{ii}}{m_i^2} + 1 \right) \, , \quad i=1,\dots,k \, ,$

teremos um vetor lognormal com os meios e covariâncias prescritos.

σ_{i j} = \log (\frac{c_{i j}}{m_{i} m_{j}} + 1), i, j = 1, \dots, k,

$\sigma_{ij} = \log\left(\frac{c_{ij}}{m_i m_j} + 1 \right) \, , \quad i,j=1,\dots,k \, ,$

A restrição em e é equivalente à condição natural . $C$ $m$ $\textrm{E}[X_i X_j]>0$

— zen
fonte

Ótimo, Paulo! Você tem uma solução funcional e a condição adequada na matriz de covariância, que também responde a essa pergunta . Log-normais são mais úteis que gammas, no final.

— Xian

Na verdade, tenho uma solução definitivamente para pedestres.

Comece com e escolha os dois parâmetros para ajustar os valores de , . $X_1\sim \text{Ga}(\alpha_{11},\beta_{1})$ $\mathbb{E}[X_1]$ $\text{var}(X_1)$
Tome e escolha os três parâmetros para ajustar os valores de , e . $X_2|X_1\sim \text{Ga}(\alpha_{21}X_1+\alpha_{22},\beta_{2})$ $\mathbb{E}[X_2]$ $\text{var}(X_2)$ $\text{cov}(X_1,X_2)$
Tome e escolha os quatro parâmetros para ajustar os valores de , , e $X_3|X_1,X_2\sim \text{Ga}(\alpha_{31}X_1+\alpha_{32}X_2+\alpha_{33},\beta_{3})$ $\mathbb{E}[X_3]$ $\text{var}(X_3)$ $\text{cov}(X_1,X_3)$ . $\text{cov}(X_2,X_3)$

e assim por diante ... No entanto, dadas as restrições sobre os parâmetros e a natureza não linear das equações de momento, pode ser que alguns conjuntos de momentos correspondam a nenhum conjunto aceitável de parâmetros.

Por exemplo, quando , termino com o sistema de equações $k=2$

β_{1} = μ_{1} / σ_{1}^{2}, α_{11} - μ_{1} β_{1} = 0

$\beta_1 =\mu_1/\sigma_1^2\,,\quad \alpha_{11}-\mu_1\beta_1 =0$

α_{22} = μ_{2} β_{2} - α_{21} μ_{1}, α_{21} = \frac{(σ_{12} + μ_{1} μ_{2} - μ_{2})}{σ_{1}^{2} + μ_{1}^{2} - μ_{1}} β_{2}

$\alpha_{22} = \mu_2\beta_2 - \alpha_{21}\mu_1\,,\quad \alpha_{21} = \dfrac{(\sigma_{12}+\mu_1\mu_2-\mu_2)}{\sigma^2_1+\mu_1^2- \mu_1}\beta_2$

\frac{(σ_{12} + μ_{1} μ_{2} - μ_{2})^{2}}{(σ_{1}^{2} + μ_{1}^{2} - μ_{1})^{2}} σ_{1}^{2} + \frac{μ_{2}}{β_{2}} = σ_{2}^{2} .

$\dfrac{(\sigma_{12}+\mu_1\mu_2-\mu_2)^2}{(\sigma^2_1+\mu_1^2- \mu_1)^2} \sigma_1^2 + \dfrac{\mu_2}{\beta_2} = \sigma^2_2\,.$ Running an R code with arbitrary (and a priori acceptable) values for

μ

$\mu$ and

Σ

$\Sigma$ led to many cases with no solution. Again, this does not mean much because correlation matrices for distributions on

R_{+}^{2}

$\mathbb{R}_+^2$ may have stronger restrictions that a mere positive determinant.

update (04/04): deinst rephrased this question as a new question on the math forum.

— Xi'an
fonte

One way of slighly extending this is to consider the natural exponential family

f (X | θ) = h (x) e^{θ^{T} X - A (θ)} .

$f(\mathbf{X}|\mathbf\theta)=h(\mathbf{x})e^{\mathbf\theta^T\mathbf{X}-A(\mathbf\theta)}.$ Then the mean and covariance are the gradient and Hessian of

A

$A$ . If

h

$h$ is a polynomial (with real exponents > -1) then

A

$A$ is the log of a polynomial (with real exponents), and the variance and Hessian are rational functions. I think this gives enough freedom to represent any mean and covariance matrix.

— deinst

@deinst: (+1) Do you have an example where this exponential family representation can be exploited straightforwardly?

— Xi'an

Maybe I don't quite understand the problem. But, consider a bivariate random vector

(X, Y)

$(X,Y)$ with the same marginal

F

$F$ with full support on

R_{+}

$\mathbb R_{+}$ and having mean

0 < μ < \infty

$0 < \mu < \infty$ . How can such a bivariate distribution have correlation

ρ

$\rho$ close to -1, for example? Heuristically, though I haven't carried this out, it seems that if

P (X > 2 μ) > 0

$\mathbb P(X > 2 \mu) > 0$ , then a contradiction regarding the support must arise. No?

— cardinal

There are certainly constraints on the covariance matrix

Σ

$\Sigma$ when the support is

R_{+}^{k}

$\mathbb{R}^k_+$ , covered via the Stieltjes moment condition. Anyway, I do not see why a correlation close to -1 is excluded a priori.

— Xi'an

Right, this is related to what I was getting at. Regarding the correlation, consider my example. If

X

$X$ and

Y

$Y$ have the same marginal

F

$F$ with mean

μ

$\mu$ and a correlation of exactly -1 and

P (X > 2 μ) > 0

$\mathbb P(X > 2 \mu) > 0$ , what must the value of

Y

$Y$ be for all such realizations of

X

$X$ ? (+1 on both question and answer. I like this.)

— cardinal

OK, this is a response to Xi'an's comment. It is too long and has to much TeX to be a comfortable comment. Caveat Lector: It is virtually certain that I have made an algebra mistake. This does not seem to be quite as flexible as I first thought.

Let us create a family of distributions in $\mathbb{R}_+^3$ of the form

f (x | θ) = h (x) e^{- θ^{T} x - A (θ)}

$f(\mathbf{x}|\mathbf\theta)=h(\mathbf{x})e^{-\mathbf\theta^T\mathbf{x}-A(\mathbf\theta)}$ Let

x = (x, y, z)

$\mathbf{x}=(x,y,z)$ and

θ = (θ_{1}, θ_{2}, θ_{3})

$\mathbf\theta=(\theta_1,\theta_2,\theta_3)$ . Let

h (x) = c x_{1}^{e_{1} - 1} x_{2}^{e_{2} - 1} x_{3}^{e_{3} - 1} + d x_{1}^{f_{1} - 1} x_{2}^{f_{2} - 1} x_{3}^{f_{3} - 1}

$h(\mathbf{x})=c x_1^{e_1-1}x_2^{e_2-1}x_3^{e_3-1}+d x_1^{f_1-1}x_2^{f_2-1}x_3^{f_3-1}$ be a two term polynomial where

e_{i}, f_{i}

$e_i, f_i$ are real numbers greater than 0 for all

i

$i$ . Then we find that

A (θ) = \log (c \frac{Γ (e_{1})}{θ_{1}^{e_{1}}} \frac{Γ (e_{2})}{θ_{2}^{e_{2}}} \frac{Γ (e_{3})}{θ_{3}^{e_{3}}} + d \frac{Γ (f_{1})}{θ_{1}^{f_{1}}} \frac{Γ (f_{2})}{θ_{2}^{f_{2}}} \frac{Γ (f_{3})}{θ_{3}^{f_{3}}}) .

$A(\mathbf\theta)=\log\left(c\frac{\Gamma(e_1)}{\theta_1^{e_1}}\frac{\Gamma(e_2)}{\theta_2^{e_2}}\frac{\Gamma(e_3)}{\theta_3^{e_3}}+d\frac{\Gamma(f_1)}{\theta_1^{f_1}}\frac{\Gamma(f_2)}{\theta_2^{f_2}}\frac{\Gamma(f_3)}{\theta_3^{f_3}}\right).$

Now, for convenience let us define

c^{'} = c Γ (e_{1}) Γ (e_{2}) Γ (e_{2}) θ_{1}^{f_{1}} θ_{2}^{f_{2}} θ_{3}^{f_{3}}

$c'=c\Gamma(e_1)\Gamma(e_2)\Gamma(e_2)\theta_1^{f_1}\theta_2^{f_2}\theta_3^{f_3}$ and

d^{'} = d Γ (f_{1}) Γ (f_{2}) Γ (f_{2}) θ_{1}^{e_{1}} θ_{2}^{e_{2}} θ_{3}^{e_{3}}

$d'=d\Gamma(f_1)\Gamma(f_2)\Gamma(f_2)\theta_1^{e_1}\theta_2^{e_2}\theta_3^{e_3}$

Now, as the mean of our distribution is the gradient of $A$ , we have $\mu_X=\frac{e_1c'+f_1d'}{\theta_1(c'+d')}$ , $\mu_Y=\frac{e_2c'+f_2d'}{\theta_2(c'+d')}$ , and $\mu_Z=\frac{e_3c'+f_3d'}{\theta_3(c'+d')}$ . And as the covariance is the Hessian of $A$ , we have

σ_{X}^{2} = \frac{(e_{1} c^{'} + f_{1} d^{'}) (c^{'} + d^{'}) + (e_{1} - f_{1})^{2} c^{'} d^{'}}{θ_{1}^{2} (c^{'} + d^{'})^{2}}

$\sigma_X^2=\frac{(e_1c'+f_1d')(c'+d')+(e_1-f_1)^2c'd'}{\theta_1^2(c'+d')^2}$ and

Cov (X, Y) = \frac{(e_{1} - f_{1}) (e_{2} - f_{2}) c^{'} d^{'}}{θ_{1} θ_{2} (c^{'} + d^{'})}

$\text{Cov}(X,Y)=\frac{(e_1-f_1)(e_2-f_2)c'd'}{\theta_1\theta_2(c'+d')}$ (the other terms of the covariance matrix obtained by changing subscripts in the obvious way).

This does not seem to be quite enough flexibility to get any covariance matrix. I need to try another term in the polynomial (but I suspect that also may not work (obviously I need to think about this more)).

— deinst
fonte

Four parameters

(θ_{1}, θ_{2}, θ_{3}, c)

$(\theta_1,\theta_2,\theta_3,c)$ for five constraints...?

— Xi'an

@xian There are the 6 exponents

e_{i}

$e_i$ and

f_{i}

$f_i$ as well.

— deinst

Estou um pouco confuso: você não processou os expoentes como parâmetros da família exponencial. Mas, de fato, você pode alterar esses poderes conforme desejar para acertar as equações dos 9 momentos.

— Xian

@Xi'an You are correct, I did not process them as parameters of the exponential family. Doing so would have made the family no longer a natural family, and including them would have just muddled the algebra for comuting the moment equations (which was muddled enough to begin with).

— deinst