パラメータ化可能な共分散行列を持つ正のk次元象限上の分布は何ですか？

負のシミュレーションに関する彼の問題に関する zzkの質問に続いて、共分散行列を設定できる正のk次元象限上の分布のパラメータ化されたファミリは何であるかと思います。 $\mathbb{R}_+^k$ $\Sigma$

zzkで説明したように、分布から開始し、線形変換しても機能しません。 $\mathbb{R}_+^k$ $X \longrightarrow\Sigma^{1/2} (X-\mu) + \mu$

distributions multivariate-analysis covariance

— 西安
ソース

回答:

我々は多変量正規乱数ベクトルを持っていると仮定しと及びフルランク対称正定行列。

(\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})

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

対数正規分布のためのその証明することは困難ではない $(X_1,\dots,X_k)$

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,

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

そして、ます。 $c_{ij}>-m_im_j$

したがって、我々は逆の質問をすることができる。所与の及び対称正定値行列を満足、、我々が許可すれば、 $m=(m_1,\dots,m_k)\in\mathbb{R}^k_+$ $k\times k$ $C=(c_{ij})$ $c_{ij}>-m_im_j$

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

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

所定の平均と共分散をもつ対数正規ベクトルを作成します。

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

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

およびの制約は、自然条件と同等です。 $C$ $m$ $\textrm{E}[X_i X_j]>0$

— 禅
ソース

すごい、パウロ！共分散行列に有効な解と適切な条件の両方があり、この質問にも答えています。最終的に、対数正規分布はガンマよりも便利です。

— 西安

実際、私は間違いなく歩行者の解決策を持っています。

でスタートとの値に合わせて2つのパラメータを選んで、。 $X_1\sim \text{Ga}(\alpha_{11},\beta_{1})$ $\mathbb{E}[X_1]$ $\text{var}(X_1)$
取るの値に適合するように3つのパラメータを選ぶ、、及び。 $X_2|X_1\sim \text{Ga}(\alpha_{21}X_1+\alpha_{22},\beta_{2})$ $\mathbb{E}[X_2]$ $\text{var}(X_2)$ $\text{cov}(X_1,X_2)$
取るの値に合わせて4つのパラメータが選択、、および $X_3|X_1,X_2\sim \text{Ga}(\alpha_{31}X_1+\alpha_{32}X_2+\alpha_{33},\beta_{3})$ $\mathbb{E}[X_3]$ $\text{var}(X_3)$ $\text{cov}(X_1,X_3)$ 。 $\text{cov}(X_2,X_3)$

など...ただし、パラメータの制約とモーメント方程式の非線形性を考えると、モーメントのいくつかのセットはパラメータの許容セットに対応しない可能性があります。

例えば、場合、Iは式のシステムで終わる $k=2$

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} .

$\dfrac{(\sigma_{12}+\mu_1\mu_2-\mu_2)^2}{(\sigma^2_1+\mu_1^2- \mu_1)^2} \sigma_1^2 + \dfrac{\mu_2}{\beta_2} = \sigma^2_2\,.$

μ

$\mu$

Σ

$\Sigma$

R_{+}^{2}

$\mathbb{R}_+^2$

更新（04/04）： deinstは、この質問を数学フォーラムの新しい質問として言い換えました。

— 西安
ソース

これをわずかに拡張する1つの方法は、自然の指数関数族を考慮することです。

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

$f(\mathbf{X}|\mathbf\theta)=h(\mathbf{x})e^{\mathbf\theta^T\mathbf{X}-A(\mathbf\theta)}.$ 次に、平均と共分散は勾配とヘッセ行列です

A

$A$ 。もし

h

$h$ 多項式（実指数> -1）である場合

A

$A$ は多項式（実数の指数を含む）の対数であり、分散とヘッセ行列は有理関数です。これにより、平均と共分散行列を表現するのに十分な自由度が得られると思います。

— -17:

@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
ソース

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

I am slightly (?) confused: you did not process the exponents as parameters of the exponential family. But indeed you can change those powers as you wish towards getting the 9 moment equations right.

— Xi'an

@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