Apa arti dan varian dari normal multivariat 0-disensor?

Pertama-tama kita dapat mengurangi ini hanya bergantung pada momen-momen tertentu dari distribusi normal univariat / bivariat terpotong: perhatikan tentu saja bahwa $\DeclareMathOperator{\E}{\mathbb E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} \newcommand{\N}{\mathcal N} \newcommand{\T}{\tilde} \newcommand{\v}{\mathcal V}$

\begin{matrix} E [Z_{+}] = {[\begin{matrix} E [(Z_{i})_{+}] \end{matrix}]}_{i} \\ Cov (Z_{+}) = {[\begin{matrix} Cov ((Z_{i})_{+}, (Z_{j})_{+}) \end{matrix}]}_{i j}, \end{matrix}

$\begin{gather} \E[Z_+] = \begin{bmatrix} \E[(Z_i)_+] \end{bmatrix}_i \\ \Cov(Z_+) = \begin{bmatrix} \Cov\left( (Z_i)_+, (Z_j)_+ \right) \end{bmatrix}_{ij} ,\end{gather}$ dan karena kami membuat transformasi bijaksana dimensi tertentu dari distribusi normal, kami hanya perlu khawatir tentang mean dan varians dari sensor normal 1d disensor dan kovarians dari dua sensor normal 1d disensor.

Kami akan menggunakan beberapa hasil dari

S Rosenbaum (1961). Momen dari Distribusi Normal Bivariat Terpotong . JRSS B, vol 23 hlm 405-408. ( jstor )

Rosenbaum menganggap dan pertimbangkan pemotongan ke acara .

[\begin{matrix} \tilde{X} \\ \tilde{Y} \end{matrix}] \sim N ([\begin{matrix} 0 \\ 0 \end{matrix}], [\begin{matrix} 1 & ρ \\ ρ & 1 \end{matrix}]),

$\begin{bmatrix}\T X \\ \T Y\end{bmatrix} \sim \N\left( \begin{bmatrix}0 \\ 0\end{bmatrix}, \begin{bmatrix}1 & \rho \\ \rho & 1\end{bmatrix} \right) ,$

V = {\tilde{X} \geq a_{X}, \tilde{Y} \geq a_{Y}}

$\v = \{ \T X \ge a_X, \T Y \ge a_Y \}$

Secara khusus, kami akan menggunakan tiga hasil berikut ini, (1), (3), dan (5). Pertama, tentukan yang berikut:

\begin{matrix} q_{x} = ϕ (a_{x}) q_{y} = ϕ (a_{y}) \\ Q_{x} = Φ (- a_{x}) Q_{y} = Φ (- a_{y}) \\ R_{x y} = Φ (\frac{ρ a_{x} - a_{y}}{\sqrt{1 - ρ^{2}}}) R_{y x} = Φ (\frac{ρ a_{y} - a_{x}}{\sqrt{1 - ρ^{2}}}) \\ r_{x y} = \frac{\sqrt{1 - ρ^{2}}}{\sqrt{2 π}} ϕ (\sqrt{\frac{h^{2} - 2 ρ h k + k^{2}}{1 - ρ^{2}}}) \end{matrix}

$\begin{gather} q_x = \phi( a_x) \qquad q_y = \phi( a_y) \\ Q_x = \Phi(-a_x) \qquad Q_y = \Phi(-a_y) \\ R_{xy} = \Phi\left( \frac{\rho a_x - a_y}{\sqrt{1 - \rho^2}} \right) \qquad R_{yx} = \Phi\left( \frac{\rho a_y - a_x}{\sqrt{1 - \rho^2}} \right) \\ r_{xy} = \frac{\sqrt{1-\rho^2}}{\sqrt{2 \pi}} \phi\left( \sqrt{\frac{h^2 - 2 \rho h k + k^2}{1 - \rho^2}} \right) \end{gather}$

Sekarang, Rosenbaum menunjukkan bahwa:

\begin{aligned} (1) & Pr (V) E [\tilde{X} ∣ V] & = q_{x} R_{x y} + ρ q_{y} R_{y x} \\ (3) & Pr (V) E [{\tilde{X}}^{2} ∣ V] & = Pr (V) + a_{x} q_{x} R_{x y} + ρ^{2} a_{y} q_{y} R_{y x} + ρ r_{x y} \\ (5) & Pr (V) E [\tilde{X} \tilde{Y} ∣ V] & = ρ Pr (V) + ρ a_{x} q_{x} R_{x y} + ρ a_{y} q_{y} R_{y x} + r_{x y} . \end{aligned}

$\begin{align} \Pr(\v) \E[\T X \mid \v] &= q_x R_{xy} + \rho q_y R_{yx} \tag{1} \\ \Pr\left(\v \right) \E\left[\T X^2 \mid \v \right] &= \Pr\left(\v \right) + a_x q_x R_{xy} + \rho^2 a_y q_y R_{yx} + \rho r_{xy} \tag{3} \\ \Pr(\v) \E\left[ \T X \T Y \mid \v \right] &= \rho \Pr(\v) + \rho a_x q_x R_{xy} + \rho a_y q_y R_{yx} + r_{xy} \tag{5} .\end{align}$

Akan berguna juga untuk mempertimbangkan kasus khusus (1) dan (3) dengan , yaitu pemotongan 1d: $a_y = -\infty$

\begin{aligned} (*) & Pr (V) E [\tilde{X} ∣ V] & = q_{x} \\ (**) & Pr (V) E [{\tilde{X}}^{2} ∣ V] & = Pr (V) = Q_{x} . \end{aligned}

$\begin{align} \Pr(\v) \E[\T X \mid \v] &= q_x \tag{*} \\ \Pr\left(\v \right) \E\left[\T X^2 \mid \v \right] &= \Pr\left(\v \right) = Q_x \tag{**} .\end{align}$

Kami sekarang ingin mempertimbangkan

\begin{aligned} [\begin{matrix} X \\ Y \end{matrix}] & = [\begin{matrix} μ_{x} \\ μ_{y} \end{matrix}] + [\begin{matrix} σ_{x} & 0 \\ 0 & σ_{y} \end{matrix}] [\begin{matrix} \tilde{X} \\ \tilde{Y} \end{matrix}] \\ \sim N ([\begin{matrix} μ_{X} \\ μ_{Y} \end{matrix}], [\begin{matrix} σ_{x}^{2} & ρ σ_{x} σ_{y} \\ ρ σ_{x} σ_{y} & σ_{y}^{2} \end{matrix}]) \\ = N (μ, Σ) . \end{aligned}

$\begin{align} \begin{bmatrix}X \\ Y\end{bmatrix} &= \begin{bmatrix}\mu_x\\\mu_y\end{bmatrix} + \begin{bmatrix}\sigma_x & 0 \\ 0 & \sigma_y\end{bmatrix}\begin{bmatrix}\T X \\ \T Y\end{bmatrix} \\&\sim \N\left( \begin{bmatrix}\mu_X\\\mu_Y\end{bmatrix}, \begin{bmatrix}\sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix} \right) \\&= \N\left( \mu, \Sigma \right) .\end{align}$

Kita akan menggunakan yang merupakan nilai dan ketika , .

a_{x} = \frac{- μ_{x}}{σ_{x}} a_{y} = \frac{- μ_{y}}{σ_{y}},

$a_x = \frac{-\mu_x}{\sigma_x} \qquad a_y = \frac{-\mu_y}{\sigma_y} ,$

\tilde{X}

$\T X$

\tilde{Y}

$\T Y$

X = 0

$X = 0$

Y = 0

$Y = 0$

Sekarang, menggunakan (*), kita memperoleh dan menggunakan keduanya (*) dan (**) menghasilkan sehingga

\begin{aligned} E [X_{+}] & = Pr (X_{+} > 0) E [X ∣ X > 0] + Pr (X_{+} = 0) 0 \\ = Pr (X > 0) (μ_{x} + σ_{x} E [\tilde{X} ∣ \tilde{X} \geq a_{x}]) \\ = Q_{x} μ_{x} + q_{x} σ_{x}, \end{aligned}

$\begin{align} \E[ X_+ ] &= \Pr(X_+ > 0) \E[X \mid X > 0] + \Pr(X_+=0) \, 0 \\&= \Pr(X > 0) \left( \mu_x + \sigma_x \E[\T X \mid \T X \ge a_x] \right) \\&= Q_x \mu_x + q_x \sigma_x ,\end{align}$

\begin{aligned} E [X_{+}^{2}] & = Pr (X_{+} > 0) E [X^{2} ∣ X > 0] + Pr (X_{+} = 0) 0 \\ = Pr (\tilde{X} \geq a_{x}) E [(μ_{x} + σ_{x} \tilde{X})^{2} ∣ \tilde{X} \geq a_{x}] \\ = Pr (\tilde{X} \geq a_{x}) E [μ_{x}^{2} + μ_{x} σ_{x} \tilde{X} + σ_{x}^{2} {\tilde{X}}^{2} ∣ \tilde{X} \geq a_{x}] \\ = Q_{x} μ_{x}^{2} + q_{x} μ_{x} σ_{x} + Q_{x} σ_{x}^{2} \end{aligned}

$\begin{align} \E[ X_+^2 ] &= \Pr(X_+ > 0) \E[X^2 \mid X > 0] + \Pr(X_+=0) 0 \\&= \Pr\left(\T X \ge a_x\right) \E\left[(\mu_x + \sigma_x \T X)^2 \mid \T X \ge a_x\right] \\&= \Pr\left(\T X \ge a_x\right) \E\left[\mu_x^2 + \mu_x \sigma_x \T X + \sigma_x^2 \T X^2 \mid \T X \ge a_x\right] \\&= Q_x \mu_x^2 + q_x \mu_x \sigma_x + Q_x \sigma_x^2 \end{align}$

\begin{aligned} Var [X_{+}] & = E [X_{+}^{2}] - E [X_{+}]^{2} \\ = Q_{x} μ_{x}^{2} + q_{x} μ_{x} σ_{x} + Q_{x} σ_{x}^{2} - Q_{x}^{2} μ_{x}^{2} - q_{x}^{2} σ_{x}^{2} - 2 q_{x} Q_{x} μ_{x} σ_{x} \\ = Q_{x} (1 - Q_{x}) μ_{x}^{2} + (1 - 2 Q_{x}) q_{x} μ_{x} σ_{x} + (Q_{x} - q_{x}^{2}) σ_{x}^{2} . \end{aligned}

$\begin{align} \Var[X_+] &= \E[X_+^2] - \E[X_+]^2 \\&= Q_x \mu_x^2 + q_x \mu_x \sigma_x + Q_x \sigma_x^2 - Q_x^2 \mu_x^2 - q_x^2 \sigma_x^2 - 2 q_x Q_x \mu_x \sigma_x \\&= Q_x (1 - Q_x) \mu_x^2 + (1 - 2 Q_x) q_x \mu_x \sigma_x + (Q_x - q_x^2) \sigma_x^2 .\end{align}$

Untuk menemukan , kita perlu $\Cov(X_+, Y_+)$

\begin{aligned} E [X_{+} Y_{+}] & = Pr (V) E [X Y ∣ V] + P r (\neg V) 0 \\ = Pr (V) E [(μ_{x} + σ_{x} \tilde{X}) (μ_{y} + σ_{y} \tilde{Y}) ∣ V] \\ = μ_{x} μ_{y} Pr (V) + μ_{y} σ_{x} Pr (V) E [\tilde{X} ∣ V] + μ_{x} σ_{y} Pr (V) E [\tilde{Y} ∣ V] \\ + σ_{x} σ_{y} Pr (V) E [\tilde{X} \tilde{Y} ∣ V] \\ = μ_{x} μ_{y} Pr (V) + μ_{y} σ_{x} (q_{x} R_{x y} + ρ q_{y} R_{y x}) + μ_{x} σ_{y} (ρ q_{x} R_{x y} + q_{y} R_{y x}) \\ + σ_{x} σ_{y} (ρ Pr (V) - ρ μ_{x} q_{x} R_{x y} / σ_{x} - ρ μ_{y} q_{y} R_{y x} / σ_{y} + r_{x y}) \\ = (μ_{x} μ_{y} + σ_{x} σ_{y} ρ) Pr (V) + (μ_{y} σ_{x} + μ_{x} σ_{y} ρ - ρ μ_{x} σ_{y}) q_{x} R_{x y} \\ + (μ_{y} σ_{x} ρ + μ_{x} σ_{y} - ρ μ_{y} σ_{x}) q_{y} R_{y x} + σ_{x} σ_{y} r_{x y} \\ = (μ_{x} μ_{y} + Σ_{x y}) Pr (V) + μ_{y} σ_{x} q_{x} R_{x y} + μ_{x} σ_{y} q_{y} R_{y x} + σ_{x} σ_{y} r_{x y}, \end{aligned}

$\begin{align} \E[X_+ Y_+] &= \Pr(\v) \E[ X Y \mid \v] + Pr(\lnot\v) \, 0 \\&= \Pr(\v) \E\left[ (\mu_x + \sigma_x \T X) (\mu_y + \sigma_y \T Y) \mid \v \right] \\&= \mu_x \mu_y \Pr(\v) + \mu_y \sigma_x \Pr(\v) \E[ \T X \mid \v] + \mu_x \sigma_y \Pr(\v) \E[ \T Y \mid \v] \\&\qquad + \sigma_x \sigma_y \Pr(\v) \E\left[ \T X \T Y \mid \v \right] \\&= \mu_x \mu_y \Pr(\v) + \mu_y \sigma_x (q_x R_{xy} + \rho q_y R_{yx}) + \mu_x \sigma_y (\rho q_x R_{xy} + q_y R_{yx}) \\&\qquad + \sigma_x \sigma_y \left( \rho \Pr\left( \v \right) - \rho \mu_x q_x R_{xy} / \sigma_x - \rho \mu_y q_y R_{yx} / \sigma_y + r_{xy} \right) \\&= (\mu_x \mu_y + \sigma_x \sigma_y \rho) \Pr(\v) + (\mu_y \sigma_x + \mu_x \sigma_y \rho - \rho \mu_x \sigma_y) q_x R_{xy} \\&\qquad + (\mu_y \sigma_x \rho + \mu_x \sigma_y - \rho \mu_y \sigma_x) q_y R_{yx} + \sigma_x \sigma_y r_{xy} \\&= (\mu_x \mu_y + \Sigma_{xy}) \Pr(\v) + \mu_y \sigma_x q_x R_{xy} + \mu_x \sigma_y q_y R_{yx} + \sigma_x \sigma_y r_{xy} ,\end{align}$ dan kemudian mengurangi kita mendapatkan

E [X_{+}] E [Y_{+}]

$\E[X_+] \E[Y_+]$

\begin{aligned} Cov (X_{+}, Y_{+}) & = (μ_{x} μ_{y} + Σ_{x y}) Pr (V) + μ_{y} σ_{x} q_{x} R_{x y} + μ_{x} σ_{y} q_{y} R_{y x} + σ_{x} σ_{y} r_{x y} \\ - (Q_{x} μ_{x} + q_{x} σ_{x}) (Q_{y} μ_{y} + q_{y} σ_{y}) . \end{aligned}

$\begin{align} \Cov(X_+, Y_+) &= (\mu_x \mu_y + \Sigma_{xy}) \Pr(\v) + \mu_y \sigma_x q_x R_{xy} + \mu_x \sigma_y q_y R_{yx} + \sigma_x \sigma_y r_{xy} \\&\qquad - (Q_x \mu_x + q_x \sigma_x) (Q_y \mu_y + q_y \sigma_y) .\end{align}$

Berikut ini beberapa kode Python untuk menghitung momen:

import numpy as np
from scipy import stats

def relu_mvn_mean_cov(mu, Sigma):
    mu = np.asarray(mu, dtype=float)
    Sigma = np.asarray(Sigma, dtype=float)
    d, = mu.shape
    assert Sigma.shape == (d, d)

    x = (slice(None), np.newaxis)
    y = (np.newaxis, slice(None))

    sigma2s = np.diagonal(Sigma)
    sigmas = np.sqrt(sigma2s)
    rhos = Sigma / sigmas[x] / sigmas[y]

    prob = np.empty((d, d))  # prob[i, j] = Pr(X_i > 0, X_j > 0)
    zero = np.zeros(d)
    for i in range(d):
        prob[i, i] = np.nan
        for j in range(i + 1, d):
            # Pr(X > 0) = Pr(-X < 0); X ~ N(mu, S) => -X ~ N(-mu, S)
            s = [i, j]
            prob[i, j] = prob[j, i] = stats.multivariate_normal.cdf(
                zero[s], mean=-mu[s], cov=Sigma[np.ix_(s, s)])

    mu_sigs = mu / sigmas

    Q = stats.norm.cdf(mu_sigs)
    q = stats.norm.pdf(mu_sigs)
    mean = Q * mu + q * sigmas

    # rho_cs is sqrt(1 - rhos**2); but don't calculate diagonal, because
    # it'll just be zero and we're dividing by it (but not using result)
    # use inf instead of nan; stats.norm.cdf doesn't like nan inputs
    rho_cs = 1 - rhos**2
    np.fill_diagonal(rho_cs, np.inf)
    np.sqrt(rho_cs, out=rho_cs)

    R = stats.norm.cdf((mu_sigs[y] - rhos * mu_sigs[x]) / rho_cs)

    mu_sigs_sq = mu_sigs ** 2
    r_num = mu_sigs_sq[x] + mu_sigs_sq[y] - 2 * rhos * mu_sigs[x] * mu_sigs[y]
    np.fill_diagonal(r_num, 1)  # don't want slightly negative numerator here
    r = rho_cs / np.sqrt(2 * np.pi) * stats.norm.pdf(np.sqrt(r_num) / rho_cs)

    bit = mu[y] * sigmas[x] * q[x] * R
    cov = (
        (mu[x] * mu[y] + Sigma) * prob
        + bit + bit.T
        + sigmas[x] * sigmas[y] * r
        - mean[x] * mean[y])

    cov[range(d), range(d)] = (
        Q * (1 - Q) * mu**2 + (1 - 2 * Q) * q * mu * sigmas
        + (Q - q**2) * sigma2s)

    return mean, cov

dan tes Monte Carlo yang berfungsi:

np.random.seed(12)
d = 4
mu = np.random.randn(d)
L = np.random.randn(d, d)
Sigma = L.T.dot(L)
dist = stats.multivariate_normal(mu, Sigma)

mn, cov = relu_mvn_mean_cov(mu, Sigma)

samps = dist.rvs(10**7)
mn_est = samps.mean(axis=0)
cov_est = np.cov(samps, rowvar=False)
print(np.max(np.abs(mn - mn_est)), np.max(np.abs(cov - cov_est)))

yang memberi 0.000572145310512 0.00298692620286, menunjukkan bahwa ekspektasi dan kovarian yang diklaim sesuai dengan perkiraan Monte Carlo (berdasarkan sampel). $10,000,000$

Dougal
sumber

dapatkah Anda meringkas apa nilai akhir itu? Apakah mereka memperkirakan parameter mu dan L yang Anda hasilkan? Mungkin mencetak nilai target itu?

AdamO

Tidak, nilai yang dikembalikan adalah dan ; apa yang saya cetak adalah jarak tak antara estimasi Monte Carlo dari jumlah-jumlah itu dan nilai yang dihitung. Anda mungkin dapat membalikkan ekspresi ini untuk mendapatkan estimator pencocokan momen untuk dan - Rosenbaum sebenarnya melakukan itu di bagian 3 dalam kasus terpotong - tetapi bukan itu yang saya inginkan di sini.

\E (Z_{+})

$\E(Z_+)$

\Cov (Z_{+})

$\Cov(Z_+)$

L_{\infty}

$L_\infty$

μ

$\mu$

Σ

$\Sigma$

Dougal

Apa arti dan varian dari normal multivariat 0-disensor?

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