Por que a matriz de informações de Fisher é semidefinida positiva?

18

Seja $\theta \in R^{n}$ . A Matriz de Informações de Fisher é definida como:

I (θ)_{i, j} = - E [\frac{\partial^{2} \log (f (X | θ))}{\partial θ_{i} \partial θ_{j}} | θ]

$I(\theta)_{i,j} = -E\left[\frac{\partial^{2} \log(f(X|\theta))}{\partial \theta_{i} \partial \theta_{j}}\bigg|\theta\right]$

Como posso provar que a Matriz de informações de Fisher é semidefinida positiva?

inference linear-algebra fisher-information

— madprob
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7

Não é o valor esperado de um produto externo da pontuação consigo?

— Neil G

19

Confira: http://en.wikipedia.org/wiki/Fisher_information#Matrix_form

A partir da definição, temos

I_{i j} = E_{θ} [(\partial_{i} \log f_{X ∣ Θ} (X ∣ θ)) (\partial_{j} \log f_{X ∣ Θ} (X ∣ θ))],

$I_{ij} = \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] \, ,$ para

i, j = 1, \dots, k

$i,j=1,\dots,k$ , em que

\partial_{i} = \partial / \partial θ_{i}

$\partial_i=\partial /\partial \theta_i$ . Sua expressão para

I_{i j}

$I_{ij}$ segue desta em condições de regularidade.

$u = (u_1,\dots,u_k)^\top\in\mathbb{R}^n$

\sum_{i, j = 1}^{k} u_{i} I_{i j} u_{j} = \sum_{i, j = 1}^{k} (u_{i} E_{θ} [(\partial_{i} \log f_{X ∣ Θ} (X ∣ θ)) (\partial_{j} \log f_{X ∣ Θ} (X ∣ θ))] u_{j}) = E_{θ} [(\sum_{i = 1}^{k} u_{i} \partial_{i} \log f_{X ∣ Θ} (X ∣ θ)) (\sum_{j = 1}^{k} u_{j} \partial_{j} \log f_{X ∣ Θ} (X ∣ θ))] = E_{θ} [{(\sum_{i = 1}^{k} u_{i} \partial_{i} \log f_{X ∣ Θ} (X ∣ θ))}^{2}] \geq 0 .

$\sum_{i,j=1}^k u_i I_{ij} u_j = \sum_{i,j=1}^k \left( u_i \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] u_j \right) \\ = \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\sum_{j=1}^k u_j \partial_j \log f_{X\mid\Theta} (X\mid\theta)\right)\right] \\ = \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right)^2 \right] \geq 0 \, .$

If this component wise notation is too ugly, note that the Fisher Information matrix $H=(I_{ij})$ can be written as $H = \mathrm{E}_\theta\left[S S^\top\right]$ , in which the scores vector $S$ is defined as

S = {(\partial_{1} \log f_{X ∣ Θ} (X ∣ θ), \dots, \partial_{k} \log f_{X ∣ Θ} (X ∣ θ))}^{⊤} .

$S = \left( \partial_1 \log f_{X\mid\Theta}(X\mid\theta), \dots, \partial_k \log f_{X\mid\Theta}(X\mid\theta) \right)^\top \, .$

Hence, we have the one-liner

u^{⊤} H u = u^{⊤} E_{θ} [S S^{⊤}] u = E_{θ} [u^{⊤} S S^{⊤} u] = E_{θ} [| | S^{⊤} u | |^{2}] \geq 0.

$u^\top H u = u^\top \mathrm{E}_\theta[S S^\top] u = \mathrm{E}_\theta[u^\top S S^\top u] = \mathrm{E}_\theta\left[|| S^\top u ||^2\right] \geq 0.$

— Zen
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3

(+1) Good answer and welcome back, Zen. I was becoming concerned we might have lost you permanently given the length of your hiatus. That would have been a real shame!

— cardinal

5

WARNING: not a general answer!

If $f(X|\theta)$ corresponds to a full-rank exponential family, then the negative Hessian of the log-likelihood is the covariance matrix of the sufficient statistic. Covariance matrices are always positive semi-definite. Since the Fisher information is a convex combination of positive semi-definite matrices, so it must also be positive semi-definite.

— gusl
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