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Is there a version of Stein's lemma for multivariate Gaussians, I am attempting to solve integrals of the form:

$$ \mathbb{E}(\mathbf{x} g(\mathbf{x})) =\int_{\mathbb{R}^p} \mathbf{x} g(\mathbf{x}) \mathcal{N}(\mathbf{x}| \mu, \Sigma) d\mathbf{x} $$

and

$$ \mathbb{E}(\mathbf{x}^T\mathbf{x} g(\mathbf{x})) $$

where $\mathbf{x} \in \mathbb{R}^p$ is a Gaussian vector with mean $\mu$, and covariance matrix $\Sigma$ and $g: \mathbb{R}^p \to\mathbb{R}$

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    $\begingroup$ Lemma 1 from Liu is probably what you want. $\endgroup$ Commented Jun 11, 2018 at 21:48
  • $\begingroup$ Yes, see lemma 6.20 (page 74) in this paper for a more general version that does not require the covariance matrix to be the identity matrix. $\endgroup$
    – Resu
    Commented Mar 15, 2022 at 15:27

1 Answer 1

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Here is an exercise from my book:

If $x\sim\mathcal{N}(\theta,1)$ and $f$ is continuous and a.e. differentiable, then $$ \mathbb{E}_\theta[(x-\theta)f(x)] = \mathbb{E}_\theta[f'(x)]. $$ Deduce that, if $x\sim \mathcal{N}_p(\theta,\Sigma)$, $\delta(x) = x+\Sigma\gamma(x)$, and $\mathrm{L}(\theta,\delta) = (\delta-\theta)^tQ(\delta-\theta)$, with $\gamma$ differentiable, then $$ R(\theta,\delta) = \mathbb{E}_\theta\left[{\mathrm{tr}}(Q\Sigma)+2\,{\mathrm{tr}}(J_\gamma(x)Q^*)+ \gamma(x)^tQ^*\gamma(x)\right], $$ where ${\mathrm{tr}}(A)$ is the trace of $A$, $Q^* = \Sigma Q\Sigma$ and $J_\gamma(x)$ is the matrix with generic element ${\partial \over \partial x_i}\gamma_j(x)$.

which implies in particular that $$\mathbb{E}_\theta[(\mathbf{X}-\theta)^\text{T} g(\mathbf{X})] =\mathbb{E}_\theta[\nabla^\text{T} g(\mathbf{X})]$$ in the case of an identity covariance matrix.

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  • $\begingroup$ Shouldn't we have the covariance matrix somewhere in the expression? Also shouldn't the expectations in the last line be with respect to $X$ rather than $\theta$ ? $\endgroup$ Commented Jan 24, 2018 at 0:11
  • $\begingroup$ I think what you've got there is the case with identity covariance $\endgroup$ Commented Jan 24, 2018 at 2:34
  • $\begingroup$ The notation $\mathbb{E}_\theta$ means that the expectation depends on $\theta$, i.e., that the distribution of the rv $\mathbf{X}$ depends on $\theta$, not that $\theta$ is the random variable. $\endgroup$
    – Xi'an
    Commented Jan 24, 2018 at 6:45
  • $\begingroup$ if you could please update to provide a version of Stein's Lemma for the general covariance case, or provide a reference, that would be appreciated $\endgroup$ Commented Jan 25, 2018 at 4:37
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    $\begingroup$ I do not understand what role $R(\theta,\delta)$ and $L(\theta,\delta)$ play here, so it is hard for me to see how the exercise implies the given relation. $\endgroup$ Commented Jun 11, 2018 at 21:24

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