Consider a multivariate Gaussian-type measure $$d\lambda(x):=\nu_{\mu,\Sigma} e^{-\langle (x-\mu), \Sigma^{-1}(x-\mu) \rangle - \vert x \vert^2} $$

with vector $\mu \in \mathbb R^n$ and $\Sigma$ positive definite and $\nu_{\mu,\Sigma}$ a normalizing constant to turn $d\lambda$ into a probability measure.

Let $m$ be the vector-valued expectation value $m:=\int_{\mathbb R^n} x d\lambda(x).$

We then consider the expectation value for $X$ distributed according to the measure $\lambda:$

$$\mathbb E \left( \langle X-m, \Sigma^{-1} y \rangle^2 \langle X-m, \Sigma^{-1} \mu \rangle \right).$$

Question: Can we say anything about the sign of this expectation value for general vectors $y \in \mathbb R^n$?-From how I obtained this expression I conjecture that this expression is never strictly positive, but I cannot see it right away.

  • $\begingroup$ Isn't $\lambda$ simply a Gaussian measure with covariance matrix $\tfrac{1}{2} (\Sigma^{-1} + \operatorname{Id})^{-1}$? $\endgroup$ Commented Jan 29, 2020 at 19:59
  • $\begingroup$ @MateuszKwaśnicki well, there is also the $\mu$... $\endgroup$
    – Sascha
    Commented Jan 29, 2020 at 20:02
  • $\begingroup$ But $\mu$ only shows up in the expression for the mean $m$, does it not? I mean, the exponent is a quadratic function of $x$ with prinicipal term $\Sigma^{-1} + \operatorname{Id}$. $\endgroup$ Commented Jan 29, 2020 at 20:04
  • $\begingroup$ I definitely agree with the last sentence you wrote. $\endgroup$
    – Sascha
    Commented Jan 29, 2020 at 20:15

1 Answer 1


The expectation is indeed never strictly positive: it is equal to zero.

The density of $\lambda$ is proportional to $\exp(-\tfrac{1}{2} \langle (x - m), A^{-1} (x - m)\rangle)$, where $A = \tfrac{1}{2} (\Sigma^{-1} + \operatorname{Id})^{-1}$ is a positive definite matrix. Thus, $Y = X - m$ is a centred Gaussian vector (with covariance matrix $A$). It follows that $$\mathbb{E} ((\langle (X - m), \Sigma^{-1} y \rangle)^2 \langle (X - m), \Sigma^{-1} \mu \rangle) = \mathbb{E} ((\langle Y, a\rangle)^2 \langle Y, b\rangle)$$ for appropriate vectors $a = \Sigma^{-1} y$, $b = \Sigma^{-1} \mu$. Now it is straightforward to note that the last expectation is zero: $Y$ is equal in distribution to $-Y$, and the expression under the expectation is an antisymmetric function of $Y$.


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