# Sign of expectation value

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.

• Isn't $\lambda$ simply a Gaussian measure with covariance matrix $\tfrac{1}{2} (\Sigma^{-1} + \operatorname{Id})^{-1}$? Commented Jan 29, 2020 at 19:59
• @MateuszKwaśnicki well, there is also the $\mu$... Commented Jan 29, 2020 at 20:02
• 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 $\Sigma^{-1} + \operatorname{Id}$. Commented Jan 29, 2020 at 20:04
• I definitely agree with the last sentence you wrote. Commented Jan 29, 2020 at 20:15

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 $$Y$$.