0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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 $Y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.