2
$\begingroup$

Assume that $(H,\langle\cdot,\cdot\rangle)$ is a real separable Hilbert space equipped with a Gaussian measure $\mu$ with a mean $m$ and a covariance operator $C$. Let $x\in H$ be a fixed vector. What are the values of the following integrals: $$\int_H\frac{|\langle x,h\rangle|}{\|h\|}\mu(\mbox{d}h)$$ and $$\int_H\frac{|\langle x,h\rangle|^2}{\|h\|^2}\mu(\mbox{d}h),$$ can we express these integrals in terms of just $x$ and an inner product or norm?

$\endgroup$
  • 4
    $\begingroup$ Do you know the result if $H$ is finite dimensional? $\endgroup$ – András Bátkai Jun 24 '16 at 18:22
2
$\begingroup$

First integral: no, this expression is not polynomial in $x$ due to the absolute value sign in the numerator, so it cannot be written as an inner product or norm of $x$. It certainly can be expressed in terms of an inner product (for instance, in the form you already have), and I argue below that any finite number of terms in its series expansion (in $C$) should be obtainable in closed form.

Second integral: this expression is a seminorm. Whether it is a norm depends on $C$. Evaluating it is also not easy to my knowledge, but should be easier than the first integral.

I cannot speak to the infinite dimensional case.

Your first integrand is not differentiable in $h$. I'd suggest integrating separately over the half-spaces separated by the plane $\langle x, h\rangle=0$. In each half space, I'd perform the Taylor expansion around the $m$, then carry out the Gaussian integrals over the dimensions orthogonal to $x$ via Isserlis's theorem. The final integration parallel to $x$ should leave you with a series of expressions in closed form in terms of incomplete Gamma functions.

My inner engineer notes that if $m$ is large compared to the root-eigenvalues of $C$, then

$$ \int_H\frac{\left|\langle x,h\rangle\right|}{\|h\|}\mu(\mbox{d}h)\approx \mathrm{sgn}(\langle x,m\rangle) \left\langle x, \int_H\frac{h}{\|h\|}\mu(\mbox{d}h) \right\rangle \\ = \mathrm{sgn}(\langle x,m\rangle) \left\langle x, \int_H\frac{m+\xi}{\|m+\xi\|}\nu(\mbox{d}\xi) \right\rangle, $$ where $\nu$ is a Gaussian measure over $H$ with mean $0$ and covariance operator $C$.

$$ = \mathrm{sgn}(\langle x,m\rangle) \left\langle x, \int_H(m+\xi)\left(\langle m,m\rangle + 2 \mbox{Re}\langle m,\xi\rangle + \langle \xi, \xi\rangle\right)^{-1/2}\nu(\mbox{d}\xi) \right\rangle \\ \approx \frac{\mathrm{sgn}(\langle x,m\rangle)}{\|m\|} \left\langle x, \int_H(m+\xi)\left(1 - \frac{\langle m,\xi\rangle}{\|m\|^2} - \frac12\frac{\langle \xi, \xi\rangle}{\|m\|^2}+\frac32\frac{\langle m,\xi\rangle^2}{\|m\|^4}\right)\nu(\mbox{d}\xi) \right\rangle \\ = \frac{\mathrm{sgn}(\langle x,m\rangle)}{\|m\|} \left[\langle x, m\rangle\left(1 - \frac{\mbox{Tr}C}{2\|m\|^2} + \frac{3\langle m, C m\rangle}{2\|m\|^4}\right) - \frac{\langle x,Cm\rangle}{\|m\|^2}\right] \\ = \left|\langle x, \hat m\rangle\right|\left(1 - \frac12\mbox{Tr}\hat C + \frac32\langle \hat m, \hat C \hat m\rangle\right) - \mathrm{sgn}(\langle x,\hat m\rangle) \langle x,\hat C\hat m\rangle, \\ \text{where }\hat C = C/\|m\|^2, \hat m = m/\|m\|. $$

For your second integral,

$$ \int_H \frac{\left|\langle x,h\rangle\right|^2}{\|h\|^2}\mu(\mathrm{d}h) = \left\langle x\right| \left(\int_H \frac{\left|h\right\rangle\left\langle h\right|}{\|h\|^2}\mu(\mathrm{d}h)\right) \left|x\right\rangle $$

The integral is quadratic in $x$. If $C$ is positive definite, then the operator in parentheses is a norm, since the integrand is a (positive semidefinite) orthogonal projector onto the subspace parallel to $h$, and the integration measure is nonzero over a full-dimensional region of $H$. In general, it looks like it is a seminorm, e.g. in the case where the kernel of $C$ includes vectors orthogonal to $m$.

The matrix elements of this operator could, in general, be evaluated by Taylor expanding around $m$, then using Isserlis' theorem to write the resulting Gaussian integrals over monomials in terms of traces of products of $C$ and $m$.

So I think the answer to your question is mostly "yes, in certain limiting cases".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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