I am trying to get my head around this question and was reading (1) which states the same a little bit more general:

Let $X$ be a separable Banach space and $X^*$ the dual space. The mean value $m \in X$ and covariance $C\colon X^*\to X$ are defined by \begin{align} x^*(m) &= \int_X x^*(x) \mu(dx),\\ y^*(Cx^*) &= \int_X x^*(x)y^*(x) \mu(dx) - x^*(m)y^*(m) \end{align} $x^*,y*\in X^*$ and measure $\mu$.


A measure $\mu$ is called Gaussian if all the functionals $x^*\in X^*$, regarded as random variables on $(X,\mu)$ are Gaussian random variables.

I considered $X=\mathbb{R}^2$ since this is easier to visualize and I can check the results with my experience of $\mathbb{R}^2$ random variables.

I generated samples $Z$ from a Gaussian distribution with mean $m = (1,1)^T$ and covariance $C = \bigl(\begin{smallmatrix} 1&0.5 \\ 0.5&0.4 \end{smallmatrix} \bigr)$.

enter image description here

Choosing a $x^* = (2,3)$ (or any other value) then $x^*(Z)$ seems to be Gaussian with $x^*(m) = (2,3)\cdot(1,1)^T = 5$ and $x^*(Cx^*) = (2,3)\cdot (C(2,3)\cdot (2,3)) = 13.6$ as in the histogram:

enter image description here

and everything seems to be fine.

My question now is, can I somehow access the density of the (multivariate Gaussian) distribution at $x^*$ if $X$ is a Banach space, to get the value of \begin{align} f(x^*) = \frac{1}{\sqrt{2\pi^2|C|}} \exp\big(-0.5(x^*-m)^TC^{-1}(x^*-m) \big) \end{align} It would be great to get an example assuming that $X$ and the measure are sufficiently "nice" (eg. the measure is not degenerate etc, X is separable or a Hilbert space, etc.).

Edit: As Nate Eldredge pointed out there is no Lebesgue measure. Let me give a possible application: Failure detection on an engine. There is some sensor delivering a curve (function) which carries information about the engine properties. So each test of an enigne delivers a curve (function) and therefore we model $X$ to be a separable Hilbert space of square-integrable functions. A malfunction is now detection if a test yields a curve outside two standard deviations.

(1) Rudnìcki, Ryszard. "Gaussian measure-preserving linear transformations." Univ. Iagel. Acta Math 30 (1993): 105-112.

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    $\begingroup$ I am not sure I understand what you are asking for when you say "density". On a finite-dimensional space such as $\mathbb{R}^2$, the density of a measure such as $\mu$ is its Radon-Nikodym derivative with respect to Lebesgue measure $m$. Note that the density is a function on the sample space, $\mathbb{R}^2$ in your example, not its dual (which confusingly in your example is also isomorphic to $\mathbb{R}^2$). $\endgroup$ Commented Oct 13, 2015 at 14:54
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    $\begingroup$ So the natural interpretation of "the density of $\mu$" is "the Radon-Nikodym derivative of $\mu$ with respect to Lebesgue measure $m$" (this would be a function on $W$, not $W^*$). But infinite-dimensional spaces don't have Lebesgue measure, so this notion doesn't make sense. This is one of the major motivations for studying Gaussian measures in infinite dimensions in the first place - there aren't so many other natural measures to study! $\endgroup$ Commented Oct 13, 2015 at 14:56
  • $\begingroup$ Thx. The density seems to be one of the most interesting quantities in finite dimensions telling me how "likely" an observation is. Is there anything similar in infinite dimensions? $\endgroup$ Commented Oct 13, 2015 at 15:22
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    $\begingroup$ There are a few possible candidates, most of them unsatisfying in one way or another. One candidate is something like $e^{-\|x\|_H^2}$, where $\|\cdot\|_H$ is the Cameron-Martin norm. This is in some sense analogous to the finite-dimensional density, except for the annoying detail that it is zero almost everywhere. Perhaps if you could elaborate on what you are trying to accomplish, it would help. $\endgroup$ Commented Oct 13, 2015 at 19:13
  • $\begingroup$ Cameron-Martin is interesting. I added a simple motivation for what I want to accomplish. $\endgroup$ Commented Oct 14, 2015 at 7:46

2 Answers 2


First of all, please note that I am not an expert for this. Nevertheless, I think what you want (in order to transfer your intuition from $\mathbb{R}^2$ to the Banach space case) is to write the integral of a suitable function $f$ over the Gaussian measure $\mu$ as $$ \int_X f(x)\,\mu(d x) \,\,\,= \,\,\, \int_X f(x)\,\exp\big(q(x)\big)\,\lambda(dx) $$ where $\lambda$ is analogous to the standard Lesbegue-meassure on $\mathbb{R}^2$ and $q$ some quadratic function on $X$. The problem then is that in order to define the standard Lesbegue measure on $\mathbb{R}^2$ one makes use of the standard basis of $\mathbb{R}^2$ in order to define the measure of a rectangle (therefor, if you transform to a different basis via a linear transformation $T$ you get a factor $\det(T)$).

Because of this, there is no analogoue of the standard Lesbegue measure on an abstract Banach space. Nevertheless, on a Hilbert space $H$ you would have a natural notion of a "length" and one might try to find an analogue in that case. But even then there are more problems arising because integration on infinite dimensional spaces is way more complicated than on finite dimensional ones. Heuristically you could say that $$ \int_H \exp\big(-\pi<x\,|\,x>\big)\,\lambda(dx) \,\,\,=\,\,\,1$$ because it is the infinite product of $1$-dimensional integrals that evaluate to $1$. But then $$ \int_H \exp\big(-2\pi<x\,|\,x>\big)\,\lambda(dx) \,\,\,=\,\,\,\infty$$ because it is the infinite product of $1$-dimensional integrals that evaluate to something larger than $1$. So you could only give a meaning to $$ \int_H c\,\exp\big(-\pi<(x-x_m)\,|\,V\,(x-x_m)>\big)\,\lambda(dx)$$ where $c\in \mathbb{R}$ is for normalisation, $x_m\in H$ is the mean and $V$ is a bounded operator on $H$ that is sufficiently close to the identity (maybe the exponential of an operator of Hilbert Schmidt type) and is of course related to the Variance. I think with some work one can get at least a formula for this heuristic -- but not necessarily a measure on $H$.


The answer is contained in your presentation. The operator $C: X^*\to X$ defines a bilinear map $\newcommand{\bR}{\mathbb{R}}$

$$ B_C:X^*\times X^*\to\bR,\;\; B_C(x^*, y^*)=\langle y^*, Cx^*\rangle, $$

where $\langle-,-\rangle: X^*\times X\to\bR$ denotes the natural pairing between a topological vector space and its dual.

The mean of the random variable $x^*\in X^*$ is $x^*(m)$ and its variance is $B_C(x^*,x^*)$. These two numbers determine the density of the Gaussian r.v. $x^*$.

  • $\begingroup$ I edited the question to clarify things. I am not interested in the density of the transformed distribution, but the density at $x^*$ $\endgroup$ Commented Oct 13, 2015 at 13:15

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