3
$\begingroup$

Let $W: \mathbb R^d \to \mathbb R^{d \times d}$ be a matrix-valued function with domain $\mathbb R^d$ and taking values in the set of $d \times d$ real matrices, such that $W(x)$ is positive-definite for every $x \in \mathbb R^d$, and such that $\int_{\mathbb{R}^d} W(x) \, dx = I_d$. Does there exist a probability density function $p$ over $\mathbb{R}^d$ such that for every appropriately integrable $f: \mathbb{R}^d \to \mathbb{R}^d$ $$\int W(x) f(x) \, dx = \int p(x) f(x) \,dx \text{ ?}$$

$\endgroup$

1 Answer 1

5
$\begingroup$

I don't think so. Choose the vector $f(x)=\{g(x),0,0\cdots 0\}$ with a function $g(x)>0$ peaked at $x=x_0$, for example, a smoothed $d$-dimensional delta function $\delta(x-x_0)$. Then $\int W(x)f(x)\,dx$ will return the first column of $W(x_0)$, which can have all nonzero elements, while the right-hand-side of the equation gives $\{p(x_0),0,0\cdots 0\}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .