6
$\begingroup$

(An earlier version of this at stackexchange got no answers.)

Bayesianism says that all uncertainties, or at least all uncertainties about the truth or falsity of propositions, can be expressed by probabilities. That would seem to suggest that the result described below might have practical implications for numerical analysis in situations where no probabilities are mentioned. Does it?

Let $B_t$, $t\ge0$ be a standard Brownian motion and suppose $0<x_1<x_2<\cdots<x_n<1$. Then the conditional expectation

$$ \mathbb E\left(\int_0^1 B_t\,dt \,\middle\vert\, B_0, B_{x_1},B_{x_2},\ldots,B_{x_n},B_1 \right) $$

is just the trapezoidal-rule approximation to the integral. Since expected values minimize mean squared errors, it follows that for every integrable function $g$,

$$ \mathbb E\left(\left(\int_0^1 B_t\,dt - E\left(\int_0^1 B_t\,dt \,\middle\vert\, B_0, B_{x_1},B_{x_2},\ldots,B_{x_n},B_1 \right) \right)^2\right) $$

$$\le \mathbb E\left(\left( \int_0^1 B_t \, dt - g(B_0, B_{x_1},B_{x_2},\ldots,B_{x_n},B_1) \right)^2\right). $$

A theorem I have seen attributed to Persi Diaconis (but is that correct?) says Simpson's rule is not admissible in the decision-theoretic sense. That means that for every convex function $L$ that we could put in place of the squaring function above, there is some measurable function $g$ such that for every probability distribution on functions $C_t$ on $[0,1]$,

$$ \mathbb E\left(L\left(\int_0^1 C_t\,dt - g(C_0,C_{x_1},C_{x_2},\ldots,C_1)\right)\right) $$

$$\le \mathbb E\left(L\left(\int_0^1 C_t\,dt - \operatorname{Simpson}(C_0, C_{x_1},C_{x_2},\ldots,C_{x_n},C_1)\right)\right), $$

and for at least one probability distribution the inequality is strict.

$\endgroup$
2

0

Your Answer

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