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Gaussian even moments for Brownian increments

Statement

Assume the Axiom of Choice. If $0 \leq s, t$ and a real random variable $X_{t} - X_{s}$ has law $N (0, ∣ t - s ∣)$ , then, for every integer $m \geq 1$ , $E ∣ X_{t} - X_{s} ∣^{2 m} = c_{m} ∣ t - s ∣^{m},$ where $c_{m} = E ∣ Z ∣^{2 m} = (2 m - 1)!! = 1 \cdot 3 \dots (2 m - 1) < \infty$ for $Z \sim N (0, 1)$ . In particular, when $m \geq 2$ , this is the moment hypothesis of the one-parameter Kolmogorov criterion with $α = 2 m$ , $β = m - 1$ , and $C_{T} = c_{m}$ .

Facts & Assumptions

Given: Times $s, t \geq 0$ , the stated increment law, and an integer $m \geq 1$ .

[F1]

Under AC, $N (0, 1)$ is the probability measure with density $ϕ (x) = e^{- x^{2} / 2} / \sqrt{2 π}$ , and $N (0, h)$ is its image under $x \mapsto \sqrt{h} x$ , including $h = 0$ . Standard normal and normal laws The Axiom of Choice

[F2]

A nonnegative expectation is the integral of the corresponding function against the random variable's law. Change of variables for expectation

[F3]

Integration against the measure with density $ϕ$ equals integration of the product with $ϕ$ against Lebesgue measure. Integrating against a density agrees with integrating the product

[F4]

Increasing nonnegative measurable functions may be passed to the limit under the integral. Monotone convergence for the integral

[F5]

Compact-interval integration by parts includes its two endpoint terms. Under countable choice, a bounded Riemann-integrable function on a compact interval is Lebesgue integrable there with the same integral. If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral

[F6]

The power, product, and chain rules, together with $(e^{x})^{'} = e^{x}$ , give $ϕ^{'} (x) = - x ϕ (x)$ . For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ The exponential function is smooth and $(\exp)^{'} = \exp$

[F7]

The exponential is its nonnegative power series, so for $R > 0$ and every integer $k \geq 0$ , $e^{R^{2} / 2} \geq (R^{2} / 2)^{k} / k!$ . The power-series, product-limit, IVP, functional-equation, and Picard definitions agree

Proof

technique · direct

1.1

Let $Z$ be the coordinate map on the canonical $N (0, 1)$ probability space. By [F1]--[F3], for each integer $r \geq 0$ , $E ∣ Z ∣^{2 r} = \int_{R} x^{2 r} ϕ (x) d x,$ with either side initially allowed to be infinite.

F1F2F3

1.2

For an integer $R \geq 1$ and $r \geq 1$ , [F5] on $[- R, R]$ , applied to $u (x) = x^{2 r - 1}$ and $v (x) = ϕ (x)$ , is legitimate by [F6] and gives $\int_{- R}^{R} x^{2 r} ϕ (x) d x = - 2 R^{2 r - 1} ϕ (R) + (2 r - 1) \int_{- R}^{R} x^{2 r - 2} ϕ (x) d x .$ The sign and factor two come from the odd power at the two endpoints and the evenness of $ϕ$ .

F5F6algebra

2.1

Taking $k = r + 1$ in [F7] shows $0 \leq R^{2 r - 1} ϕ (R) \leq \frac{2^{r + 1} (r + 1)!}{\sqrt{2 π} R^{3}} ⟶ 0.$ The compact bridge in [F5] identifies every Riemann integral in step 1.2 with the corresponding Lebesgue integral. The truncated nonnegative integrands then increase to their whole-line counterparts, so [F4], step 1.1, and $\int ϕ = 1$ from [F1] yield recursively $c_{0} = 1, c_{r} = (2 r - 1) c_{r - 1} .$ Thus $c_{m} = (2 m - 1)!!$ and every $c_{m}$ is finite.

step 1.1step 1.2F1F4F5F7algebra

3.1

Put $h = ∣ t - s ∣$ . By [F1], the law $N (0, h)$ is that of $\sqrt{h} Z$ ; applying [F2] to the nonnegative function $x \mapsto ∣ x ∣^{2 m}$ and using step 2.1 gives $E ∣ X_{t} - X_{s} ∣^{2 m} = E ∣ \sqrt{h} Z ∣^{2 m} = h^{m} c_{m} .$ This includes $h = 0$ , when both sides vanish and the law is the Dirac mass at zero.

givenstep 2.1F1F2algebra

4.1

If $m \geq 2$ , set $α = 2 m$ and $β = m - 1 > 0$ . Then $m = 1 + β$ , so step 3.1 reads $E ∣ X_{t} - X_{s} ∣^{α} = c_{m} ∣ t - s ∣^{1 + β} .$ The constant $C_{T} = c_{m}$ is finite and independent of $T$ . AC is used through [F1], which supplies the normal-law probability measure, and through the countable-choice hypothesis of the compact bridge used in step 2.1; all truncations and the recurrence are canonical.

step 2.1step 3.1F1algebra∎

Source notes

Durrett, Section 7.1, printed p. 358, uses the finite even moments of a normal increment in the Brownian continuity argument. Steps 1.1--2.1 supply the full compact-truncation integration-by-parts calculation, including the boundary term and its limit.

Depends on

Used by

Brownian paths are locally Holder below one half Corollary
Existence of continuous Brownian motion Theorem

Dependency tree · two levels

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Sources

Rick Durrett, Probability: Theory and Examples, Section 7.1 (standard reference, not scraped)