Continuity of Covariance Operator

April 28, 2023 · #Gaussian random field

We start with Kolmogorov’s continuity theorem which lies at the heart of continuous martingale theory.

Theorem 1 (Kolmogorov). Let ${𝑋_{𝑡}}_{𝑡 \geq 0}$ be a real-valued stochastic process such that there exist $𝛼, 𝛽, 𝐾$ satisfying

𝔼 [{| 𝑋_{𝑡} - 𝑋_{𝑠} |}^{𝛼}] \leq 𝐾 {| 𝑡 - 𝑠 |}^{1 + 𝛽},

for any $𝑡, 𝑠 \geq 0$ . Then $𝑋$ has a $𝛾$ –Hölder continuous modification for any $0 < 𝛾 < \frac{𝛽}{𝛼}$ .

By noticing the ultracontractivity of Gaussian measures, we immediately derive that fractional Brownian motion $𝐵^{𝐻}$ has a $𝛾$ –Hölder modification where $0 < 𝛾 < 𝐻$ . More generally, we can show that any centered Gaussian random field ${𝑋 (𝑥)}_{𝑥 \in ℝ^{𝑑}}$ with covariance $𝐶$ has a $𝛾$ –Hölder modification where $0 < 𝛾 < 𝐻$ if

𝐶 (𝑥, 𝑥) + 𝐶 (𝑦, 𝑦) - 2 𝐶 (𝑥, 𝑦) \leq 𝐾 {| 𝑥 - 𝑦 |}^{2 𝐻} .

(1)

Here, $𝐶$ is given by $𝐶 (𝑥, 𝑦) = 𝔼 [𝑋 (𝑥) 𝑋 (𝑦)]$ . This, of course, implies the continuity of $𝐶$ . But, can we circumvent using Kolmogorov’s theorem?

Proposition 2. Let $𝐶 : ℝ^{𝑑} \times ℝ^{𝑑} \to ℝ$ be symmetric, locally bounded, and positive definite in the sense that for any finite collection ${𝑥_{𝑖}}_{𝑖 = 1}^{𝑚}$ in $ℝ^{𝑑}$ , the matrix $𝑀_{𝑖 𝑗} = 𝐶 (𝑥_{𝑖}, 𝑥_{𝑗})$ is positive definite. Then Equation 1 implies the joint continuity of $𝐶$ .

Proof. By assumption we take the positive definite matrix

𝑀 = (\begin{matrix} 𝐶 (𝑦, 𝑦) & 𝐶 (𝑦, 𝑥_{1}) & 𝐶 (𝑦, 𝑥_{2}) \\ 𝐶 (𝑥_{1}, 𝑦) & 𝐶 (𝑥_{1}, 𝑥_{1}) & 𝐶 (𝑥_{1}, 𝑥_{2}) \\ 𝐶 (𝑥_{2}, 𝑦) & 𝐶 (𝑥_{2}, 𝑥_{1}) & 𝐶 (𝑥_{2}, 𝑥_{2}) \end{matrix})

and vector

𝛼 = (\begin{matrix} 𝐶 (𝑥_{1}, 𝑥_{1}) + 𝐶 (𝑥_{2}, 𝑥_{2}) - 2 𝐶 (𝑥_{1}, 𝑥_{2}) \\ 𝐶 (𝑥_{2}, 𝑦) - 𝐶 (𝑥_{1}, 𝑦) \\ 𝐶 (𝑥_{1}, 𝑦) - 𝐶 (𝑥_{2}, 𝑦) \end{matrix}) .

Then we notice $𝛼^{⊤} 𝑀 𝛼 \geq 0$ is equivalent to

{[𝐶 (𝑥_{1}, 𝑦) - 𝐶 (𝑥_{2}, 𝑦)]}^{2} \leq 4 𝐶 (𝑦, 𝑦) [𝐶 (𝑥_{1}, 𝑥_{1}) + 𝐶 (𝑥_{2}, 𝑥_{2}) - 2 𝐶 (𝑥_{1}, 𝑥_{2})] .

By the triangle inequality we conclude the proof.⁠∎