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Kexing Ying

March 2024

Contents

An application of the Cameron-Martin theorem

15 May. 2024

7 Mar. 2024

27 Oct. 2023

30 Jul. 2022

15 May. 2024. An application of the Cameron-Martin theorem

The Cameron-Martin theorem provides a characterization of the Gaussian measure on infinite dimensional spaces. While I first saw this definition in Martin's lectures on SPDEs, it was mostly presented with context with its application to SPDEs. However, while reading Li and Sieber's paper, I saw a neat application of it in bounding the probability that the fractional Brownian motion stays near a given function. In this entry, I'll record this application in the simple case of the Brownian motion.

The Cameron-Martin theorem is the following.

Theorem. Let \(\mu \in \mathcal{M}(X)\) be a Gaussian and denote \(\mathcal{H}_\mu\) for its Cameron-Martin space. For \(h \in X\), denoting \(\tau_h : x \in X \mapsto x + h\), we have that \(\tau_h^* \mu \ll \mu\) if and only if \(h \in \mathcal{H}_\mu\). In this case, \[\frac{d\tau_h^* \mu}{d\mu}(x) = \exp\left(h^*(x) - \frac{1}{2}\|h\|^2_\mu\right).\]

Let \(\mu\) be the Gaussian measure on \(\mathcal{C}([0, 1], \mathbb{R})\) with covariance \(C_\mu(\delta_t, \delta_s) = s \wedge t\). \(\mu\) corresponds to the law of the Brownian motion on \([0, 1]\) and we call it the Wiener measure. We have that \(\mathcal{H}_\mu = H^{1, 2}_0([0, 1])\) - the space of all absolutely continuous functions \(h\) with \(h(0) = 0\) and \(\dot h(s) \in L^2([0, 1])\). Moreover, for \(h \in \mathcal{H}_\mu\), \(h^*(k) = \int_0^1 \dot h(s) \dot k(s) d s\). As this defines a bounded linear functional from \(\mathcal{H}_\mu\), we can almost surely uniquely extend \(h^*\) to a measurable element of \(C_\mu(\delta_t, \delta_s)^*\). To see this, it suffices to check that \[h_t = \delta_t(h) = C_\mu(\delta_t, h^*) = h^*(C_\mu(\delta_t)).\] We see that \[C_\mu(\delta_t)_s = \left(\int \omega \delta_t(\omega) \mu(d \omega)\right)_s = \int \omega_s \omega_t \mu(d \omega) = C_\mu(\delta_t, \delta_s) = s \wedge t.\] Thus, \(\dot{C}_\mu(\delta_t)_s = \mathbb{1}_{[0, t]}(s)\) and hence, \[h^*(C_\mu(\delta_t)) = \int_0^t \dot f_s d s = f_t\] as required.

As an application, the Cameron-Martin theorem allows us to bound the probability that a Brownian motion stays near a given function. In particular, for \(f \in H^{1, 2}_0([0, 1])\), denoting \(\mu_W\) for the Wiener measure, we have that \[\mathbb{P}(|B_t - f|_\infty \le \epsilon) = \tau_{-f}^* \mu_W(\overline B_\epsilon(0)) = e^{-\frac{1}{2}\|f\|^2_{H^{1, 2}_0([0, 1])}} \int_{\overline B_\epsilon(0)} e^{- f^*(\omega)} \mu_W(d \omega).\] This is then bounded above by \[e^{-\frac{1}{2}\|f\|^2_{H^{1, 2}_0([0, 1])} + \epsilon \|f^*\|_{X^*}} \mathbb{P}(|B_t|_\infty \le \epsilon).\] On the other hand, using the inequality \(e^x \ge 1 + x\), we have that \[\int_{\overline B_\epsilon(0)} e^{- f^*(\omega)} \mu_W(d \omega) \ge \int_{\overline B_\epsilon(0)} (1 + - f^*(\omega)) \mu_W(d \omega) \ge \mathbb{P}(|B_t|_\infty \le \epsilon) - \int_{\overline B_\epsilon(0)} f^*(\omega) \mu_W(d \omega)\] Now, since \(f^*_* \mu_W \sim \mathcal{N}(0, \|f\|_{\mathcal{H}_W}^2)\), the last integral is zero by spherical symmetry. Consequently, have the bound \[\mathbb{P}(|B_t - f|_\infty \le \epsilon) \in e^{-\frac{1}{2}\|f\|^2_{H^{1, 2}_0([0, 1])}} \left[\mathbb{P}(|B_t|_\infty \le \epsilon), e^{\epsilon \|f^*\|_{X^*}}\mathbb{P}(|B_t|_\infty \le \epsilon)\right]\] and thus, \[\lim_{\epsilon \downarrow 0} \frac{\mathbb{P}(|B_t - f|_\infty \le \epsilon)}{\mathbb{P}(|B_t|_\infty \le \epsilon)} = e^{-\frac{1}{2}\|f\|^2_{H^{1, 2}_0([0, 1])}}.\]

7 Mar. 2024. Precompactness in the Total Variational Topology

There is the following classical result which relates the Wasserstein distance to the weak topology on the space of measures:

Let \(X\) be a Polish space and let \(d\) be a metric compatible with the topology of \(X\). By replacing \(d\) by \(d \wedge 1\), we may assume that \(d \le 1\). Denoting \(\mathcal{M}(X)\) the space of probability measures on \(X\), for a sequence of measures \((\mu_n)_{n \in \mathbb{N}} \subseteq \mathcal{M}(X)\) and \(\mu \in \mathcal{M}(X)\), we have that \begin{equation}\label{eq:wasserstein-weak} d_1(\mu_n, \mu) \to 0 \iff \mu_n \rightharpoonup \mu. \end{equation} where \(d_1\) denotes the Wasserstein distance of order 1 corresponding to the metric \(d\) and \(\rightharpoonup\) denotes weak convergence.

While playing around with the proof of this result, I noticed that by assuming a slightly stronger condition (which is reminiscent of tightness), the result can be strengthened to convergence in total variation. In this note I record these relations.

Recall that Lusin's theorem for Radon measures states that: Let \(X\) be a locally compact Polish space and take \(\mu \in \mathcal{M}(X)\) (so \(\mu\) is automatically a Radon measure), \(f : X \to \mathbb{R}\) a bounded measurable function and \(\epsilon > 0\). Then, there exists a compact set \(K \subseteq X\) such that \(f\) is continuous on \(K\) and \(\mu(K) \ge 1 - \epsilon\). This motivates the following definition.

Definition. Given a set of measures \(M \subseteq \mathcal{M}(X)\), we say that \(M\) is strongly tight if for all \(f : X \to \mathbb{R}\) bounded measurable and \(\epsilon > 0\), there exists a compact set \(K \subseteq X\) such that \(f\) is continuous on \(K\) and for all \(\mu \in M\), \(\mu(K) \ge 1 - \epsilon\).

We remark that, by applying Lusin's theorem to \(\epsilon / n\), it is clear that any finite family of measures is automatically strongly tight. Moreover, we see that if \(\mu_n \to \mu\) in total variation, then \(\{\mu_n, \mu\}_n\) is also strongly tight. Finally, as the name suggests, it is clear that if \(M \subseteq \mathcal{M}(X)\) is strongly tight, then it is also tight.

By slightly modifying the proof of convergence in Wasserstein distance implies weak convergence, we observe the following relation between convergence in Wasserstein distance and convergence in total variation.

Proposition. Let \((\mu_n)_{n \in \mathbb{N}} \subseteq \mathcal{M}(X)\) and \(\mu \in \mathcal{M}(X)\). Then, \(\mu_n \to \mu\) in total variation if and only if \((\mu_n)_n\) is strongly tight and \(d_1(\mu_n, \mu) \to 0\).

Proof. The forward direction is obvious so we only prove the reverse. Suppose that \((\mu_n)_n\) is strongly tight and \(d_1(\mu_n, \mu) \to 0\). Then, taking \(f : X \to \mathbb{R}\) be a bounded measurable function, we will show that \[\left|\int f d \mu_n - \int f d \mu\right| \to 0.\] Fixing \(\epsilon > 0\), by strong tightness, there exists some \(K\) compact such that \(f\) is uniformly continuous on \(K\) and \(\mu(K^c), \mu_n(K^c) < \epsilon\) for all \(n\). Thus, there exists some \(\delta > 0\) such that for all \(x, y \in \Delta_\delta := \{(x, y) \in K^2 : d(x, y) < \delta\}\), we have \(|f(x) - f(y)| < \epsilon\).

Moreover, since \(d_1(\mu_n, \mu) \to 0\), there exists a sequence of couplings \(\Pi_n \in \mathcal{C}(\mu_n, \mu)\) such that \[\int d(x, y) \Pi_n(d x, d y) \to 0.\] So, \[0 \leftarrow \int d(x, y) \Pi_n(d x, d y) \ge \int_{\Delta_\delta^c \cap K^2} d(x, y) \Pi_n(d x, d y) \ge \delta \Pi_n(\Delta_\delta^c \cap K^2) \ge 0\] implying \(\Pi_n(\Delta_\delta^c \cap K^2) < \epsilon\) for sufficiently large \(n\). Thus, for sufficiently large \(n\), we have that \begin{align*} \left|\int f d \mu_n - \int f d \mu\right| & = \left|\int (f(x) - f(y)) \Pi_n(d x, d y)\right|\\ & \le \left|\int_{\Delta_\delta} (f(x) - f(y)) \Pi_n(d x, d y)\right| + \left|\int_{\Delta_\delta^c \cap K^2} (f(x) - f(y)) \Pi_n(d x, d y)\right|\\ & + \left|\int_{X \times K^c} (f(x) - f(y)) \Pi_n(d x, d y)\right| + \left|\int_{K^c \times X} (f(x) - f(y)) \Pi_n(d x, d y)\right|\\ & \le \epsilon + 2|f|_\infty \epsilon + 2|f|_\infty \mu_n(X)\mu(K^c) + 2|f|_\infty \mu_n(K^c)\mu(K)\\ & \le \epsilon(1 + 6|f|_\infty). \end{align*}

Consequently, taking \(\epsilon \to 0\) gives the desired limit.

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Alternatively, one can also obtain the above by using the equivalence of weak convergence and convergence is Wasserstein distance. In particular, we note that for a given bounded measurable function \(f : X \to \mathbb{R}\) and \(\mu \in \mathcal{M}(X)\), there exists a continuous function \(\tilde f : X \to \mathbb{R}\) such that \[\left| \int f d \mu - \int \tilde f d \mu \right| < \epsilon.\] This follows by Lusin's theorem and Tietze extension. Hence, in a similar vein, in the case where we have strong tightness, applying Tietze extension and triangle inequality provides convergence in total variation given weak convergence.

In any case, this provides a characterization of precompactness in the total variational topology.

Proposition. Let \(M \subseteq \mathcal{M}(X)\) be a family of measures. Then \(M\) is strongly tight if and only if it is precompact in the total variational topology.

Proof. The forward direction is simple. Indeed, if \(M\) is strongly tight, then it is tight and so any sequence of measures \((\mu_n)\) in \(M\) has a weakly convergent subsequence \(\mu_{n_k} \to \mu \in \mathcal{M}(X)\). However, as \((\mu_{n_k})\) is also strongly tight, the previous lemma show that this convergence can be taking to be instead convergence in total variation. Thus, any sequence in \(M\) has a convergent subsequence in total variation which implies that \(M\) is precompact.

For the converse, we will prove the contrapositive, i.e. not strongly tight implies the existence of a sequence of measures which does not have a convergence subsequence. In particular, we will construct a sequence of measures \((\mu_n) \subseteq M\) and compact sets \((K_n)\) such that for all \(m < n\), \(\mu_n(K_n^c) - \mu_m(K_n^c) > \epsilon / 2\). Thus, as \[\|\mu - \nu\|_{TV} = 2\sup_{S \in \mathcal{B}(X)} |\mu(S) - \nu(S)|,\] we have that \(\|\mu_n - \mu_m\|_{TV} > \epsilon\) for all \(m < n\) implying \((\mu_n)\) has no convergence subsequence.

Assuming \(M\) is not strongly tight, then there exists some \(\epsilon > 0\) and a bounded measurable function \(f : X \to \mathbb{R}\) such that for all compact sets \(K \subseteq X\), \(f\) is continuous on \(K\) implies the existence of some \(\mu \in M\) such that \(\mu(K^c) \ge \epsilon\). So, picking \(\mu_0 \in M\) and supposing that we are given \(\mu_1, \dots, \mu_n\), as a finite family of measures is strongly tight, we can choose a compact set \(K_n \subseteq X\) such that \(f\) is continuous on \(K_n\) and for all \(m \le n\), \(\mu_m(K_n^c) \le \epsilon / 2\). Then, by the lack of strong tightness, we can choose \(\mu_{n + 1} \in M\) such that \(\mu_{n + 1}(K_n^c) \ge \epsilon\). Consequently, \[\mu_{n + 1}(K_n^c) - \mu_m(K_n^c) \ge \epsilon - \frac{\epsilon}{2} = \frac{\epsilon}{2}\] as desired.

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To conclude, we remark the following criterion for strong tightness.

Proposition. Let \(M \subseteq \mathcal{M}(X)\) be a family of measures. Suppose that there exists a measure \(\nu \in \mathcal{M}(X)\) and \(g \in L^1(\nu)\) such that for all \(\mu \in M\), \(\mu \ll \nu\) and \(d \mu / d \nu \le g\). Then, \(M\) is strongly tight.

Proof. Define \(\eta \in \mathcal{M}(X)\) by \(d \eta = \frac{g}{\|g\|_{L^1(\nu)}} d \nu\). Then, for all \(\epsilon > 0\) and \(f : X \to \mathbb{R}\) bounded measurable, there exists some compact set \(K\) such that \(f\) is continuous on \(K\) and \(\eta(K^c) \le \|g\|_{L^1(\nu)}^{-1}\epsilon\). Then, for all \(\mu \in M\), we have that \[\mu(K^c) = \int_{K^c} \frac{d \mu}{d \nu} d \nu \le \int_{K^c} g d \nu = \|g\|_{L^1(\nu)}\eta(K^c) \le \epsilon\] as desired.

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Consequently, recall that if \((\mu_n) \subseteq \mathcal{M}(X)\) is a sequence of measures which are absolutely continuous with respect to some \(\nu \in \mathcal{M}(X)\), then \(\mu_n\) converges to \(\mu\) in total variation if and only if \(d \mu_n / d \nu \to d \mu / d \nu\) in \(L^1\). Thus, the above allows us to recover the dominated convergence theorem.

27 Oct. 2023.   Random Compact Sets in Polish Spaces

Here is a fun puzzle I thought about today: Let \(X\) be a polish space and \(K\) a random compact set of \(X\) (i.e. a random variable taking values in compact subsets of \(X\)). Fix \(\epsilon > 0\), does there necessarily exists a deterministic compact set \(A\) such that \(\mathbb{P}(K \subseteq A) \ge 1 - \epsilon\)?

This question seems quite similar to the notion of tightness for measures and it would not come as a shock that the solution to this question uses tightness of a single measure on Polish spaces.

Define \(\mu\) such that \[\mu(B) = \mathbb{P}(K \subsetneq B),\] it is not difficult to see that \(\mu\) defines a measure on \(\mathcal{B}(X)\) (note that strict subset is necessary as \(K\) might achieve \(\varnothing\) with positive probability). Thus, as a measure on \(X\) is tight, there exists a compact set \(A\) such that \[\mathbb{P}(K \subseteq A) \ge \mathbb{P}(K \subsetneq A) = \mu(A) \ge 1 - \epsilon\] which is precisely what we needed.

30 Jul. 2022.   Order Connected Sets are Measurable in \(\mathbb{R}^n\)

Yaël Dillies was reading a book by Bollobas in which he claimed without proof that an order connected set (with respect to the pointwise ordering) in \(\mathbb{R}^n\) is measurable. Yaël asked on Discord if anyone knows how to prove it. This short note contains my proof of this fact.

Definition. Given a partially ordered set \((X, (\le))\), \(S \subseteq X\) is said to be order connected if for all \(x, y \in S\), \[[x, y] := \{z \mid x \le z \le y\} \subseteq S.\]

Equipping \(\mathbb{R}^n\) with the point-wise ordering (i.e. for \(x := (x_i)_{i = 1}^n, y := (y_i)_{i = 1}^n \in \mathbb{R}^n\), \(x \le y\) if and only if \(x_n \le y_n\) for all \(n = 1, \cdots, n\)), \(\mathbb{R}^n\) form a partially ordered set and we can therefore talk about its order connected subsets.

Theorem. \(S \subseteq \mathbb{R}^n\) is Lebesgue measurable if \(S\) is order connected with respect to the point-wise ordering on \(\mathbb{R}^n\).

Proof. Defining \(S^+ := \{x \mid \exists \ y \in S, y \le x\}\) and \(S^- := \{x \mid \exists \ y \in s, x \le y\}\), we observe \(S = S^+ \cap S^-\). Therefore, it is sufficient to show that \(S^+\) and \(S^-\) are measurable. We will show measurability for \(S^+\) while the case for \(S^-\) is similar.

It is clear that for all \(x \in \mathbb{R}^n\), \(\{x\}^+\) is measurable. So, defining \(Q := S^+ \cap \mathbb{Q}^n\), \[Q^+ = \bigcup_{q \in Q} \{q\}^+ \subseteq S^+\] is measurable. Furthermore, \(S^+ \setminus Q^+ \subseteq \partial S^+\) since for all \(x \in S^\circ\), there exists some open neighborhood \(B \subseteq S^\circ\) of \(x\); so, as \(Q\) is dense in \(S^+\), there exists an element \(q \in Q \cap B\) such that \(q \le x\) and hence, \(x \in \{q\}^+ \subseteq Q^+\). Now, as the Lebesgue \(\sigma\)-algebra is complete and \(S^+ = Q^+ \cup (S^+ \setminus Q^+)\), it suffices to show that \(\partial S^+\) is a null-set.

To show \(\text{Leb}(\partial S^+) = 0\) we will invoke the Lebesgue density theorem. Namely, by showing \[\partial S^+ \subseteq \{x \in \overline{S^+} \mid d(x) \notin \{0, 1\}\}\] where \[d(x) := \lim_{\epsilon \to 0} d_\epsilon(x) = \lim_{\epsilon \to 0} \frac{\text{Leb}(\overline{S^+} \cap B_\epsilon(x))}{\text{Leb}(B_\epsilon(x))}\] as \(\overline{S^+}\) is closed and hence measurable, the Lebesgue density theorem tells us \[\text{Leb}(\partial S^+) \le \text{Leb}(\{x \in \overline{S^+} \mid d(x) \notin \{0, 1\}\}) = 0.\]

Indeed, for all \(x \in \partial S^+, \epsilon > 0\), \(\overline{S^+} \cap B_\epsilon(x)\) must contain the with the right-upper quadrant of the ball by the very definition of \(S^+\), implying \(\text{Leb}(\overline{S^+} \cap B_\epsilon(x)) \gtrapprox 2^{-n} \text{Leb}(B_\epsilon(x))\) bounding \(d(x)\) from below. Similarly, \(\overline{S^+}^c \cap B_\epsilon(x)\) must contain the left-bottom quadrant of the ball as otherwise \(x\) is contained by some \(\{y\}^+\) for some \(y \in \overline{S}^+\) contradicting \(x \in \partial S^+\). Hence, \(\text{Leb}(\overline{S^+} \cap B_\epsilon(x)) \lessapprox (1 - 2^{-n}) \text{Leb}(B_\epsilon(x))\) bounding \(d(x)\) from above and so, proving the required inclusion.

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3 Apr. 2022.   Preorder Cannot Generate the Cofinite Topology on Infinite Sets

Kevin Buzzard asked on Discord whether or not the topology generated by any preorder on \(\mathbb{C}\) can equal to either the usual topology or the cofinite topology. The first question is rather easy and one can take the base of the usual topology to be the boxes and of which is generated by the preorder \[z < w \iff \text{Re}(z) < \text{Re}(w) \text{ and } \text{Im}(z) < \text{Im}(w).\] The second part is a bit more tricky and by playing around, I found that the following is true.

Theorem. Given a infinite set \(X\), there does not exist a preorder on \(X\) which generates the cofinite topology.

We will in this short article provide a proof for this theorem. For reference, let us first quickly recall some basic definitions from lattice theory.

Definition. For a given set \(X\), a preorder \((\le)\) on \(X\) is a binary relation such that it is reflexive: for all \(x \in X\), \(x \le x\); and it is transitive: for all \(x, y, z \in X\), if \(x \le y\) and \(y \le z\), then \(x \le z\).

Definition. For a given set \(X\), a strict order \((<)\) on \(X\) is a binary relation such that it is irreflexive: for all \(x < X\), \(\neg x < x\); and it is transitive: for all \(x, y, z \in X\), if \(x \le y\) and \(y \le z\), then \(x \le z\).

It is clear that every preorder has an associated strict order by setting \(x < y\) whenever \(x \le y\) and \(\neg y \le x\). With this in mind, the topology generated the preorder \((\le)\) is defined to be is the smallest topology such that all sets of the form \(\{y \in X \mid y < x\}\) and \(\{y \in X \mid x < y\}\) are open for any \(x \in X\) where \((<)\) is the strict order associated with the preorder.

On the other hand, the cofinite topology on \(X\) is the topology in which the open sets are precisely the sets whose complements are finite or the empty set.

Proposition. Given a infinite set \(X\), there does not exist a preorder on \(X\) which generates the cofinite topology.

Proof. Suppose there exists a preorder \((\le)\) with the associated strict order \((<)\) such that the order topology generated by it is the same as the cofinite topology. For the cofinite topology, any non-empty open sets must intersect, and thus, since \(\{z \mid x < z\}\) and \(\{z \mid z < x\}\) are open in the order topology, their intersection is nonempty for all \(x\) if neither sets are empty. But, if this is the case, there exists some \(z\) such that \(z < x < z\) implying \(z < z\) which is a contradiction. Hence, at least one of the set is empty.

Now, denoting \(U_+\) the set of \(x\) for which \(\{z \mid z < x\}\) is non-empty (so \(\{z \mid x < z\} = \varnothing\) for all \(x \in U_+\)), as \(\{z \mid z < x\}\) is open, \(\{z \mid \neg z < x\}\) is finite for all \(x \in U_+\). As for all \(x, y \in U_+\), if \(y < x\), then \(x \in \{z \mid y < z\}\) contradicting \(\{z \mid y < z\} = \varnothing\), \(\neg y < x\) for all \(x, y \in U_+\). Hence, \(y \in \{z \mid \neg z < x\}\) for all \(y \in U_+\). Thus, as \(\{z \mid \neg z < x\}\) is finite, so is \(U_+\).

Similarly, defining \(U_-\) the set of \(x\) for which \(\{z \mid x < z\}\) is non-empty, by the same argument, \(U_-\) is finite. Hence, \(\{z \mid x < z\}\) and \(\{z \mid z < x\}\) is empty for all but finitely many \(x\). But then, the topology generated by the preorder has a finite base, implying only finitely many sets are open which is a contradiction.

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