tcs math – some mathematics of theoretical computer science

February 18, 2013

Talagrand’s Bernoulli Conjecture, resolved.

Filed under: Math — Tags: , , — James Lee @ 5:34 pm

Apparently, Bednorz and LataƂa have solved Talagrand’s $5,000 Bernoulli Conjecture. The conjecture concerns the supremum of a Bernoulli process.

Consider a finite subset {T \subseteq \ell^2} and define the value

\displaystyle  b(T) = \mathbb{E} \max_{t \in T} \sum_{i \geq 1} \varepsilon_i t_i\,,

where {\varepsilon_1, \varepsilon_2, \ldots} are i.i.d. random {\pm 1} . This looks somewhat similar to the corresponding value

\displaystyle  g(T) = \mathbb{E} \max_{t \in T} \sum_{i \geq 1} g_i t_i\,,

where {g_1, g_2, \ldots} are i.i.d. standard normal random variables. But while {g(T)} can be characterized (up to universal constant factors) by the Fernique-Talagrand majorizing measures theory, no similar control was known for {b(T)} . One stark difference between the two cases is that {g(T)} depends only on the distance geometry of {T} , i.e. {g(A(T))=g(T)} whenever {A} is an affine isometry. On the other hand, {b(T)} can depend heavily on the coordinate structure of {T} .

There are two basic ways to prove an upper bound on {b(T)} . One is via the trivial bound

\displaystyle  b(T) \leq \max_{t \in T} \|t\|_1\,. \ \ \ \ \ (1)

The other uses the fact that the tails of Gaussians are “fatter” than those of Bernoullis.

\displaystyle  b(T) \leq \sqrt{\frac{\pi}{2}} g(T)\,. \ \ \ \ \ (2)

This can be proved easily using Jensen’s inequality.

Talagrand’s Bernoulli conjecture is that {b(T)} can be characterized by these two upper bounds.

Bernoulli conjecture: There exists a constant {C > 0} such that for every {T \subseteq \ell^2} , there are two subsets {T_1, T_2 \subseteq \ell^2} such that

\displaystyle  T \subseteq T_1 + T_2 = \{ t_1 + t_2 : t_1 \in T_1, t_2 \in T_2 \}\,,

and

\displaystyle  g(T_1) + \sup_{t \in T_2} \|t\|_1 \leq C b(T)\,.

Note that this is a “characterization” because given such sets {T_1} and {T_2} , equations (1) and (2) imply

\displaystyle  b(T) \leq \sqrt{\frac{\pi}{2}} g(T_1) + \sup_{t \in T_2} \|t\|_1\,.

The set {T_1} controls the “Gaussian” part of the Bernoulli process, while the set {T_2} controls the part that is heavily dependent on the coordinate structure.

This beautiful problem finally appears to have met a solution. While I don’t know of any applications yet in TCS, it does feel like something powerful and relevant.

July 18, 2010

The majorizing measures theorem

Filed under: lecture, Math — Tags: , , , — James Lee @ 4:05 am

We will now prove Talagrand’s majorizing measures theorem, showing that the generic chaining bound is tight for Gaussian processes. The proof here will be a bit more long-winded than the proof from Talagrand’s book, but also (I think), a bit more accessible as well. Most importantly, we will highlight the key idea with a simple combinatorial argument.

First, let’s recall the bound we proved earlier.

Theorem 1 Let {\{X_t\}_{t \in T}} be a Gaussian process, and let {T_0 \subseteq T_1 \subseteq \cdots \subseteq T} be a sequence of subsets such that {|T_0|=1} and {|T_n| \leq 2^{2^{n}}} for {n \geq 1}. Then,

\displaystyle   \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \, d(t, T_n). \ \ \ \ \ (1)

In order to make things slightly easier to work with, we look at an essentially equivalent way to state (1). Consider a Gaussian process {\{X_t\}_{t \in T}} and a sequence of increasing partitions {\{\mathcal A_n\}_{n \geq 0}} of {T}, where increasing means that {\mathcal A_{n+1}} is a refinement of {\mathcal A_n} for {n \geq 0}. Say that such a sequence {\{\mathcal A_n\}} is admissible if {\mathcal A_0 = \{T\}} and {|\mathcal A_n| \leq 2^{2^n}} for all {n \geq 1}. Also, for a partition {P} and a point {t \in T}, we will use the notation {P(t)} for the unique set in {P} which contains {t}.

By choosing {T_n} to be any set of points with one element in each piece of the partition {\mathcal A_n}, (1) yields,

\displaystyle  \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \,\mathrm{diam}(\mathcal A_n(t)). \ \ \ \ \ (2)

We can now state our main theorem, which shows that this is essentially the only way to bound {\mathop{\mathbb E} \sup_{t \in T} X_t}.

Theorem 2 There is a constant {L > 0} such that for any Gaussian process {\{X_t\}_{t \in T}}, there exists an admissible sequence {\{\mathcal A_n\}} which satisfies,

\displaystyle  \mathop{\mathbb E} \sup_{t \in T} X_t \geq L \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} \,\mathrm{diam}(\mathcal A_n(t)). \ \ \ \ \ (3)

Recall that for a subset {A \subseteq T}, we defined {g(A) = \mathop{\mathbb E} \sup_{t \in A} X_t}, and in the last post, we proved the following “Sudakov inequality.”

Theorem 3 For some constants {\kappa > 0} and {r \geq 4}, the following holds. Suppose {\{X_t\}_{t \in T}} is a Gaussian process, and let {t_1, t_2, \ldots, t_m \in T} be such that {d(t_i,t_j) \geq \alpha} for {i \neq j}. Then,

\displaystyle   g(T) \geq \kappa \alpha \sqrt{\log_2 m} + \min_{i=1,2,\ldots,m} g(B(t_i, \alpha/r)). \ \ \ \ \ (4)

We will use only Theorem 3 and the fact that {g(A) \leq g(B)} whenever {A \subseteq B} to prove Theorem 2 (so, in fact, Theorem 2 holds with {\mathop{\mathbb E} \sup_{t \in T} X_t} replaced by more general functionals satisfying an inequality like (4)).

The partitioning scheme

First, we will specify the partitioning scheme to form an admissible sequence {\{\mathcal A_n\}}, and then we will move on to its analysis. As discussed in earlier posts, we may assume that {T} is finite. Every set {C \in \mathcal A_n} will have a value {\mathrm{rad}(C)} associated with it, such that {\mathrm{rad}(C)} is always an upper bound on the radius of the set {C}, i.e. there exists a point {x \in C} such that {C \subseteq B(x, \mathrm{rad}(C))}.

Initially, we set {\mathcal A_0 = \{T\}} and {\mathrm{rad}(T) = \mathrm{diam}(T)}. Now, we assume that we have constructed {\mathcal A_n}, and show how to form the partition {\mathcal A_{n+1}}. To do this, we will break every set {C \in \mathcal A_n} into at most {2^{2^n}} pieces. This will ensure that

\displaystyle |\mathcal A_{n+1}| \leq 2^{2^n} \cdot |\mathcal A_n| \leq 2^{2^n} \cdot 2^{2^n} = 2^{2^{n+1}}.

Let {r} be the constant from Theorem 3. Put {m = 2^{2^n}}, and let {\Delta = \mathrm{rad}(C)}. We partition {C} into {m} pieces as follows. First, choose {t_1 \in C} which maximizes the value

\displaystyle  g(B(t_1, \Delta/r^2) \cap C).

Then, set {C_1 = B(t_1, \Delta/r) \cap C}. We put {\mathrm{rad}(C_1) = \Delta/r}.

A remark: The whole idea here is that we have chosen the “largest possible piece,” (in terms of {g}-value), but we have done this with respect to the {\Delta/r^2} ball, while we cut out the {\Delta/r} ball. The reason for this will not become completely clear until the analysis, but we can offer a short explanation here. Looking at the lower bound (4), observe that the balls {B(t_i, \alpha/3)} are disjoint under the assumptions, but we only get “credit” for the {B(t_i, \alpha/r)} balls. When we apply this lower bound, it seems that we are throwing a lot of the space away. At some point, we will have to make sure that this thrown away part doesn’t have all the interesting stuff! The reason for our choice of {\Delta/r} vs. {\Delta/r^2} is essentially this: We want to guarantee that if we miss the interesting stuff at this level, then the previous level took care of it. To have this be the case, we will have to look forward (a level down), which (sort of) explains our choice of optimizing for the {\Delta/r^2} ball.

Now we continue in this fashion. Let {D_{\ell} = C \setminus \bigcup_{i=1}^{\ell-1} C_i} be the remaining space after we have cut out {\ell-1} pieces. For {\ell \leq m}, choose {t_{\ell} \in C} to maximize the value

\displaystyle  g(B(t_{\ell}, \Delta/r^2) \cap D_{\ell}).

For {\ell < m}, set {C_{\ell} = B(t_{\ell}, \Delta/r) \cap D_{\ell}}, and put {\mathrm{rad}(C_{\ell}) = \Delta/r}.

So far, we have been chopping the space into smaller pieces. If {D_{\ell} = \emptyset} for some {\ell \leq m}, we have finished our construction of {\mathcal A_{n+1}}. But maybe we have already chopped out {m-1} pieces, and still some remains. In that case, we put {C_m = D_m}, i.e. we throw everything else into {C_m}. Since we cannot reduce our estimate on the radius, we also put {\mathrm{rad}(C_m) = \Delta}.

We continue this process until {T} is exhausted, i.e. eventually for some {n} large enough, {\mathcal A_n} only contains singletons. This completes our description of the partitioning.

The tree

For the analysis, it will help to consider our partitioning process as having constructed a tree (in the most natural way). The root of the the tree is the set {T}, and its children are the sets of {\mathcal A_1}, and so on. Let’s call this tree {\mathcal W}. It will help to draw and describe {\mathcal W} in a specific way. First, we will assign values to the edges of the tree. If {C \in \mathcal A_n} and {C_j} is a child of {C} (i.e., {C_j \in \mathcal A_{n+1}} and {C_j \subseteq C}), then the edge {(C,C_j)} is given value:

\displaystyle  \kappa \cdot \frac{\mathrm{rad}(C)}{r} \cdot 2^{n/2}, \ \ \ \ \ (5)

where {\kappa} and {r} are the constants from Theorem 3.

If we define the value of a root-leaf path in {\mathcal W} as the sum of the edge lengths on that path, then for any {t \in T},

\displaystyle  \sum_{n \geq 0} 2^{n/2}\,\mathrm{diam}(\mathcal A_n(t)) \leq 2 \frac{r}{\kappa} \left(\textrm{value of the path from the root to } t\right),

simply using {\mathrm{diam}(\mathcal A_n(t)) \leq 2\, \mathrm{rad}(\mathcal A_n(t))}.

Thus in order to prove Theorem 2, which states that for some {L > 0},

\displaystyle  g(T) \geq L \sup_{t \in T} \sum_{n \geq 0} 2^{n/2}\,\mathrm{diam}(\mathcal A_n(t)),

it will suffice to show that for some (other) constant {L > 0}, for any root-leaf path {P} in {\mathcal W}, we have

\displaystyle  g(T) \geq L \cdot \mathrm{value}(P). \ \ \ \ \ (6)

Before doing this, we will fix a convention for drawing parts of {\mathcal W}. If a node {C \in \mathcal A_n} has children {C_1, C_2, \ldots, C_m}, we will draw them from left to right. We will call an edge {(C,C_m)} a right turn and every other edge will be referred to as a left turn. Note that some node {C} may not have any right turn coming out of it (if the partitioning finished before the last step). Also, observe that along a left turn, the radius always drops by a factor of {r}, while along a right turn, it remains the same.

We now make two observations about computing the value {\mathrm{value}(P)} up to a universal constant.

Observation (1): In computing the value of a root-leaf path {P}, we only need to consider right turns.

To see this, suppose that we have a right turn followed by a consecutive sequence of left turns. If the value of the right turn is {\frac{\kappa}{r} \Delta 2^{n/2}}, then the value of the following sequence of left turns is, in total, at most

\displaystyle  \frac{\kappa}{r} \sum_{j=1}^{\infty} 2^{(n+j)/2} \frac{\Delta}{r^j} \leq O(1) \frac{\kappa}{r} \Delta 2^{n/2}.

In other words, because the radius decreases by a factor of {r} along every left turn, their values decrease geometrically, making the whole sum comparable to the preceding right turn. (Recall that {r \geq 4}, so indeed the sum is geometric.)

If the problem of possibly of having no right turn in the path {P} bothers you, note that we could artificially add an initial right turn into the root with value {\mathrm{diam}(T)}. This is justified since {g(T) \geq \frac12 \mathrm{diam}(T)} always holds. A different way of saying this is that if the path really contained no right turn, then its value is {O(\mathrm{diam}(T))}, and we can easily prove (6).

Observation (2): In computing the value of a root-leaf path {P}, we need only consider the last right turn in any consecutive sequence of right turns.

Consider a sequence of consecutive right turns, and the fact that the radius does not decrease. The values (taking away the {\kappa/r} factor) look like {\Delta 2^{n/2}, \Delta 2^{(n+1)/2}, \Delta 2^{(n+2)/2}, \ldots}. In other words, they are geometrically increasing, and thus using only the last right turn in every sequence, we only lose a constant factor.

We will abbreviate last right turn to LRT, and write {\mathrm{value}_{\mathrm{LRT}}(P)} to denote the value of {P}, just counting last right turns. By the two observations, to show (6) (and hence finish the proof), it suffices to show that, for every root-leaf path {P} in {\mathcal W},

\displaystyle  2 \cdot g(T) \geq \mathrm{value}_{\mathrm{LRT}}(P). \ \ \ \ \ (7)

The analysis

Recall that our tree {\mathcal W} has values on the edges, defined in (5). We will also put some natural values on the nodes. For a node {C} (which, recall, is just a subset {C \subseteq T}), we put {\mathrm{value}(C) = g(C)}. So the edges have values and the nodes have values. Thus given any subset of nodes and edges in {\mathcal W}, we can talk about the value of the subset, which will be the sum of the values of the objects it contains. We will prove (7) by a sequence of inequalities on subsets.

Fix a root-leaf path {P}, for which we will prove (7). Let’s prove the fundamental inequality now. We will consider two consecutive LRTs along {P}. (If there is only one LRT in {P}, then we are done by the preceding remarks.) See the figure below. The dashed lines represent a (possibly empty) sequence of left turns and then right turns. The two LRTs are marked.

We will prove the following inequality, which is the heart of the proof. One should understand that the inequality is on the values of the subsets marked in red. The first subset contains two nodes, and the second contains two nodes and an edge.

Figure A.

Figure A.

With this inequality proved, the proof is complete. Let’s see why. We start with the first LRT. Since g(T) \geq g(C) for any node C in \mathcal W, we have the inequality:

This gets us started. Now we apply the inequality of Figure A repeatedly to each pair of consecutive LRTs in the path {P}. What do we have when we’ve exhausted the path P? Well, precisely all the LRTs in {P} are marked, yielding {2 \cdot g(T) \geq \mathrm{value}_{\mathrm{LRT}}(P)}, as desired.

The LRT inequality

Now we are left to prove the inequality in Figure A. First, let’s label some of the nodes. Let {\Delta = \mathrm{rad}(C)}, and suppose that {C \in \mathcal A_n}. The purple values are not the radii of the corresponding nodes, but they are upper bounds on the radii, recalling that along every left turn, the radius decreases by a factor of {r}. Since there are at least two left turns in the picture, we get a {\Delta/r^2} upper bound on the radius of {J}.

Part of the inequality is easy: We have {g(A) \geq g(B)} since {B \subseteq A}. So we can transfer the red mark from {A} to {B}. We are thus left to prove that

\displaystyle  g(C) \geq \frac{\kappa}{r} \Delta 2^{n/2} + g(J). \ \ \ \ \ (8)

This will allow us to transfer the red mark from {C} to the LRT coming out of {C} and to {J}.

When {C} was partitioned into {m = 2^{2^n}} pieces, this was by our greedy partitioning algorithm using centers {t_1, t_2, \ldots, t_m}. Since we cut out the {\Delta/r} ball around each center, we have {d(t_i, t_j) \geq \Delta/r} for all {i \neq j}. Applying the Sudakov inequality (Theorem 3), we have

\displaystyle  \begin{array}{rl} g(C) &\displaystyle \geq \kappa \frac{\Delta}{r} \sqrt{\log_2 m} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2)) \\ \\ &\displaystyle = \frac{\kappa}{r} \Delta 2^{n/2} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2)) \\ \\ &\displaystyle \geq \frac{\kappa}{r} \Delta 2^{n/2} + \min_{i=1,\ldots,m} g(B(t_i, \Delta/r^2) \cap D_i) \\ \\ &\displaystyle = \frac{\kappa}{r} \Delta 2^{n/2} + g(B(t_m, \Delta/r^2) \cap D_m), \\ \end{array}

where the last line follows from the greedy manner in which the {t_i}‘s were chosen.

But now we claim that

\displaystyle  g(B(t_m, \Delta/r^2) \cap D_m) \geq g(J). \ \ \ \ \ (9)

This follows from two facts. First, {J \subseteq D_m} (since {D_m=C_m} actually). Secondly, the radius of {J} is at most {\Delta/r^2}! But {t_m} was chosen to maximize the value of {g(B(t_m, \Delta/r^2) \cap D_m)} over all balls of radius {\Delta/r^2}, so in particular its {g}-value is at least that of the {\Delta/r^2} ball containing {J}.

Combining (9) and the preceding inequality, we prove (8), and thus that the inequality of Figure A is valid. This completes the proof.

July 8, 2010

Majorizing measures: Gaussian tools

Filed under: lecture, Math — Tags: , , , — James Lee @ 2:33 am

In order to prove that the chaining argument is tight, we will need some additional properties of Gaussian processes. For the chaining upper bound, we used a series of union bounds specified by a tree structure. As a first step in producing a good lower bound, we will look at a way in which the union bound is tight.

Theorem 1 (Sudakov inequality) For some constant C > 0, the following holds. Let {\{X_t\}_{t \in T}} be a Gaussian process such that for every distinct {s,t \in T}, we have {d(s,t) \geq \alpha}. Then,

\displaystyle  \mathop{\mathbb E} \sup_{t \in T} X_t \geq C \alpha \sqrt{\log |T|}.

The claim is an elementary calculation for a sequence of i.i.d. {N(0,1)} random variables {g_1, g_2, \ldots, g_n} (i.e. {\mathop{\mathbb E} \sup_i g_i \geq C\sqrt{\log n}}). We will reduce the general case to this one using Slepian’s comparison lemma.

Lemma 2 (Slepian’s Lemma) Let {\{X_t\}_{t \in T}} and {\{Y_t\}_{t \in T}} be two Gaussian processes such that for all {s,t \in T},

\displaystyle   \mathop{\mathbb E} \,|X_s - X_t|^2 \geq \mathop{\mathbb E} \,|Y_s - Y_t|^2. \ \ \ \ \ (1)

Then {\mathop{\mathbb E} \sup_{t \in T} X_t \geq \mathop{\mathbb E} \sup_{t \in T} Y_t}.

There is a fairly elementary proof of Slepian’s Lemma (see, e.g. the Ledoux-Talagrand book), if one is satisfied with the weaker conclusion {2\, \mathop{\mathbb E}\,|X_s-X_t|^2 \geq \mathop{\mathbb E}\,|Y_s-Y_t|^2}, which suffices for our purposes.

To see that Lemma 2 yields Theorem 1, take a family {\{X_t\}_{t \in T}} with {d(s,t) \geq \alpha} for all {s \neq t \in T} and consider the associated variables {Y_t = \frac{\alpha}{\sqrt{2}} g_t} where {\{g_t\}_{t \in T}} is a family of i.i.d. {N(0,1)} random variables. It is straightforward to verify that (1) holds, hence by the lemma, {\mathop{\mathbb E} \sup_{t \in T} X_t \geq \frac{\alpha}{\sqrt{2}} \mathop{\mathbb E} \sup_{t \in T} g_t}, and the result follows from the i.i.d. case.

The Sudakov inequality gives us “one level” of a lower bound; the following strengthening will allow us to use it recursively. If we have a Gaussian process {\{X_t\}_{t \in T}} and {A \subseteq T}, we will use the notation

\displaystyle  g(A) = \mathop{\mathbb E} \sup_{t \in A} X_t.

For {t \in T} and {R \geq 0}, we also use the notation

\displaystyle  B(t,R) = \{ s \in T : d(s,t) \leq R \}.

Here is the main theorem of this post; its statement is all we will require for our proof of the majorizing measures theorem:

Theorem 3 For some constants C > 0 and {r > 1}, the following holds. Suppose {\{X_t\}_{t \in T}} is a Gaussian process, and let {t_1, t_2, \ldots, t_m \in T} be such that {d(t_i,t_j) \geq \alpha} for {i \neq j}. Then,

\displaystyle  g(T) \geq C \alpha \sqrt{\log m} + \min_{i=1,2,\ldots,m} g(B(t_i, \alpha/r)).

The proof of the preceding theorem relies on the a strong concentration property for Gaussian processes. First, we recall the classical isoperimetric inequality for Gaussian space (see, for instance, (2.9) here).
We remind the reader that for a function {F : \mathbb R^n \rightarrow \mathbb R},

\displaystyle \|F\|_{\mathrm{Lip}} = \sup_{x \neq y \in \mathbb R^n} \frac{|F(x)-F(y)|}{\|x-y\|}.

Theorem 4 Let {F : \mathbb R^n \rightarrow \mathbb R}, and let {\mu = \int F \,d\gamma_n}, where {\gamma_n} is the standard {n}-dimensional Gaussian measure. Then,

\displaystyle   \gamma_n\left(x \in \mathbb R^n : |F(x)-\mu| > \lambda\right) \leq 2\,\exp\left(\frac{-\lambda^2}{2 \|F\|_{\mathrm{Lip}}}\right). \ \ \ \ \ (2)

Using this, we can prove the following remarkable fact.

Theorem 5 Let {\{X_t\}_{t \in T}} be a Gaussian process, then

\displaystyle   \mathbb P\left(\left|\sup_{t \in T} X_t - \mathop{\mathbb E} \sup_{t \in T} X_t\right| > \lambda\right) \leq 2\,\exp\left(\frac{-\lambda^2}{2 \sup_{t \in T} \mathop{\mathbb E}(X_t^2)}\right). \ \ \ \ \ (3)

A notable aspect of this statement is that only the maximum variance affects the concentration, not the number of random variables. We now prove Theorem 5 using Theorem 4.

Proof: We will prove it in the case {|T|=n}, but of course our bound is independent of {n}. The idea is that given a Gaussian process {\{X_1, X_2, \ldots, X_n\}}, we can write

\displaystyle  X_i = a_{i1} \,g_1 + a_{i2}\, g_2 + \cdots + a_{in}\, g_n,

for {i=1,2,\ldots, n}, where {\{g_i\}_{i=1}^n} are standard i.i.d. normals, and the matrix {A=(a_{i,j})} is a matrix of real coefficients. In this case, if {g = (g_1, g_2, \ldots, g_n)} is a standard {n}-dimensional Gaussian, then the vector {Ag} is distributed as {(X_1, X_2, \ldots, X_n)}.

If we put {F(x)=\max \{ (Ax)_i : i=1,\ldots,n\}}, then Theorem 4 yields (3) as long as {\|F\|_{\mathrm{Lip}} \leq \max_i \sqrt{\mathop{\mathbb E}(X_i^2)}}. It is easy to see that

\displaystyle  \|F\|_{\mathrm{Lip}} = \|A\|_{2 \rightarrow \infty} = \sup_{\|x\|_2 = 1} \|A x\|_{\infty}.

But {\|A\|_{2 \rightarrow \infty}} is just the maximum {\ell_2} norm of any row of {A}, and the {\ell_2} norm of row {i} is

\displaystyle \sqrt{\sum_{j=1}^n a_{ij}^2} = \sqrt{\mathop{\mathbb E}(X_i^2)}.

\Box

Using this theorem, we are ready to prove Theorem 3. I will only give a sketch here, but filling in the details is not too difficult.

Assume that the conditions of Theorem 3 hold. Pick an arbitrary {t_0 \in T}, and recall that we can write

\displaystyle g(T) = \mathop{\mathbb E} \sup_{t \in T} X_t = \mathop{\mathbb E} \sup_{t \in T} (X_t - X_{t_0})

since our gaussians are centered.

Now, by Theorem 1,

\displaystyle \mathop{\mathbb E} \max_{i=1,\ldots,m} \left(X_{t_i} - X_{t_0}\right) \geq C \alpha \sqrt{\log m}.

Suppose that {t_1} achieves this. By definition,

\displaystyle g(B(t_1, \alpha/r)) = \mathop{\mathbb E} \sup_{t \in B(t_1, \alpha/r)} \left(X_t - X_{t_1}\right),

so we could hope that for some {t \in B(t_1,\alpha/r)}, we simultaneously have {X_t - X_{t_1} \geq g(B(t_1,\alpha/r))}, yielding

\displaystyle  X_t - X_{t_0} = (X_t - X_{t_1}) + (X_{t_1} - X_{t_0}) \geq C\alpha \sqrt{\log m} + g(B(t_1, \alpha/r)). \ \ \ \ \ (4)

The problem, of course, is that the events we are discussing are not independent.

This is where Theorem 5 comes in. For any {i}, all the variances of the variables {\{X_t - X_{t_i} : t \in B(t_i,\alpha/r)\}} are bounded by {d(t,t_i)^2 \leq (\alpha/r)^2}. This implies that we can choose a constant {c_0 > 0} such that

\displaystyle  \mathbb P\left(\left|\sup_{t \in B(t_i,\alpha/r)} X_t - g(B(t_i, \alpha/r))\right| > c_0 (\alpha/r) \sqrt{\log m}\right) \leq m^{-2}.

So, in fact, we can expect that none of the {m} random variables {\sup_{t \in B(t_i,\alpha/r)} X_t} will deviate from its expected value by more than {c_0 (\alpha/r) \sqrt{\log m}}. Which means we can (morally) replace (4) by

\displaystyle  \begin{array}{rl} X_t - X_{t_0} &= (X_t - X_{t_1}) + (X_{t_1} - X_{t_0}) \\ &\geq C\alpha \sqrt{\log m} + g(B(t_1, \alpha/r)) - c_0 (\alpha/r) \sqrt{\log m}. \end{array}

But now by choosing {r = 2 C c_0}, the error term is absorbed.

June 15, 2010

The generic chaining

Filed under: lecture, Math — Tags: , , , — James Lee @ 11:27 am

In the last post, we considered a Gaussian process {\{X_t\}_{t \in T}} and were trying to find upper bounds on the quantity {\mathop{\mathbb E}\sup_{t \in T} X_t}. We saw that one could hope to improve over the union bound by clustering the points and then taking mini union bounds in each cluster.

Hierarchical clustering

To specify a clustering, we’ll take a sequence of progressively finer approximations to our set {T}. First, recall that we fixed {t_0 \in T}, and we have used the observation that {\mathop{\mathbb E}\sup_{t \in T} X_t = \mathop{\mathbb E}\sup_{t \in T} (X_t-X_{t_0})}.

Now, assume that {T} is finite. Write {T_0 = \{t_0\}}, and consider a sequence of subsets {\{T_n\}} such that {T_0 \subseteq T_1 \subseteq T_2 \subseteq \cdots \subseteq T}. We will assume that for some large enough {m}, we have {T_n = T} for {n \geq m}. For every {n \geq 0}, let {\pi_n : T \rightarrow T_n} denote a “closest point map” which sends t \in T to the closest point in T_n.

The main point is that we can now write, for any {t \in T},

\displaystyle   X_t - X_{t_0} = \sum_{n \geq 1} X_{\pi_n(t)} - X_{\pi_{n-1}(t)}. \ \ \ \ \ (1)

This decomposition is where the term “chaining” arises, and now the idea is to bound the probability that {X_t - X_{t_0}} is large in terms of the segments in the chain.

What should {T_n} look like?

One question that arises is how we should think about choosing the approximations {T_n}. We are trading off two measures of quality: The denser {T_n} is in the set {T} (or, more precisely, in the set {T_{n-1}}) the smaller the variances of the segments {X_{\pi_n(t)}-X_{\pi_{n-1}(t)}} will be. On the other hand, the larger {T_n} is, the more segments we’ll have to take a union bound over.

So far, we haven’t used any property of our random variables except for the fact that they are centered. To make a more informed decision about how to choose the sets {\{T_n\}}, let’s recall the classical Gaussian concentration bound.

Lemma 1 For every {s,t \in T} and {\lambda > 0},

\displaystyle   \mathop{\mathbb P}(X_s - X_t > \lambda) \leq \exp\left(-\frac{\lambda^2}{2\, d(s,t)^2}\right). \ \ \ \ \ (2)

This should look familiar: {X_s-X_t} is a mean-zero Gaussian with variance {d(s,t)^2}.

Now, a first instinct might be to choose the sets {T_n} to be progressively denser in {T}. In this case, a natural choice would be to insist on something like {T_n} being a {2^{-n}}-net in {T}. If one continues down this path in the right way, a similar theory would develop. We’re going to take a different route and consider the other side of the tradeoff.

Instead of insisting that {T_n} has a certain level of accuracy, we’ll insist that {T_n} is at most a certain size. Should we require {|T_n| \leq n} or {|T_n| \leq 2^n}, or use some other function? To figure out the right bound, we look at (2). Suppose that {g_1, g_2, \ldots, g_m} are i.i.d. {N(0,1)} random variables. In that case, applying (2) and a union bound, we see that to achieve

\displaystyle  \mathop{\mathbb P}(\exists i : g_i > B) \leq m \mathop{\mathbb P}(g_1 > B) < 1,

we need to select {B \asymp \sqrt{\log m}}. If we look instead at {m^2} points instead of {m} points, the bound grows to {\sqrt{2 \log m}}. Thus we can generally square the number of points before the union bound has to pay a constant factor increase. This suggests that the right scaling is something like {|T_{n+1}| = |T_n|^2}. So we’ll require that {|T_n| \leq 2^{2^n}} for all {n \geq 1}.

The generic chaining

This leads us to the generic chaining bound, due to Fernique (though the formulation we state here is from Talagrand).

Theorem 2 Let {\{X_t\}_{t \in T}} be a Gaussian process, and let {T_0 \subseteq T_1 \subseteq \cdots \subseteq T} be a sequence of subsets such that {|T_0|=1} and {|T_n| \leq 2^{2^{n}}} for {n \geq 1}. Then,

\displaystyle   \mathop{\mathbb E}\sup_{t \in T} X_t \leq O(1) \sup_{t \in T} \sum_{n \geq 0} 2^{n/2} d(t, T_n). \ \ \ \ \ (3)

Proof: As before, let {\pi_n : T \rightarrow T_n} denote the closest point map and let {T_0 = \{t_0\}}. Using (2), for any {n \geq 1}, {t \in T}, and {u > 0}, we have

\displaystyle  \mathop{\mathbb P}\left(|X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| > u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right) \leq \exp\left(-\frac{u^2}{2} 2^n\right).

Now, the number of pairs {(\pi_n(t),\pi_{n-1}(t))} can be bounded by {|T_n| \cdot |T_{n-1}| \leq 2^{2^{n+1}}}, so we have

\displaystyle   \mathop{\mathbb P}\left(\exists t : |X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| > u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right) \leq 2^{2^{n+1}} \exp\left(-\frac{u^2}{2} 2^n\right). \ \ \ \ \ (4)

If we define the event

\displaystyle  \Omega_u = \left\{ \forall n \geq 1, t \in T : |X_{\pi_n(t)} - X_{\pi_{n-1}(t)}| \leq u 2^{n/2} d(\pi_n(t),\pi_{n-1}(t))\right\},

then summing (4) yields,

\displaystyle   \mathop{\mathbb P}(\overline{\Omega_u}) \leq \sum_{n \geq 1} 2^{2^{n+1}} \exp\left(-\frac{u^2}{2} 2^n\right) \leq O(1)\, e^{-u^2} \ \ \ \ \ (5)

for {u \geq 4}, since we get geometrically decreasing summands.

Write

\displaystyle  S = \sup_{t \in T} \sum_{n \geq 1} 2^{n/2} d(\pi_n(t), \pi_{n-1}(t)).

Note that if {\Omega_u} occurs, then {\sup_{t \in T} (X_t - X_{t_0}) \leq uS}. Thus (5) implies that for u \geq 4,

\displaystyle  \mathop{\mathbb P}(\sup_{t \in T} X_t - X_{t_0} > uS) \leq O(1) \, e^{-u^2},

which implies that

\displaystyle   \mathop{\mathbb E} \sup_{t \in T} X_t \leq O(S) \leq O(1) \sup_{t \in T} \sum_{n \geq 1} 2^{n/2} d(\pi_n(t), \pi_{n-1}(t)). \ \ \ \ \ (6)

Finally, by the triangle inequality,

\displaystyle d(\pi_n(t), \pi_{n-1}(t)) \leq d(t, T_n) + d(t, T_{n-1}) \leq 2\,d(t,T_{n-1}).

Plugging this into (6) recovers (3). \Box

Theorem 1.2 gives us a fairly natural way to upper bound the expected supremum using a hierarchical clustering of {T}. Rather amazingly, as we’ll see in the next post, this upper bound is tight. Talagrand’s majorizing measure theorem states that if we take the best choice of {\{T_n\}} in Theorem 1.2, then the upper bound in (3) is within a constant factor of {\mathop{\mathbb E} \sup_{t \in T} X_t}.

Majorizing measures

Filed under: Math — Tags: , , , , — James Lee @ 12:44 am

At STOC 2010 last week, Talagrand gave a presentation on some of his favorite open problems, which included a quick review of Gaussian processes and the majorizing measures theory. In joint work with Jian Ding and Yuval Peres, we recently showed how the cover time of graphs can be characterized by majorizing measures.

While I’ll eventually try to give an overview of this connection, I first wanted to discuss how majorizing measures are used to control Gaussian processes. In the next few posts, I’ll attempt to give an idea of how this works. I have very little new to offer over what Talagrand has already written; in particular, I will be borrowing quite heavily from Talagrand’s book (which you might read instead).

Gaussian processes

Consider a Gaussian process {\{X_t\}_{t \in T}} for some index set {T}. This is a collection of jointly Gaussian random variables, meaning that every finite linear combination of the variables has a Gaussian distribution. We will additionally assume that the process is centered, i.e. {\mathbb E(X_t) = 0} for all {t \in T}.

It is well-known that such a process is completely characterized by the covariances {\{\mathop{\mathbb E}(X_s X_t)\}_{s,t \in T}}. For {s,t \in T}, consider the canonical distance,

\displaystyle  d(s,t) = \sqrt{\mathbb E\,|X_s-X_t|^2},

which forms a metric on {T}. (Strictly speaking, this is only a pseudometric since possibly {d(s,t)=0} even though {X_s} and {X_t} are distinct random variables, but we’ll ignore this.) Since the process is centered, it is completely specified by the distance {d(s,t)}, up to translation by a Gaussian (e.g. the process {\{X_t + X_{t_0}\}_{t \in T}} will induce the same distance for any {t_0 \in T}).

A concrete perspective

If the index set {T} is countable, one can describe every such process in the following way. Let {\{g_i\}_{i=1}^{\infty}} be a sequence of i.i.d. standard Gaussians, let {T \subseteq \ell^2}, and put

\displaystyle  X_t = \sum_{i \geq 1} g_i t_i.

In this case, it is easy to check that {d(s,t) = \|s-t\|_2} for {s,t \in T}. (That this construction is universal follows from the fact that every two separable Hilbert spaces are isomorphic.)

Random projections

If {T} is finite, then we can think of {T \subseteq \mathbb R^n} for some {n \in \mathbb N}. In this case, if {g} is a standard {n}-dimensional Gaussian, then

\displaystyle  X_t = \langle g, t \rangle,

and we can envision the process as the projection of {T} onto a uniformly random direction.

Studying the maxima

We will be concerned primarily with the value,

\displaystyle \mathbb E \sup_{t \in T} X_t.

(I.e. the expected value of the extremal node circled above.) One may assume that {T} is finite without losing any essential ingredient of the theory, in which case the supremum can be replaced by a maximum. Note that we are studying the tails of the process. Dealing with these extremal values is what makes understanding the above quantity somewhat difficult.

As some motivation for the classical study of this quantity, one has the following.

Theorem 1 For a separable Gaussian process {\{X_t\}_{t \in T}}, the following two assertions are equivalent.

  1. The map {t \mapsto X_t(\omega)} is uniformly continuous (as a map from {(T,d)} to {\mathbb R)} with probability one.
  2. As {\varepsilon \rightarrow 0},

    \displaystyle \mathop{\mathbb E}\sup_{d(s,t) \leq \varepsilon} |X_s-X_t| \rightarrow 0.

However, from our viewpoint, the quantitative study of {\mathbb E\sup_{t \in T} X_t} in terms of the geometry of {(T,d)} will play the fundamental role.

Bounding the sup

We will concentrate first on finding good upper bounds for {\mathop{\mathbb E}\sup_{t \in T} X_t}. Toward this end, fix some {t_0 \in T}, and observe that

\displaystyle  \mathop{\mathbb E}\sup_{t \in T} X_t = \mathop{\mathbb E}\sup_{t \in T} (X_t - X_{t_0}).

Since {\sup_{t \in T} (X_t - X_{t_0})} is a non-negative random variable, we can write

\displaystyle  \mathop{\mathbb E} \sup_{t \in T} (X_t - X_{t_0}) = \int_{0}^{\infty} \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right)\,du,

and concentrate on finding upper bounds on the latter probabilities.

Improving the union bound

As a first step, we might write

\displaystyle  \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right) \leq \sum_{t \in T} \mathop{\mathbb P}\left(X_t - X_{t_0} > u\right).

While this bound is decent if the variables {\{X_t-X_{t_0}\}_{t \in T}} are somewhat independent, it is rather abysmal if the variables are clustered.

Since the variables in, e.g. {S_1}, are highly correlated (in the “geometric” language, they tend to project close together on a randomly chosen direction), the union bound is overkill. It is natural to choose representatives {t_1 \in S_1} and {t_2 \in S_2}. We can first bound {X_{t_1}-X_{t_0}} and {X_{t_2}-X_{t_0}}, and then bound the intra-cluster values {\{X_t - X_{t_1}\}_{t \in S_1}} and {\{X_t - X_{t_2}\}_{t \in S_2}}. This should yield better bounds as the diameter of {S_1} and {S_2} are hopefully significantly smaller than the diameter of {T}.

Formally, we have

\displaystyle  \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right) \leq

\displaystyle  \mathop{\mathbb P}(X_{t_1} - X_{t_0} > u/2) + \sum_{t \in S_1} \mathop{\mathbb P}(X_t - X_{t_1} > u/2)

\displaystyle  + \mathop{\mathbb P}(X_{t_2} - X_{t_0} > u/2) + \sum_{t \in S_2} \mathop{\mathbb P}(X_t - X_{t_2} > u/2).

Of course, there is no reason to stop at one level of clustering, and there is no reason that we should split the contribution {u = u/2 + u/2} evenly. In the next post, we’ll see the “generic chaining” method which generalizes and formalizes our intuition about improving the union bound. In particular, we’ll show that every hierarchical clustering of our points offers some upper bound on \mathbb E \sup_{t \in T} X_t.

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