Yeah, thank you. So the topic of this talk is

a generalized central limit conjecture for convex bodies. And this is joint work

with Yin Tat and Santosh. So let’s start with the

central limit theorem that we all learned

in probability. And so this is the textbook

version of central limit theorem that I

learned, that suppose we were given n IID

samples, X1 to Xn, from some good enough

distribution with 0 mean. Then the sum of these

random variables, Xi, scaled by a factor of

1 over square root of n, this thing converges

in distribution to a Gaussian distribution. So in the theorem I learned

from probability course, this “good” means that you

have some bounded, say, second moment. So another way to

interpret this result is suppose we write this X1

to Xn as n-dimensional vector. Then this vector has a

product distribution. And so equivalently, the

textbook central limit theorem can be stated as

the inner product of this n-dimensional

vector, X1 to Xn, with a unit vector

where each coordinate is 1 over square root of n. This inner product is

close to a Gaussian. So natural question to

ask is, what if we have other distributions other than

probability as our product distribution, and

what if we look at the marginal distribution

of this distribution in other directions? And this is the motivation

for the central limit theorem for convex bodies. And before we can talk

about the central limit theorem for convex

bodies, we need to first introduce

some notion called isotropic log-concave measures. So a density function p

is called log-concave. It takes the form exponential

of a negative convex function f. And a probability measure

is called log-concave if it has a log-concave density. The typical examples

of log-concave measures include the uniform

measure on convex bodies and the Gaussian measure. So let’s see why these

measures are log-concave. So I guess for the Gaussian

measure, it’s straightforward. But for the uniform

measure on the convex body, we have the density

inside the convex body is 1 over the volume

of the convex body. And the density outside is 0. So that function f would take a

constant inside the convex body and infinity outside. So it’s crucial that

we allow the function f to take value infinity. And at the point where

it takes value infinity, it just means the density

at that point is 0. So now a probability

measure is called isotropic if it has 0 mean and

identity covariance. And the canonical example

of an isotropic log-concave distribution is the

standard Gaussian measure in Rn, which I will

write as N 0, identity. So now we are given

these definitions, we are ready to state

the central limit theorem for convex bodies. And roughly, it is as follows. So the central limit theorem

for convex bodies states that suppose we are given an

isotropic log-concave measure p in Rn, and we sample

random vector y from p. Then we have this guarantee,

which is a little bit hard to explain. So roughly, it says

that along most of the directions

on the unit sphere, the marginal distribution of the

isotropic log-concave measure on that direction is

close to a Gaussian. So more concretely, let’s try

to look at this long formula. So this probability is

taken only with respect to the vector x which is sampled

randomly on the unit sphere. And this total variation

distance is only with respect to the randomness of

the vector y, which is from the isotropic

log-concave distribution. So what this is

saying is that most of the direction,

all but cn fraction of the direction

on the unit sphere, if we project the isotropic

log-concave measure p until the direction x,

then we have something that’s close to the standard

normal distribution where this notion of close is

saying that its total variation distance is bounded

by some cn where cn is the sequence of

constants that goes to 0 as n goes to infinity. So there is the– yeah? AUDIENCE: So this doesn’t

mention convex body just because log-concave measure, so

it’s really stronger than that. Is that the– HAOTIAN JIANG: Yeah. I guess originally,

this thing is saying that if you have

a uniform distribution over isotropic convex bodies,

then you have this guarantee. But it turns out

it can generalize to any isotropic

log-concave distributions. So yeah, any other

questions, or are we all happy with this

mysterious definition? AUDIENCE: Is this

equivalent, by the way, to say that [INAUDIBLE]

log-concave distribution that are convex bodies? HAOTIAN JIANG: OK, at

that, I’m not sure. I always think intuitively

that it’s equivalent, but I don’t know a

formal statement of that. So there’s a long

line of work that– yeah? AUDIENCE: The probability’s

over x and y, right? HAOTIAN JIANG: So this

probability, it’s only over x. So you uniformly sample a

direction on the unit sphere and then fix this direction. You look at the projection

of your isotropic log-concave measure onto this

direction and ask, is this distribution close

to the standard normal distribution? Yeah, so there’s a long

line of work that– AUDIENCE: [INAUDIBLE]

doesn’t make sense, right? I guess it’s a– AUDIENCE: You’re looking at

a distribution of a line. AUDIENCE: Yes. HAOTIAN JIANG: Yeah, [INAUDIBLE]

one-dimensional margin of my isotropic

log-concave distribution. AUDIENCE: [INAUDIBLE]

y is a random variable. HAOTIAN JIANG: Oh, no. I guess that only comes in

this total variation distance. AUDIENCE: So this

inner product, x and y, is really this distribution

projection, which is just– HAOTIAN JIANG: Yeah, exactly. AUDIENCE: That’s the

part that’s confusing. HAOTIAN JIANG: Yeah,

just, I’m writing it this way [INAUDIBLE]. AUDIENCE: It’s really

p projected on x. HAOTIAN JIANG: Yeah, exactly. Exactly, yeah. So yeah, maybe that’s a

better way of writing it. So right. So there’s the list of

work that gets the improved bounds on [INAUDIBLE]

constant cn or how fast this converges to 0. The first one of

these is by Klartag, which proves that this

constant cn is bounded by 1 over square root of log n. And this is the work that

makes this central limit theorem for convex bodies. And I guess in the

same year, there is also a weaker bound,

like cn less than or equal to 1 over

log n to the 1 over 6. And later on, this is

improved by Klartag to 1 over some poly n and

a few follow-up work. So this leads to current

best bound on cn, which is 1 over n

to the one quarter. This is by Lee and

Vempala in 2017. So in these bounds, there

might be some log factors that I’m hiding. So how do we derive

these bounds? [INAUDIBLE] that

all these bounds are derived by a

notion of thin shell. So the motivation for thin

shell is the following. Suppose I randomly

sample a vector x from an isotropic

log-concave distribution. Then the expectation

of the Euclidean norm square of x is n. So throughout this talk, I will

use this magnitude notation to denote the Euclidean norm. It just looks simpler. So if I sample random

isotropic log-concave vector, then the expectation of

its Euclidean norm square is going to be n. And moreover, there’s

this concentration of the Euclidean norm of

x around square root of n. And the thin-shell

constant captures how much it concentrates

around square root of n. So the thin-shell constant

is defined as sigma n is the thin [INAUDIBLE]. Sigma n squared is defined

to be the expectation of the x minus square root

of n squared expectation. And moreover, you take the sup

over all isotropic log-concave measures in Rn. So roughly, this quantity

on the right-hand side is close to the variance

of the Euclidean norm of x. They’re close up to

a constant factor. But this whole thing–

so in some literature, you see this as the

definition of sigma n squared. They are equivalent up

to a constant factor. But the important thing

is that all these things are much smaller

than the expectation of the Euclidean norm of x,

which is square root of n. So that’s why I say

it concentrate around square root of n. And the first result that

proves this concentration is by Klartag, which

shows that sigma n, the thin-shell constant,

is bounded by square root of n over square root of log n. And the current best result

on the thin-shell constant is by Lee and

Vempala, which proves that sigma n is less than or

equal to n to the one quarter. So yeah, the famous

thin-shell conjecture by Anttilla-Ball-Parissinaki

and Bobkov-Koldobsky states that there should be

a universal constant such that the thin-shell constant

is bounded by C for all n. So what is the– I mentioned earlier that all

these results on central limit theorem is via the

thin-shell constant. So what is the connection

between the central limit theorem and thin shell? Let’s recall the

central limit theorem. And suppose you give me an

isotropic log-concave measure in Rn. Then along most of the

directions on the unit sphere, the marginal distribution of

this isotropic log-concave measure onto that unit direction

x is close to a Gaussian where the quantitative

convergence is cn. And it turns out that,

by a theorem of Bobkov, the connection is

that the cn is bounded by the thin-shell constant

divided by square root of n. And if we go back to

the previous slides, this would suggest that

this bound by Klartag gives us a bound on cn

less than or equal to 1 over square root of log n,

which is what I showed earlier. So it turns out that the

thin-shell and central limit theorem are all related to this

notion called isoperimetry. The isoperimetry

question asks, what’s the least surface area of a

set with a certain volume? And so it turns out, for

the Lebesgue measure in Rn, we know that the

answer is the ball. And how about the

probability measures? So to answer this

question, we first need to define surface area

for a probability measure. So suppose we are given

a probability measure p. We define the

surface area, which is p plus A. We

define the surface area as how fast

the measure of A expand when I include

all the points that’s close to its boundary. So this notion, A epsilon,

is defined as all the points that is epsilon close, in

Euclidean norm, to the set A. And the surface area or

boundary measure of A is defined as the lim inf of– you include epsilon-close

points and how much the measure increases. What’s the rate of the

increase of the volume when you increase epsilon? And the KLS constant for

measure p is defined as follows. So KLS constant is this psi p. So the reciprocal

of that is defined as the lowest expansion–

sorry, lowest expansion of all the sets with

measure at most 1/2. And it turns out,

by log concavity– so if p’s an isotropic

log-concave measure, then this inf is taken for a

set with measure exactly 1/2. So up to a constant

factor, we can write it just as the infimum of the

surface area or boundary measure over all sets

with measure 1/2. And the KLS constant

for psi n is defined as the

supremum for psi p over all isotropic

log-concave measure p in Rn. AUDIENCE: [INAUDIBLE] HAOTIAN JIANG: It’s

not that easy to see, but it’s just using the

definition of log concavity. And you can do some

calculation to get that. Yeah, but I agree. It’s not that straightforward. AUDIENCE: Probably not

essential either, right? I mean, you will not try

to prove the KLS conjecture by using exactly 1/2. So it’s not– HAOTIAN JIANG: I think

the Lee-Vempala paper uses this fact, that you look

at the measure 1/2 set and show that it’s

not going [INAUDIBLE].. AUDIENCE: [INAUDIBLE] so yeah. [LAUGHTER] So then– no, but is

that symmetry essential? Like, if [INAUDIBLE] HAOTIAN JIANG: Yeah,

it’s not too essential. AUDIENCE: OK. HAOTIAN JIANG: OK. Yeah, so the KLS conjecture

basically states that if p is an isotropic log-concave

distribution– oh, sorry. So first, let’s talk

about all the sets formed by hyperplane cut. So it turns out, if p is

isotropic log-concave measure and H is a hyperplane,

then we have the surface area of H is constant. So therefore, the KLS constant

for all the hyperplane cut is constant. And the KLS conjecture, by

Kannan, Lovász, and Simonovits, states that there exists a

universal constant such that the KLS is bounded by

a constant for all n. So this, together

with our observation that the KLS constant

for hyperplane is constant, basically states

that up to a constant factor, the worst isoperimetric

sets are half-spaces. AUDIENCE: Is there

a good intuition as to why the answer is– the y is lower-bounded by

a constant for hyperplanes? HAOTIAN JIANG: I

always imagine if you look at Gaussian measure. If you look at Gaussian

measure, then you look at the hyperplane that

cuts a Gaussian measure that is a constant. And it turns out, then, I

think, in Gaussian measure, even the worst– without this constant factor,

the worst isoperimetric sets are half-spaces. But this is stating a

generalization of that, saying that for general

isotropic log-concave measures, then you are worse up

to a constant factor by a hyperplane cut. AUDIENCE: So these are

hyperplanes through the origin? HAOTIAN JIANG: Uh– AUDIENCE: For the

original– yeah. HAOTIAN JIANG: Yeah, because

we restrict our attention, say, only on the set

with measure half. Then for– oh, you mean for a

general isotropic log-concave measure, it’s not

necessary, yeah. But for a Gaussian, I’m saying

then it’s through the origin. So I’ve talked about KLS and

thin shell and central limit theorem. Let’s try to summarize

the connections between these notions. So we have a centralized limit

theorem, thin shell, KLS. Recall the definition. Central limit theorem

basically says the margin of p, on most of direction,

is close to Gaussian. And thin shell basically

states that the variance of– thin shell, the

definition of that is the variance of the

Euclidean norm of x. And KLS is the

measure of s divided by the boundary measure of s. And so the conjectured bound

for these three quantities is 1 over square root of n. AUDIENCE: [INAUDIBLE] HAOTIAN JIANG: Oh, I

defined 1 over psi p as the boundary measure

divided by the measure. I guess this is– AUDIENCE: [INAUDIBLE] at least? It is at most a constant? HAOTIAN JIANG: So the reciprocal

of this– so the boundary measure divided by the measure,

this is at least something. So if you flipped

that, then it’s going to be at most something. So I guess this is

just for convenience. So any other questions so far? AUDIENCE: [INAUDIBLE]

boundary is very small. If the boundary is– AUDIENCE: Any way of

splitting the measure in half as a large expansion? [INAUDIBLE] expansion is large. AUDIENCE: [INAUDIBLE] HAOTIAN JIANG: So

I guess maybe– AUDIENCE: The universe is small. [INAUDIBLE] AUDIENCE: [INAUDIBLE] is small. This is– HAOTIAN JIANG: Yeah, so my

definition is 1 over psi p is this thing. So take the reciprocal. You get the psi p

upper-bounded by something. Yeah, so these are all

the conjecture bounds. So what’s the connection

between all these quantities? So it turns out that the

KLS constant, the psi n, upper-bounds the thin-shell

constant, sigma n. And this is by inequality

of Cheeger, which I’m not going to talk in detail. And the reverse direction

was recently proved by Eldan, stating that the thin-shell

constant implies the KLS constant up to a log factor. And so as mentioned earlier,

the thin-shell constant will give some bound on

the central limit theorem. And this is by Bobkov. And it turns out the

reverse direction is also fine by some calculations. So all these are

classical results. AUDIENCE: Which ones

are [INAUDIBLE]?? What about the two on the left? HAOTIAN JIANG: So basically,

if thin shell is constant, then his result

implies KLS is log n. AUDIENCE: Yeah, I know. But the one on the left, is

there a loss on the left, or is it– HAOTIAN JIANG: So you

mean this one there? I think this one, there is

a log n factor loss, yeah. Yeah, but because the

current bound of these are all polynomials,

so that log factor is not crucial at the moment. AUDIENCE: [INAUDIBLE]

also related to the isoperimetry

conjecture, that you have a convex set of volume

1, it’s isotropic, [INAUDIBLE] through the origin, [INAUDIBLE]? HAOTIAN JIANG: It’s just

called the slicing conjecture. AUDIENCE: Oh, I think

that’s slicing conjecture. HAOTIAN JIANG: Yeah,

I think that constant is bounded by KLS constant. So the reverse direction,

I don’t think it’s no. AUDIENCE: [INAUDIBLE]

it’s just the– I mean, what you said is just

about whether the expansion can be 0, basically. Like, it’s– OK, what

you said was just– like, isn’t the slicing

problem just about the– well, I guess I don’t know. OK. Yeah, we maybe can

chat offline for that if there’s some interest. OK, so yeah, this

is all known before. And now, to motivate this notion

of generalized central limit theorem, let’s come

back to central limit theorem for convex bodies. So it basically states this. As I mentioned many times,

it’s like, the marginal of this isotropic

log-concave measure on most of the directions is

close to a Gaussian. And so if we think of

this uniform distribution on the sphere as being close

to the standard Gaussian, because there is those standard

concentration of Gaussian measure, then we can restate

this central limit theorem for convex bodies as saying

that the inner product of an isotropic log-concave

vector with a standard Gaussian vector, this is close

to a Gaussian for most of the Gaussian direction. And so a natural question to

ask is, how about you take– I give you two isotropic

log-concave measures. And you sample random

vector from each one and take the inner product. Is this also close

to a Gaussian? Basically, can we generalize

this Gaussian vector here to any isotropic

log-concave measure? Yeah, and we formulate

this question as the generalized

central limit theorem. And so this

definition of this is saying that suppose

we are given two isotropic log-concave

measures in Rn. Sample x, y independently

from p and q and a Gaussian random

variable, G. So notice here, I put variance is n. It’s not a standard

Gaussian random variable. And then we define

this, the W2 distance between the inner

product of x and y and G. Let’s call

this kappa p, q. So recall the

Wasserstein distance here between two measures, mu and nu. It’s basically the infimum

over all coupling of the ls norm under this coupling. And again, we

define this kappa n to be the largest kappa p, q

over all isotropic log-concave measures in Rn. And in the later

part of this talk, I will call this kappa n as

the generalized central limit constant. Yeah? AUDIENCE: So how

should we think of pi? HAOTIAN JIANG: How should

we think of pi in which one? AUDIENCE: So you’ve got– so in your definition. So by the coupling,

you’re mapping the space? I just– yeah. HAOTIAN JIANG: So– AUDIENCE: [INAUDIBLE]

mu to nu, and you want to minimize the

total [INAUDIBLE] distance that the particles move. HAOTIAN JIANG: So we

put them on the same– AUDIENCE: So this is– AUDIENCE: [INAUDIBLE] AUDIENCE: So this

is [INAUDIBLE] type? HAOTIAN JIANG: Yeah,

[INAUDIBLE],, yeah. So I guess the formal

definition of this is you create the product

distribution– sorry, not the product. You create a distribution on

the product space of mu and nu. And so that it’s

marginal, it’s mu and nu on the corresponding space

and over all such kind of distribution, yeah. AUDIENCE: I mean,

the generalized CLT does not imply the CLT, is

weaker because you used W2. HAOTIAN JIANG:

Yeah, yeah, yeah,. So the classical [INAUDIBLE] for

a kind of [INAUDIBLE] can be, think of it as like a high

dimension– sorry, high probably estimate. Like over most directions,

it’s something. This is, I’m saying that

I’m joining vectors, and then look at the distance,

without any like high probability statement. Yeah so, right,

so the naive bond on the generalized

central limit conjecture is that this kappa n is

bounded by square root of n, and this follows

from the observation that the expectation of the

inner product square, this is n. And also, the expectation of the

Gaussian random variable is n. Because the variance is n. And you just take the

square root on both sides, and you get that naive bound

kappa n is square root of n So the conjecture we made about

the generalized centroid limit is that there exists a

universal constant such that kappa n is bounded by

that constant for all n. So what is the

connection between the generalized central

limit conjecture and the KLS conjecture? And so the main theorem we

proved in the paper is that, basically, the central

limit, generalized central limit

constant is roughly the KLS constant squared. So this, roughly,

this approximation is hiding, like, log factors. But since the

current bound of KLS constant is still

polynomial, so I would just ignore that in this talk. So at the current

best bound on the KLS constant is beside and

bounded by n to the 1/4. And this gives,

using the theorem, that kappa n is founded

by square root of n, which is the naive bound on

generalized central limit theorem. So this result doesn’t– using the current

on KLS conjecture, doesn’t really give

some, like, a non-trivial bound on the generalized

central limit theorem. But if the KLS conjecture

is true, [INAUDIBLE] bonded by constant,

then we have a kappa n is bounded by constant, and so

the generalized central limit conjecture is also true. So kind of this theorem is

kind of a conceptual thing, saying that any quantitative

improvement each one of these conjectures leads

to similar improvements in each other. OK so, in the rest

of the talk, I will describe how to

prove this connection. And so our proof

of the connection between generalized centroid

limit conjecture and KLS conjecture is

through this notion called third moment bound. And so the third

moment is defined as, like, if I’m given p q,

two isotopic log concave measures, p q, and r n,

then the third moment is defined as the expectation

of the third movement of their inner product. It turns out the

Lee-Vempala bound of psi n bounded by 1 to

the 1/4 on KLS constant is essentially saying

the third moment is bounded by n to the 1.5. And we’ll build our

equivalence between generalized central

conjecture and KLS conjecture through this third moment bound. So roughly our picture

proof is the following. So we show that generalized

central limit theorem implies the third moment, and then show

that the third moment implies KLS, and then the KLS implies

the generalized central limit theorem. And moreover, we proved this

quantitative implication that, if the generalized central

limit constant is n to the 0.5 minus epsilon, then the

corresponding third moment bound is going to be n

to the 1.5 minus epsilon. And the KLS bound is going

to be n to the 1/4 minus 1/2 of epsilon. So we start by showing that

generalized central limit constant being bounded

by this, implies the third moment being bounded

by n to the 1.5 minus epsilon. So formally speaking, the

theorem we prove is this. So suppose we fix epsilon,

which is between 0 and 1/2, and let p be an isotropic

log-concave measuring our n. And we draw independent

random vectors x, y from p, and Gaussian random variable. Also notice here,

I only have one– we only have the same

isotropic log concave measure. So recall, in generalize

central limit theorem, we have two isotropic

log concave random– two isotopic log-concave

distributions, p and q, and we draw vectors from each one. But here, it turns

out that you just give me one isotropic

log-concave measure, and tell me this, then

it’s enough to prove the bound on the third moment. So if the W-2 distance

between x, y, and G, that this is bounded by

into the 0.5 minus epsilon, then we have the third

moment being bounded by n to the 1.5 minus epsilon. Any question about

the statement of this? OK. So now let’s try to prove this. And so basically the previous

just by the definition of these notions. So let’s call pi 2 the best

coupling in this W2 distance. And then it follows

by definition that, under the coupling

pi 2, the expectation is n to the 1 minus 2 epsilon. This is just definition. So it turns out, that

by the monotone property of Watt’s sustained

distance, we can show something about

the third moment under this coupling, pi 2. So the third moment

is founded by n to the 1.5 minus 2

epsilon [INAUDIBLE].. The way to understand this

is to do some normalization. So basically, this

g has various n. So let’s try to divide

both sides by n. And then this will become big O

of n to the negative 2 epsilon. And then do the same

normalization here. This will become big O of n

the negative 2 epsilon log n. So kind of the

Monotone property is saying that, after

normalization, then this is bounded by this

up to a log n factor. AUDIENCE: [INAUDIBLE] HAOTIAN JIANG: Which one? AUDIENCE: [INAUDIBLE]

What was the proof? HAOTIAN JIANG: I

didn’t give a proof. I’m saying the way to

kind of understand this. So because the tricky

part is this 1.5. This is 2 epsilon. This is 2 epsilon. This is 1. This is 1.5. But I’m just claiming that,

after normalization, this 1.5 will be gone. This 1 will be gone. And kind of the statement is

that, after the normalization, third moment is bounded

by the second moment times some poly log factor. AUDIENCE: I guess that

the way you’re using it. I mean, in general,

you’re not going to bound a third moment

by a second moment? That’s not right. But here, you’re using the fact

that there is some kind of– HAOTIAN JIANG: Yeah,

I’m using the fact that you are like having x,

y both isotropic log-concave. AUDIENCE: So you’re

using tails of some sort. HAOTIAN JIANG: Yes. This is exactly a tail bound. AUDIENCE: The second

moment, and you can try to apply to

the third moment, you won’t get

something interesting. HAOTIAN JIANG: That’s true. That’s more generalized. AUDIENCE: Then you will. HAOTIAN JIANG: Yeah, so

essentially the proof is, like, you use the

tail bound and say that x, y are, like, at most,

some constant times square root of n. Yeah, this is not

true in general. OK, so. OK yeah, so using this

and the coupling pi 2, we can write the third

moment bound as– so we put a negative

g and a plus g inside and then

expand this equation. So the first term,

expectation with G Q, this is 0 because G is symmetric. And the last term, x1 is GE

cubed using the bound above. This is bounded by n to

the 1.5 minus epsilon. So we’re also good about that. So the remaining two terms

that we need to bound is the second and

the third terms. And so basically, these are

just like using Cauchy Schwarz inequality. So for the second term,

we separate this G squared and the x, y minus G as

using Cauchy Schwarz. And so the first term, the

fourth moment of the Gaussian variables, the square root of

that, that is bounded by order n. The second term, so that is

bounded by the definition of the W2 distance. This bounded by n to

the 0.5 minus epsilon. So this term, we are done. For the other term, we

also use Cauchy Schwarz, but with some different p and q. And we can show this

is, again, bounded by n to the 1.5 minus epsilon. So I’m not going to go through

the calculation carefully. OK. So therefore, we

conclude that we have– the third moment is bounded

by into the 1.5 minus epsilon. Yeah, so we are done with

this first implication. So this second implication

is a bit complicated, and I’m not going

to talk about it. So let’s look at this. So I’ll briefly

sketch what happens with this third

implication, basically how KOS bound implies

place n to the 0.5 minus epsilon bound of the generalized

central limit theorem. So formally, this is the

theorem we want to prove. So assume the KLS is bounded

by into the 1/4 minus 1/2 of epsilon, for some

constant epsilon. And let PQ be isotropic

log concave measures in RN and independent

vectors x, y from p, and G from the variance

n normal distribution. Then we have this bound under

generalized central limit constant. So the proof idea

is the following. So here we have two different

isotropic log-concave measures. So ideally, you want to make

one of these a Gaussian, and then use classical

central limit theorem for convex bodies. So roughly, what we want to

show is that the W2 distance between this thing, and being

a product of x with a Gaussian vector, with a standard

Gaussian vector– this is small. And then, if we can show this,

I’m claiming this is done. Because somehow we

can kind of believe that the inner product of the

isotropic log-concave measure and a Gaussian vector should

be close to a real Gaussian random variable by classical

central limit theorem. I’m not going to do

that calculation, but this bound is pretty small. And so let’s see how to prove

the prove the W2 distance between these two quantities. So roughly, the idea is to use

the stochastic localization scheme, which I’m not

going to go into details, but roughly, what

is this doing is, it kind of embeds

the random variable x into a Brownian motion using

Stochastic localization so that we can write x

as the integration of some covariance matrix

times the Brownian motion. And then we can express the– so using this formula, we

can express the inner product of x and y as a

one-dimensional Brownian motion with this factor,

squared off y transpose 80y. 80 is some x some

covariance matrix of the stochastic

localization scheme. I didn’t– yeah I kind

of hide those details. Yeah it’s a bit hard to

explain that carefully. Yeah, so the rough

idea is that– so what’s the

advantage of writing something really simple as

a very complicated thing? So the advantage it is that,

we observe that we can also write x, the inner product

x, and the Gaussian vector g in the same form where

we’ve placed y by g. And the observation

is that, since y is isotropic

log-concave, Gaussian’s isotropic log-concave,

both of these quantities are concentrated around

this mean process, with control factor square

root of trace of 80 squared. So then this gives us a

way to compare to compare both these quantities. So we want to

compare them, right? It’s hard to compare

with them directly. The way to compare them is to

compare both with this quantity L, which is kind of the

mean Gaussian Process. And yeah so, roughly,

what we can do is, we compare the inner product

of x and y with this random variable L. And we can use some

theorem from the Lee-Vempala result on KLS constant

to show that the– like the operator

norm of this matrix is bounded up to some time t. And so the right

hand side we can bound as the fourth moment

of the KLS constant. AUDIENCE: Remind me what

the stochastic localization scheme is? HAOTIAN JIANG: So

roughly, you start with– you start with this, so we have

a target, probably, measure. So you start with

this measure and apply a kind of linear steps, random

linear steps, to this measure. Yeah kind of, it’s a

bit hard to describe. AUDIENCE: [INAUDIBLE] HAOTIAN JIANG: But essentially,

what you are doing is, you are introducing a Gaussian

factor inside the density function. AUDIENCE: So you’re

interpolating– are you interpolating

between this measure and the Gaussian measure? HAOTIAN JIANG: Yes. Yeah, kind of. AUDIENCE: So at time

0, you’re [INAUDIBLE] HAOTIAN JIANG: I time 0

is your target measure. And at time infinity, you

have a huge Gaussian term, which makes you a needle. AUDIENCE: [INAUDIBLE]

you now go back to this, and the expression of your inner

product as a previous slide? HAOTIAN JIANG: OK. AUDIENCE: OK, so the

first line, so now x here, you’re writing x as the integral

from 0 to infinity of At Wt. HAOTIAN JIANG: Right, yeah. AUDIENCE: So I’m

just saying, when you sum up all these

increments, you get a sample from redistribution? HAOTIAN JIANG: Sorry, when I? AUDIENCE: When you sum up all

increments and integrate them? You get a sample

from redistribution. HAOTIAN JIANG: Yes, yeah. Exactly. AUDIENCE: I’m just

trying to square that with what you said about

the localization scheme. HAOTIAN JIANG: OK, OK, Yeah. If you go to

infinity, then you get a sample of your distribution. AUDIENCE: How is this

n dimensional thing become a one-dimensional thing? HAOTIAN JIANG: Oh, so basically,

all these are kind of, like, so they have

the correct variance. They’re all Brownian motion. If you think about

one incremental step– let’s say 80 is

fixed, like, basically then the variance

of this is exactly this square of this times dt. AUDIENCE: So what I was

trying to say [INAUDIBLE].. If you look at the

first integral, there’s an inner product

between y and att yt. At is self-adjoint. So move it over, so

it’s multiplying the y. And now you’re just taking

some fixed vector inner product with the Gaussian vector. And so the law only

depends on the norm. HAOTIAN JIANG:

Sorry, I feel this– I made a mistake here. This should be At squared. Yeah. It should be At squared. All those should be At squared. Yeah, so this is [INAUDIBLE]. This is At squared, At

squared, At squared. And then you get this trace

of [INAUDIBLE] squared. Yeah sorry, I apologize. AUDIENCE: James [INAUDIBLE] HAOTIAN JIANG:

And then, this, we use some bound from

Lee-Vempala to show that, kind of, this integration you

can bound the operator norm up to some time t. And then you can bound the

right hand side by psi n to the fourth moment

of KLS constant. And that’s how you get

this O to the n to the n to the 1 minus 2 epsilon. This is the KLS constant– yeah. Yeah, so OK. So to summarize, we

introduced this notion of generalized

central limit theorem. And recall what this is saying. It’s saying that

the inner product of isotropic log-concave

vectors are W2 close to Gaussian distribution. And we showed the

quantitative of connection that the generalized central

limit constant is roughly the KLS constant squared. And kind of the current KLS

bound doesn’t really improve, doesn’t really give

anything non-trivial about generalized central limit. So it’s kind of still open. So kind of a future

direction is how we get some better bound on

the generalized central limit theorem to make it a real

generalized central limit theorem. But it turns out that,

by our connection, this is really asking how you

prove the KLS bound, which is pretty hard. OK, that’s everything. Thank you. Any questions? AUDIENCE: So I’m just

asking about KLS conjecture. If you look at the Markov

chain, and, let’s say, I have a convex body. You look at the Markov chain

maybe from every point you, and go in and ball around it,

and move at a random point or something like. Does it– HAOTIAN JIANG: [INAUDIBLE] AUDIENCE: Yeah probably,

but does it imply that– does that Markov

chain conjecture to have a second eigenvalue

spectral constant or something like that? HAOTIAN JIANG: Oh. Yeah sorry, I’m

not very familiar with that random log

on the convex bodies. Ask the internet. AUDIENCE: If you took

the right [INAUDIBLE].. AUDIENCE: [INAUDIBLE] AUDIENCE: I’m not

sure I think that. AUDIENCE: You need to make

the body isotropic first, and then just do a

random [INAUDIBLE].. Over the steps– AUDIENCE: How about

the cube root? AUDIENCE: It will be almost

the same as this [INAUDIBLE].. Because if you sample

from [INAUDIBLE].. AUDIENCE: From the cube side. Because all I want

to do is, I want to– we go only around standard

basis vectors in a cube. AUDIENCE: If your step

size is small enough, then the two processes

is almost equivalent, so. AUDIENCE: But the thing about

this which is interesting is, this is like

this some sort of– like it’s not [INAUDIBLE]. It’s sort of a

[INAUDIBLE] of this. OK. So then we get a [INAUDIBLE]. AUDIENCE: As Jim pointed

out, the statements of the generalized one,

because it’s using [INAUDIBLE],, doesn’t itself locally

get the total variation. But, so am I correct in saying

that, the bounds that you get for the two, you

can prove that they are related as a result of this

whole chain of equivalences? HAOTIAN JIANG: Sorry,

which page are you? AUDIENCE: So I mean,

the original one is just stated in terms

of total variation where one must [INAUDIBLE]

Gaussian, basically. And one was an

arbitrary log-concave. So this one doesn’t– Wasserstein doesn’t– I mean,

it’s not saying something total variation, or does it imply

locally on a per-distribution basis, or? AUDIENCE: So I’m not sure

I understand the question. I think the generalized wouldn’t

be true for total variation. The same [INAUDIBLE]. AUDIENCE: So for the

concave distribution, almost all these things

is equivalent, up to log. AUDIENCE: And you guys stated

in Wasserstein [INAUDIBLE].. AUDIENCE: It’s just

because we tried to make the proofs simpler. HAOTIAN JIANG: I

guess we can proved that for like– so the

covariance distance version is saying that I first sample from

one distribution, fix that one, and then look at the marginal

of the other to that direction. Yeah, like there’s

something– we can prove something similar for that. AUDIENCE: Oh, you can. OK. HAOTIAN JIANG: Yeah, yeah. But I forgot what

the bound is there. Yeah, any other questions? Good. Thank you.