Sure you a few examples of
hypothesis testing problems in which

there is some kind of a computer
a structure which makes it interesting so

we saw one the then click problem was so
that these are some

related problems in some sense so
here's the first one.

This is a hypothesis what does
the problem which one of serves and

I mention the vector and this is.

This is a Gaussian vector OK And and
it's a very simple problem on the dyno

hypothesis this is just noise right there
are the components are independent.

Stand their normal random variables and
the alternative hypothesis is that

there exists a small set of
the components so as that the means

over those components are different or
non-zero OK.

So there's a small set of size K..

Of the components that are different let's
say let's say this new is positive I can.

Say we know this meal and you want to
detect Now what makes this problem

a little bit interesting is that we know
that there is a class of possible subsets

of size that can happen OK so
not not not a little sub all

subsets of not necessarily all subsets
of the can be we call this contaminated.

But we know the there's some
kind of a structure it's OK So

this is the hypothesis
testing problem we want to be

one such want to know is
this just pure noise or

is there a sparse signal somewhere
where we know what the signal could be.

And the question is given such a class
of sets for what values of knew it.

Is that action possible.

Right is this clear so that the problem.

Let's hear everything so
the distribution when when the said this

is the is the content here to us and this
distribution is denoted by piece of S.

OK.

Other Westby subzero is the hypothesis so
here are some examples so

the when there is no structure all
possible subsets of sites possible

this is that this is a classical multiple
distinct problem and then the statistical

history teacher this has been studied
very thorough it is very well understood.

And may be.

A richer or more and more interesting.

Example where there's some kind of
structure is when when they have a grid

OK And every edge of the grid
there is a normal random variable

sitting and there is we know that
the other defy POTUS is that there is

a path from this corner of this corner
on one of the bath over which the.

The means are shifted again this was
studied by various Gus through and

Candice and then the whole OK another.

There are all kinds of other
examples of this type so

we will see a few words about for example
the an interesting example is when you

we have a complete graph on every edge
of the complete gov There is there is.

There is a independent
normal random variable and

there is one spanning tree over which the.

The means are different and

we want to detect if if there
is such a spanning tree looking.

For.

For each example I can I can
invent the story but there's no.

Some of them are actually
meaningful sort of for

example when we have bike Leaks You
can have patience and jeans and.

If there is they're matching is that or

by click that some genes are expressed or
exposed in some group of people and so

there are papers to come from
actual motivations of this.

OK So so I test is just a function
that off of such a vector and

then the outputs zero one zero is really
iffy if we think that there is no signal

and the one otherwise and exactly the same
way as before as in the first lecture we

can we can define the risk
as the sum of the two.

Types of errors OK So

the this is the if if the true
distribution is zero what artists

thinks that there is signal this is the
type one error and the type two where is.

The same way as before we find is
the average of the of the probabilities so

this is when.

The set so there's a kind of for
this base in.

Set up OK So
this is the average of the of the errors

over all possible subsets
that may be quantum in it.

And writing writing it this way is
is nice because because we know.

Exactly what the optimal test is the
optimal test is the likelihood ratio first

so we look at for each possible
observation vector we can compute

the density under this
mixture distribution and

the density which is just the standard
normal density and we look at the ratio so

the likelihood ratio is written
here it's the average over all

all subsets and

this is the mew mew is is
the strength of the signal and X.

of us is just the sum of the components
over that subset of the likelihood.

Sure as this nice form looks
like a partition function OK so

the optimal desk is the one that
compares this to one if it's bigger

than one then the accept the alternative
hypothesis in the likelihood ratio is

smaller than one then the effects
of the the null hypothesis and

the risk of the optimal of the optimal
test is this expression this is

a standard statistic exactly like in
the first lecture this one minus one half

times the expected value on their denial
so the under the stand their normal

of the absolute value up so the difference
between the likelihood ratio and

one OK and so so
we are interested in the study of this.

Saw in particularly if new is very
small then of course there's naught

test that can tell the two
distributions apart so when new is

zero then this equals one and this new
increases the problem becomes easier and

easier so this is the decreasing function.

We want to.

We want to understand
when is this small for

what values of Mueller's
this becomes more.

And this would depend on
the structure of this class.

Sorry k's the size caves the size
of the of the sets OK So we have N.

components and keys which is typically
much smaller is the size is there.

Is the size of each I assume that
in every element of this class C.

all the sets have the same size just.

This is not necessary with makes
things a little bit simpler OK so

so an obvious test is just.

We just some the wit
this is really easy and

if new is large then then this should
already detect the signal right because if

on their done all hypothesis this is
a normal with me and zero on their

the alternative hypothesis this is
a normal with me with me in new time ski

OK the variances are the same so so so
this test will be able to detect them over

me was bigger the square of the off and
over squared so that's it's trivial.

And it's interesting that in
many cases this is already

pretty close to the best best one.

Another very simple test is the sort so

called Scan statistic this so
in the scans that this think we look at.

We look at.

We scan through all possible candidates
and take the sum of the components over

all those candidates so for example in
the example of the spanning trees for

each span it is possible spanning through
there are many of them we computed the sum

over the edges we maximize this actually
can we can compute it quickly for

the case of spanning trees and
that and we threshed it so

we're do with Threshold it well on their
demo hypothesis is this is what we expect

is the expected value of the of
the mags in one of these standard.

There's this is just the maximum of a
Gaussian process that has some actual And

under the alternative hypothesis Well
this there exists one of these sets

where there is signal so one of them
should be at least new time ski so

we just look at what we just threshold
halfway between these two OK.

So if new timescales much bigger than this
thing then this will this test will work.

And it will work because.

Because this random variable is very
concentrated because of the Gaussian.

Concentration in equality OK so

won't won't get sure that if new is
bigger than this quantity the ratio

of the of the expected magazine
will of the skull same process over

then then this test tube work
is that this is the this this

is the scan statistic just
OK Now interestingly.

In all all examples I know what holds
these tools to do simple tests is

is near optimal and
you often mock up the cost of factors.

So it's if you really want to know the
optimal does then of course you hope we

have to be a little bit
more sophisticated but

with one of these two thirst counting
just just taking the average or

looking at the overall possible subsets
one of these two is it is usually good and

so how do we know that well we can
you can look at lower balance.

And we do it exactly the same way as we
did for the hidden click problem so so

the the true risk the best risk right we
had this expression in terms of the in

terms of the likelihood ratio Well this
is a complicated random variable but but

Chevy shows in the car story a question
what's in the quality wheel will help

because this guy is computable very nicely
this is just this is just the expected

value of the of the likelihood
ratio squared minus one and we can

calculate that and we get something nice
which only depends on the coming out or X.

of this class so
what do we get there the risk is

greater than one minus one half
times the square root of and

what's this is the is the momentary think
function of this random variables Z.

which we often by taking our class
of sets the possible sets big to at.

And look at the look at the overlap.

OK so we choose to sit independently
at Randall look at look at the overlap

and though we should understand
the the woman generating function of that.

It's really nice because it's
purely coming after a quantity or

that all the governors are gone and

this is really easy you just write down
what this is and that's what pops out.

Yes.

See is that C.

is our class of all possible sets of Z.

is so from a class of all possible sets I
choose to work independently at random and

I look at the over.

The cardinality of the I count how many.

The intersection.

So that the random variable so
now we can play right now we can we can

we can try to understand
the behavior of this.

So here is a little curiosity.

Which.

So somehow we've come back here in the
second half of my thought the sometimes

when we when we study hypothesis testing
problems we can see something above.

Probably random quantities that we may
be difficult to analyze that way so

here's here's something go straight so
they take new C.

to B..

To B.
this critical games

when this equals to this.

Right when when in this moment of
the function equal to then this is one so

this is just one half right
it's greater than one half so

that value there is at least one half.

But we also know that the scant
artistic statistic test

performs better of them but they.

This whenever a new is greater than that.

So we have a lower bound but and we have
an upper bunk so these are in conflict and

therefore we get a lower bound for
the expected magazine or for

gas in process in terms
of the strange quantity.

So I don't know but there are people
here who know about the gas in process

I haven't seen this example and

then there's some cases of course
the expected magazine for girls in process

this is very well understood telegram
wrote think books about this but but but

but the but computing what
this is is not always easy and

with this little trick sometimes we
get things I will show you how one can

get interesting balance for this yeah.

I.

Know there is or there's always
some chance that you pick the same.

So that it's it's always about the.

Right so
in normal clearly not always sharp but

in a computer in some cases and in some
non-trivial cases it seems to be sure.

No but maybe maybe vice a versa it
would Burke but it's not so clear so

I it's not so clear OK so
the floor bonding these things

is what is this mythical packing of
the of the set of this class and

work over that the gods were bound.

This is somehow in a very different spirit
in which we think we think we think random

subsets we look at the overlap compute
the moment I think function and.

So one way of when when one can
get very nice lower bounds is.

Is when we have a negative association.

So suppose.

Suppose that the class a symmetry just for
the case of discussion so

that means that when I when I pick
two sets I can fix the first set and

the distribution is not going to depend on
what the First Cities So let's take one of

the first set that we want to carry
the first key components and take an S.

that's random that that's chosen randomly.

So then then the size of the overlap is
just the number of these indicators right.

OK Now in some cases one can prove that
these these indicators are negatively

associated with the IF IF IF IF
the art then it's very easy

to do to get this lower bound so
the risk is great going to be greater

than one health whenever a new is smaller
and the square root of flog over squared.

This just happens to come out I'm
not going to show you the proof but

it's really simple it's just
comes comes out from the writing.

So that negative association
of these indicators or

characterize it in terms of the class
I don't I don't know I don't that.

I haven't thought about it but
here is a nice example.

Spanning trees so in the case of Spanish
This is true this negative association

through Does this goes back to the case
there and how they prove that the if

we take a random spanning tree then and
we look at two edges.

Where the indicator the two
that two edges are in

the both of them are in the in the same
spending through a negative we associate.

So in this case well N.

is so if you have a complete graph on M.

vertices we have M.

choose two coordinates the cake was and

minus one because that's how many edges
each a spanning tree has there are lots of

lots of spanning trees right and
the a minus to the scale is formula and

and from what I showed you
we get we get this Lord and.

In particular we get we get the Lord so
this is one example in which it

seems difficult to prove a lower bound for
the expected magazine of the process but

with this we would be good that
it's lower bounded by a constant.

OK So it's interesting that.

In this particular case
the likelihood ratio can

test the optimal test can be
computed efficiently by because of

because there's this nice algorithm of
proppant Wilson that does you how to.

Take the really simple algorithm
to generate a random spanning tree

such that the edges such that
the probably of the tree is

proportional to two to the product
of the weight of the edges and

here the weights will be just eat
of the commute times the times of

the normal on the edges so in this case
there is a fast algorithm to compute

the true well fast randomized
algorithm to compute the true.

True.

Likelihood issued just.

Here's another example.

Perfect matchings saw So
here here we have.

What produces a permutation OK.

Right.

Imagine the complete bipartite graph on M.

vertices maybe maybe there's another
slight before you here complete by part

that graph there is a Gaussian
sitting on every age and

and we have to decide whether there
is somewhere a perfect matching

on which the Muse are greater so
there again M.

squared is the is
the number of variables M.

is the size of each perfect matching
there are many of them OK And

in this case in this
case it's the the the.

The size of the of the overlap is just
the number of fixed points in a around

the permit issue so this is very well
understood and wanking won't get Sure.

It's a lower bound so our star is going
to be greater in the one half whenever

music is smaller than this constant tenth
and the optimal test is also of the cost

of the simple averaging test
also gives some other cost and.

So these methods because I think we would
be losing question SCHWARTZ So we don't

really get We never get the right constant
but we usually get the right order.

And in this case again the.

The optimal test well is just amounts
to computing the permanent of

of Imitrex which is which is non-negative
entries and there we have this famous

people a general leader of the chosen hold
one this get that can be computed in the.

Polynomial time with
a randomized algorithm OK so

that's another case when when
these complicated likelihood ratio

this can actually actually
be computed efficiently.

OK.

There are some other examples but
I will skip them.

So here's a here's the problem that.

For all of you who look for
hard and maybe hopeless problems so

so the optimal risk is
a decreasing function right as new

becomes larger the problem becomes
the the testing problems becomes easier

it starts at one of goes down to zero
now is it true that it all it almost

always under mild conditions on what this
set is it should go down like it's big and

boom go down to zero is
there a sharp the short.

Sharp threshold I mean that
if if new sub epsilon is.

Is where we're this function equals
epsilon where there is risk equals epsilon

So we look we look at the the.

The location where where the risk
is already down to Epsilon and

we look at the location of
very still one minus epsilon

is it true that these two
are really close to each other

how close were closer to them than this
let's see the size of the middle point.

And the No Yeah no no no I don't but when
kids large I think this should be true.

So yeah so.

It should be the case that if the class is
some are sufficiently symmetric then and

then this goes down to zero one.

We spend some time on it but
when get anywhere.

We'll be looked at theorems of that state
but but the it's a Gaussian problem so

it's not really in that framework so

there are notions of geometry
influences and stuff.

Maybe there's the way but we can work it.

OK for I think for the cases I
think this is known because because

you can really study what
the optimal tests look like so

when there's no communitarian structure
but in general this could be interest.

OK.

So in the second half of the talk
we look at a different kind of

very related but
different distinct problem so here we will

test not so
again we will observe vectors but

here the the alternative hypothesis is not
going to be that that the mean is shifted

the meaning will be ZERO always but that
the sum complements would be correlated

OK so here is here's
a motivating story we have a and

sensors out there in the outer space and
all of these sensors.

Collect signals.

And the Eye sensors receives some
signal during the periods of time and.

And under the law have thought this is
this is all pure noise everything is

standard normal but
under the alternative hypothesis

there are there is a small subset of
the sensors that pick up a common signal.

But embedded in hidden in additive
noise so this this Whitey at time T.

goes from one to D..

We have a common signal that
the sum of the sensor speak up.

Now can we for what values and this is
just normalizing are my thing so that the.

So that the the variance is one.

From the start point of it all
because we just normalize so

that everything has the same.

Yeah it's in that it doesn't really matter
it makes the math a little bit simpler

it doesn't really matter of course it it
can matter if by simply looking at the but

there then the problem
is easy Anyways OK so

if it matters then if it
matters then it doesn't matter.

So they're just OK you're right.

OK forget the story look at them.

OK.

Or it.

So these white These are normal with
mean zero and variance rule and

destroys we want to understand
all can be and now.

Again this is possible class of
sets may have some structure

I'm just going to know look at the case
when the when there's no structure or

all but you can you can look at
the same story as before but

let's just suppose that there is
no structure or sets of size K.

are possible.

So here's the same thing written in
a kind of different way we observe

the independent vectors OK And under
the law have offices all the components

of these vectors are independent but under
the alternative there is a small subset

of components that are correlated the
Paradise correlations are equal to rule OK

so the the chorus matrix is
looks like there are some.

Some.

Key variable size of
the pair of us Croatians or

rules otherwise you have zero OK So
these are these vectors.

OK So again a test takes this whole
matrix these is the observations of N.

vectors and and again this is the same
story as before we look at the sum of the.

Of the two types of errors now we
have into ski possible sets OK are.

There is a likelihood ratio There's
that we can write down it looks ugly it

was a big sum.

But we can we can try to
obtain a lower bounds

now the usual technique that we used to
using a question doesn't work this way

I remember that our star is is
one minus one half times the L.

one distance between the between elephants
and one and we be bounded it by but

it doesn't work because this expected
value in this case is infinite so

we have to do something in
the way to do it is that

is that the condition on the on
the value of the signal of

the of those whiteys and then if we
condition then it becomes something

like the shifted mean problem that
with on the right if we condition on.

As it is if we condition on this then then
this is exactly like as it was before but

now the means are these whiteys
of course these are random but

with one one can't work it
out there behave nicely so so

that's how we get this Lord done
don't look at the form that's not.

Too interesting.

Just that there is
an inequality of that sort.

And they get if we if we have now the
class of if we have a structured class of

sets then again this can be exposed in the
in terms of the moment generating function

of the exact same random variables
before we can get lower bounds for.

For the optimal risk in terms of the same
coming out through a quantity as before.

So one can read off some some
immediate inequalities from these and

there are some nice.

Transitions So if if these small
the number of observations

is much smaller than a law again
over rule then then the Lord

that this cannot ever go to zero unless
keys are very large are most true then.

But if the is a little bit larger than
this just a constant times larger than

that then then the risk will go to zero
Dr Mariska even if it is a constant.

So so if these smaller than Logan
were going to overrule then the then

even if if large almost true then
there's nothing you can do but

if we go a little bit above then
all of a sudden even fast and

look OK and that constant this upper bunk
can be achieved by just a scan statistic.

So what's a scan statistic Well
we we we go through all possible

sets and we compute within that so
that we compute there was correlations.

So we are all pairs the product
because we know that

over the counter mean the set the expected
value of this is real otherwise it's zero.

So so under the alternative hypothesis
when there exists such a set this should

be at least well this is wrong times
Kate use two times the something like.

So if we threshold at half of that
then we get we get that upper OK so

this of course has to be worked out we
have to understand the concentration

behavior of such a quadratic forms or
this is not too difficult.

Yeah.

That's a good question we haven't
looked at that yeah yeah it's

it's and it's of course it's
a natural question yeah yeah.

OK so so
here is one way off of constructing get

this OK so
we have all these vectors OK we have.

No investors in the dimension of actors
OK And on the dyno hypothesis they

are a pair of eyes correlations are zero
on their the alternative hypothesis there

exists a small number of them so that
the pair wise correlations are larger.

So we might set up a graph in which
these vectors are diverted says OK and

I would draw an edge between two
vertices if the if the pair was imperial

correlation is larger than some threshold
OK And now we will look at this graph.

Under the hypothesis these
points will be randomly.

So the probity of each edge let's
see if I threshold this at zero

then the probably an edge
would be one half.

Under the.

Alternative Hypothesis there
will be a small subgraph So

is that such that the the the pairwise

correlations are larger so
they will be more likely to be connected.

So this sounds a little bit
like that hidden click problem.

We are trying to find.

A denser subgraph But this is not not and
there Additionally nigger I

because because now we have
geometry all these points so the X.

I divided by the norm of excite

them what is the eyes they are uniformly
distributed in the sphere.

On the dimension a sphere so
now we have these the points on the sphere

on there are the not
hypotheses they are random so

it's just that this graph is just
a so-called random geometry graph on this

on the sphere and
under the alternative hypothesis.

As we we have a sort small subset of
points that are somehow tend to be

closer to each other.

So in that your thirst is just a look
at the size of the largest click

in this graph OK.

So OK good so this is a nice object
it's a random generator graph and

no one can try to skip these two
slides it's kind of nice to understand

what happens when when when the connection
when we don't pressure that zero but

with somebody else but OK let's just
let's just concentrate on the case when.

When peak was one half OK so
what we do again we have these endpoints

on the on the sphere and we connect
to if the probability that they are.

If if they are if they fall if
the if there angle is less than

ninety degrees OK.

Right that means that the inner
product is greater than zero OK so

any two point there will be an edge
between two points of the inner product is

is positive OK now what
happens to this graph

Well if these small of the if
the damage is really small then

then this will behave this way around
the geometric grab but as the increases.

Then then this will become in there.

And that's that's not difficult to see.

As the dimension increases
as we add more and

more components to these Gaussians
then then then what happens well.

It's here it's written here it can
represent this random grass so

that I do know this that by G.

bar in D.

and P..

Is going to be one half it's just the
probably three that they're connected and

is the number of points and
the is the the is the dimension so again.

Here we are on the sphere units fearing
the dimensions be thrown down and

points at random we connect two
points if they're if they're angle is

is less than ninety degrees OK but
we can represent

this is how we represented know
to be these is zero in this case.

But these guys.

This is just the sum of
independent random variables.

And.

And if N.

is fixed the number of points is
fixed by the dimension increases then

one can use the multifarious multivariate
centrally me theorem to show

that these guys will converge to normals
and and they will become independent.

So the distribution of this graph
converges to that it should be

some of the graph OK So
that's that's written here

the thought of variation this tense
between these two goes to zero.

Now of course you can ask hold
large does the dimension to be

such that this is small and
there's a recent work of.

Done where the prove that
there's a nice threshold so

if if he is that much larger than
square then in cubed then this

would be very small but if these smaller
than cubes the this would be still large.

Piece one half of a piece Constance
because of costs of a constant P.

Yeah if the graphs we
don't understand anything.

But for those graphs and
P's one half were a constant then

the threshold is somewhere when
the dimension is and cubed.

Right.

Yeah that's not how they do it.

They are.

The of and
the lies the Richard matrix and.

Do some hard work yeah and for
in they can in they construct tests so

in on one for the lower bands because
struck best clever tests and for

the upper buns they analyzed matrices
instead so I don't know if and

by the look at the stands
Methot its interest.

OK OK But what we are interested in as I
said that the for our testing problem what

we are interested in is understanding is
the is the click number right because our

we construct this graph and we want to see
where there where their signal or not.

If their signal then then then B.

expect that there will be a large click
in this graph right if there are points

that are more correlate there's a small
subset of ones that are growth in that So

in order to do that we want we want to
understand the quick number off of this

random geometry graph particular it would
be nice to understand what how does

this behave in terms of the dimension
it is the dimension grows what happens.

If the measure is fixed if for
the damages to more or

three or something like that OK then.

Then the quick number will
be large right because in

that in the three dimensional sphere
if I throw down and points then

there will be many of them that will be
the third pair wise orthogonal I mean.

Their distance their pair
ways angle is less than

ninety degrees right because simply
everything that falls in a sphere or

coke out of angle of forty five degrees
there they will form a click OK.

So so I can't so
the number is linear when these fixed

then the click number of the size
of the largest click is linear and

then there are cliques of linear size but
when the goes to infinity this.

It will converge to the to the quick
number of the additional in your graph and

we know that that's only
logarithmic in that.

So we wanted to understand what
happens as we increase the right from

constant to infinity How does the this
drawdown from linear to two logarithmic.

And this is important for understanding
the testing problem because this

this will tell us how how long you have
you have to wait until we can confidently.

Say that there is signal.

And so here are here's a list of
of bounds will that go through

what happened so as I said already in
these constant when the constant and

goes to infinity then we
have cliques of linear size.

And.

It's easy to see that is soon as the goes
to infinity the clicks are not going to

billion years that's very
easy to see OK now if the is

much smaller the law again then it's
still easy to see that the same argument

will tell you that there are still cliques
not quite linear but almost linear.

But then something happens
when these look squared of.

So when these look square defining
We can only prove this for

the case of B.'s less than one health or
but it should be true also for

peace then these look squared so

if these look squared then then then the
quick numbers already Polly La Barre for

me so there's a big jump between
between those then log and

look squared and it goes down
from here and now if the damages

look you'll offend then it really starts
looking like the click number of the or

the shrinking of it's to
lock to the when thing and

if we go a little bit further then we
really nail down what happens with

up to the top of the constant the other
to fast and the click numbers.

And there's this yeah sorry.

Logon.

Thing happens OK so so
here this think those

you something about what happens so
here's a here's a lower bound for

example it does you that if the is
all going to the two minus epsilon.

Then this is much bigger than
anything probably logarithmic.

So so there's a big jump between
the two the log to minus Epsilon and

and and look square and then it goes
from some something like into the.

Log.

Through to some power.

To do to two to pull you over
think to look squared so

there's a big jump here it
also does you well here D.

has to be at least a constant times law
again but if it's a big constant term

slogan then then we see that it's still
it's still the polynomial it's and

to some power so there seem to be two
different kind of jumps one that long and

one it long and when it goes from
polynomial to two to sup on the normal and

then the clock squared moment he goes from
kind of stop on the two to put it over it.

And then then it's movies out.

So in the remaining six minutes I
will prove all of these things.

So maybe I can see a few
words about how we prove this

because it's kind of interesting.

So we found this theorem by young from.

One thousand the one.

That for every set.

Of diameter at most one.

Right if we have a set of them
there we can find a closer

a closed ball of this reviewer's in
which we can include those that.

Are.
And this is just a little bit better than

the trick completely trivial.

But that is that is enough for

us because because we can use
the the picture because there.

So.

It's so here.

I'm not going to go through but
basically basically what happens is that

is that the click is just the set of
of of a certain time it is right that

that's what it means and now we can
include it in a ball which is just.

A little bit better than the trivial
right and and then we can use and

now the size of the click is just
the it's just the empirical measure.

Off of such a satirical cap right so

the size of click is not upper bounded
by by the theory that the measure of

the sphere of your cup of which is this
ball that contains our click OK but

we know what the measure of such a severe
cap is because we can just compute it

and the end we can use the V.C. theorem
that says that in that in the space.

The.

Difference between
the empirical measure and

the true measure cannot be depends on
on the ratio of the mansion and and.

Put these things together
that's how we get.

This light OK.

Now.

These guys.

That say that if these large
larger than look you burn or

look at the fifth and
this is this goes with the.

This is just the first moment method.

So.

So if the number of clicks of
size scale this is the expected

then the expected number is inch of ski

sorry this is in the in the other strain
your graph times two to the minus K.

half.

And.
What we can prove is that the expected

number of clicks of size ski is bounded
by ensure ski OK that's the same and this

thing is some number which is not one half
to the power of Kate used to essentially

this number this is just the this is
just the standard normal distribution

function at Delta OK so if that is
small then this is close to one half.

And this is if there were whole small this
does the world there is we have this this

condition here and
we believe that OK And how do you prove

such a thing where this is concentration
OK so if we have if you have independent.

Independent points of the sphere Well
these are basically normalized independent

gal since right and we want to see what is
the probability that if we throw them K.

Galson as them the pair wise there will be
pair wise better than orthogonal can and

you can do this way by induction doing
a kind of Graeme Smith orthogonal Isaacson

at the step and then taking taking care of
the of the errors little bit technical but

not this is the the principle So
what one can get an upper bound for

the expected number of expected
number of clicks I ski and

then by the first moment
method that we saw last time

gives us a upper bound for
the clique member.

Again.

Right.

And yeah.

It doesn't work is just falls short of for

some of the Spirit to a cap that we get
is just everything it's just nothing but

yeah sure the morally there should be
no difference but this prove that.

And finally I want to see how do we get

how do we prove a lower
bound of this sort of gains.

We have a.

We have this random geometric graph.

We have this random chance or graph and.

Sure that the largest Click has
to be large if the dimension is.

Sorry what does it say.

Yeah so if the damage is small then it
has to be higher so how do we prove such

a thing if I hadn't told you the story
of testing this would be hard but

we have all of us there so this is
another example where we can we can draw

something from our hypothesis testing
formulation and see something about.

What something complicated probably
stick quantities Why because because

if the largest clique was small suppose
that the largest group under the the.

Under this model is small Well that means
that the detection problem becomes easy.

Right because because now if we don't
put in signal that we will that we

induce a large click and

then these this two distributions can
be can be pulled apart very easy.

So if their largest click was small
then there would be a test but

we know we have looked at this universal
or bounce for the risk of the best test

that see that that is impossible so
we just put these things together.

So just from from from the impossibility
of finding good hypothesis tests

one can prove lower months for
the click number of such a graph.

And the only little ingredient this is
easy to needs is that it's a performance

for that this under the of then that
if I prophesy is the graph does contain

a large click OK if the if are always
sufficiently large though is this thing.

Sufficiently large and and not that much.

So that's it.

Everything is written up here so
you go to my homepage and open this and

there's all the proofs are there and these
are some of the references that they've

gone to the results that
I mentioned OK Thank you.

You.

Know well why because we are not
good enough not skilled enough but.

It might it might but
we do it by induction and somehow Yeah.

It might work if you're more
clever than us but that would be.

Of the click number.

Apart from so human feeling this holds.

So the conjecture that the conjecture
is something like this is that

up to Logan we have these almost an.

And then shortly after logon it will
drop down to enter the epsilon game so

if it's a little bit bigger than Logan
it's going to be already into the absolute

Well it has to be at least that that
comes from here and then then air and

look squared then it it will go down up
to look squared then it's still large.

OK.

Not not quite polynomial but but the but
much bigger than the logarithmic and

then then it drops down here to
put a low earth make and and.

Yeah there are horses here so I.

Don't quite know all that it.