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PHILIPPE RIGOLLET: --bunch
of x's and a bunch of y's.
The y's were univariate,
just one real
valued random variable.
And the x's were vectors that
described a bunch of attributes
for each of our individuals
or each of our observations.
Let's assume now that we're
given essentially only the x's.
This is sometimes referred
to as unsupervised learning.
There is just the x's.
Usually, supervision
is done by the y's.
And so what you're trying to do
is to make sense of this data.
You're going to try to
understand this data,
represent this data,
visualize this data,
try to understand
something, right?
So, if I give you a
d-dimensional random vectors,
and you're going to have
n independent copies
of this individual-- of
this random vector, OK?
So you will see that
I'm going to have--
I'm going to very quickly
run into some limitations
about what I can actually
draw on the board
because I'm using
[? boldface ?] here.
I'm also going to use the
blackboard [? boldface. ?]
So it's going to
be a bit difficult.
So tell me if you're actually
a little confused by what
is a vector, what is a
number, and what is a matrix.
But we'll get there.
So I have X in Rd, and
that's a random vector.
And I have X1 to
Xn that are IID.
They're independent
copies of X. OK,
so you can think
of those as being--
the realization
of these guys are
going to be a cloud of
n points in R to the d.
And we're going to think
of d as being fairly large.
And for this to
start to make sense,
we're going to think of d
as being at least 4, OK?
And meaning that you're
going to have a hard time
visualizing those things.
If it was 3 or 2, you would
be able to draw these points.
And that's pretty
much as much sense
you're going to be
making about those guys,
just looking at the [INAUDIBLE]
All right, so I'm going to
write each of those X's, right?
So this vector, X,
has d coordinate.
And I'm going to write
them as X1, to Xd.
And I'm going to stack
them into a matrix, OK?
So once I have those guys,
I'm going to have a matrix.
But here, I'm going
to use the double bar.
And it's X1 transpose,
Xn transpose.
So what it means is that
the coordinates of this guy,
of course, are X1,1.
Here, I have--
I'm of size d, so I have X1d.
And here, I have Xn1.
Xnd.
And so the i-th, j-th--
i-th row and j-th column
is the matrix, Xij, right--
is the entry, Xi to-- sorry.
OK, so each-- so the rows
here are the observations.
And the columns are the
covariance over attributes.
OK?
So this is an n by d matrix.
All right, this is really
just some bookkeeping.
How do we store
this data somehow?
And the fact that we use a
matrix just like for regression
is going to be convenient
because we're going to able
to talk about projections--
going to be able to talk
about things like this.
All right, so everything
I'm going to say now
is about variances
or covariances
of those things, which means
that I need two moments, OK?
If the variance does
not exist, there's
nothing I can say
about this problem.
So I'm going to assume
that the variance exists.
And one way to
just put it to say
that the two norm
of those guys is
finite, which is another
way to say that each of them
is finite.
I mean, you can think
of it the way you want.
All right, so now,
the mean of X, right?
So I have a random vector.
So I can talk about
the expectation of X.
That's a vector that's in Rd.
And that's just taking
the expectation entrywise.
Sorry.
X1, Xd.
OK, so I should say it out loud.
For this, the purpose
of this class,
I will denote by
subscripts the indices that
corresponds to observations.
And superscripts, the
indices that correspond to
coordinates of a variable.
And I think that's the
same convention that we
took for the regression case.
Of course, you could
use whatever you want.
If you want to put
commas, et cetera,
it becomes just a
bit more complicated.
All right, and so
now, once I have this,
so this tells me where my cloud
of point is centered, right?
So if I have a bunch of points--
OK, so now I have a
distribution on Rd,
so maybe I should
talk about this--
I'll talk about
this when we talk
about the empirical version.
But if you think
that you have, say,
a two-dimensional
Gaussian random variable,
then you have a center
in two dimension, which
is where it peaks, basically.
And that's what we're
talking about here.
But the other thing
we want to know
is how much does it spread
in every direction, right?
So in every direction of
the two dimensional thing,
I can then try to understand
how much spread I'm getting.
And the way you measure this
is by using covariance, right?
So the covariance
matrix, sigma--
that's a matrix which is d by d.
And it records-- in
the j, k-th entry,
it records the covariance
between the j-th coordinate
of X and the k-th
coordinate of X, OK?
So with entries--
OK, so I have sigma, which is
sigma 1,1, sigma dd, sigma 1d,
sigma d1.
OK, and here I have
sigma jk And sigma jk
is just the covariance between
Xj, the j-th coordinate
and the k-th coordinate.
OK?
So in particular, it's
symmetric because the covariance
between Xj and Xk is the same
as the covariance between Xk
and Xj.
I should not put those
parentheses here.
I do not use them in this, OK?
Just the covariance matrix.
So that's just something
that records everything.
And so what's nice about
the covariance matrix
is that if I actually
give you X as a vector,
you actually can
build the matrix just
by looking at vectors
times vectors transpose,
rather than actually
thinking about building
it coordinate by coordinate.
So for example, if you're
used to using MATLAB,
that's the way you want to
build a covariance matrix
because MATLAB is good
at manipulating vectors
and matrices rather than just
entering it entry by entry.
OK, so, right?
So, what is the covariance
between Xj and Xk?
Well by definition, it's
the expectation of Xj and Xk
minus the expectation of Xj
times the expectation of Xk,
right?
That's the definition
of the covariance.
I hope everybody's seeing that.
And so, in particular,
I can actually
see that this thing
can be written as--
sigma can now be written
as the expectation
of XX transpose minus
the expectation of X
times the expectation
of X transpose.
Why?
Well, let's look at the jk-th
coefficient of this guy, right?
So here, if I look at the
jk-th coefficient, I see what?
Well, I see that
it's the expectation
of XX transpose jk, which is
equal to the expectation of XX
transpose jk.
And what are the
entries of XX transpose?
Well, they're of the
form, Xj times Xk exactly.
So this is actually equal to
the expectation of Xj times Xk.
And this is actually not
the way I want to write it.
I want to write it--
OK?
Is that clear?
That when I have a rank 1 matrix
of this form, XX transpose,
the entries are of
this form, right?
Because if I take--
for example, think
about x, y, z, and then
I multiply by x, y, z.
What I'm getting here is x--
maybe I should actually
use indices here.
x1, x2, x3.
x1, x2, x3.
The entries are x1x1, x1x2,
x1x3; x2x1, x2x2, x2x3; x3x1,
x3x2, x3x3, OK?
So indeed, this is exactly of
the form if you look at jk,
you get exactly Xj times Xk, OK?
So that's the beauty
of those matrices.
So now, once I have this, I
can do exactly the same thing,
except that here, if I
take the jk-th entry,
I will get exactly
the same thing,
except that it's not going to be
the expectation of the product,
but the product of the
expectation, right?
So I get that the jk-th entry
of E of X, E of X transpose,
is just the j-th entry of E of X
times the k-th entry of E of X.
So if I put those two together,
it's actually telling me
that if I look at the
j, k-th entry of sigma,
which I called
little sigma jk, then
this is actually equal to what?
It's equal to the first
term minus the second term.
The first term is the
expectation of Xj, Xk
minus the expectation of Xj,
expectation of Xk, which--
oh, by the way, I forgot
to say this is actually
equal to the expectation of
Xj times the expectation of Xk
because that's just the
definition of the expectation
of random vectors.
So my j and my k are now inside.
And that's by definition the
covariance between Xj and Xk,
OK?
So just if you've seen those
manipulations between vectors,
hopefully you're bored
out of your mind.
And if you have not,
then that's something
you just need to get
comfortable with, right?
So one thing that's
going to be useful
is to know very
quickly what's called
the outer product of a
vector with itself, which
is the vector of times
the vector transpose, what
the entries of these things are.
And that's what we've been using
on this second set of boards.
OK, so everybody
agrees now that we've
sort of showed that the
covariance matrix can
be written in this vector form.
So expectation of XX
transpose minus expectation
of X, expectation
of X transpose.
OK, just like the covariance
can be written in two ways,
right we know that the
covariance can also
be written as the expectation
of Xj minus expectation of Xj
times Xk minus
expectation of Xk, right?
That's the-- sometimes, this
is the original definition
of covariance.
This is the second
definition of covariance.
Just like you have
the variance which
is the expectation of the
square of X minus c of X,
or the expectation X squared
minus the expectation of X
squared.
It's the same thing
for covariance.
And you can actually see this
in terms of vectors, right?
So this actually implies that
you can also rewrite sigma
as the expectation of X
minus expectation of X
times the same thing transpose.
Right?
And the reason is because if
you just distribute those guys,
this is just the
expectation of XX transpose
minus X, expectation of X
transpose minus expectation
of XX transpose.
And then I have plus
expectation of X,
expectation of X transpose.
Now, things could go wrong
because the main difference
between matrices slash
vectors and numbers is
that multiplication
does not commute, right?
So in particular, those two
things are not the same thing.
And so that's the main
difference that we have before,
but it actually does not
matter for our problem.
It's because what's
happening is that if when
I take the expectation
of this guy, then
it's actually the same as the
expectation of this guy, OK?
And so just because the
expectation is linear--
so what we have
is that sigma now
becomes equal to the
expectation of XX transpose
minus the expectation
of X, expectation
of X transpose minus
expectation of X,
expectation of X transpose.
And then I have--
well, really, what
I have is this guy.
And then I have
plus the expectation
of X, expectation
of X transpose.
And now, those three things are
actually equal to each other
just because the
expectation of X transpose
is the same as the
expectation of X transpose.
And so what I'm
left with is just
the expectation of XX transpose
minus the expectation of X,
expectation of X transpose, OK?
So same thing that's
happening when
you want to prove
that you can write
the covariance either
this way or that way.
The same thing happens for
matrices, or for vectors,
right, or a covariance matrix.
They go together.
Is there any questions so far?
And if you have some, please
tell me, because I want to--
I don't know to which extent you
guys are comfortable with this
at all or not.
OK, so let's move on.
All right, so of
course, this is what
I'm describing in terms of
the distribution right here.
I took expectations.
Covariances are
also expectations.
So those depend on some
distribution of X, right?
If I wanted to compute
that, I would basically
need to know what the
distribution of X is.
Now, we're doing
statistics, so I
need to [INAUDIBLE] my question
is going to be to say, well,
how well can I estimate the
covariance matrix itself,
or some properties of
this covariance matrix
based on data?
All right, so if I
want to understand
what my covariance matrix
looks like based on data,
I'm going to have
to basically form
its empirical
counterparts, which
I can do by doing the age-old
statistical trick, which
is replace your expectation
by an average, all right?
So let's just-- everything
that's on the board,
you see expectation, just
replace it by an average.
OK, so, now I'm going
to be given X1, Xn.
So, I'm going to define
the empirical mean.
OK so, really, the idea
is take your expectation
and replace it by 1
over n sum, right?
And so the empirical
mean is just 1 over n.
Some of the Xi's--
I'm guessing everybody knows
how to average vectors.
It's just the average
of the coordinates.
So I will write this as X bar.
And the empirical covariance
matrix, often called
sample covariance matrix,
hence the notation, S.
Well, this is my
covariance matrix, right?
Let's just replace the
expectations by averages.
1 over n, sum from i equal 1 to
n, of Xi, Xi transpose, minus--
this is the expectation
of X. I will replace it
by the average, which I just
called X bar, X bar transpose,
OK?
And that's when I
want to use the--
that's when I want
to use the notation--
the second definition,
but I could actually
do exactly the same thing
using this definition here.
Sorry, using this
definition right here.
So this is actually
1 over n, sum from i
equal 1 to n, of Xi minus X
bar, Xi minus X bar transpose.
And those are actually--
I mean, in a way,
it looks like I
could define two
different estimators,
but you can actually check.
And I do encourage
you to do this.
If you're not comfortable
making those manipulations,
you can actually check that
those two things are actually
exactly the same, OK?
So now, I'm going to want
to talk about matrices, OK?
And remember, we defined
this big matrix, X,
with the double bar.
And the question
is, can I express
both X bar and the
sample covariance matrix
in terms of this big matrix, X?
Because right now,
it's still expressed
in terms of the vectors.
I'm summing those vectors,
vectors transpose.
The question is, can I just
do that in a very compact way,
in a way that I can actually
remove this sum term,
all right?
That's going to be the goal.
I mean, that's not
a notational goal.
That's really something
that we want--
that's going to be
convenient for us
just like it was convenient
to talk about matrices when
we did linear regression.
OK, X bar.
We just said it's 1 over
n, sum from I equal 1 to n
of Xi, right?
Now remember, what does
this matrix look like?
We said that X bar--
X is this guy.
So if I look at X transpose,
the columns of this guy
becomes X1, my first
observation, X2,
my second observation, all the
way to Xn, my last observation,
right?
Agreed?
That's what X transpose is.
So if I want to
sum those guys, I
can multiply by the
all-ones vector.
All right, so that's what the
definition of the all-ones 1
vector is.
Well, it's just a bunch of
1's in Rn, in this case.
And so when I do X transpose 1,
what I get is just the sum from
i equal 1 to n of the Xi's.
So if I divide by n,
I get my average, OK?
So here, I definitely
removed the sum term.
Let's see if with the covariance
matrix, we can do the same.
Well, and that's actually a
little more difficult to see,
I guess.
But let's use this
definition for S, OK?
And one thing that's
actually going to be--
so, let's see for
one second, what--
so it's going to be
something that involves X,
multiplying X with itself, OK?
And the question is,
is it going to be
multiplying X with X transpose,
or X tranpose with X?
To answer this
question, you can go
the easy route, which says,
well, my covariance matrix is
of size, what?
What is the size of S?
AUDIENCE: d by d.
PHILIPPE RIGOLLET: d by d, OK?
X is of size n by d.
So if I do X times
X transpose, I'm
going to have something
which is of size n by n.
If I do X transpose
X, I'm going to have
something which is d by d.
That's the easy route.
And there's basically
one of the two guys.
You can actually open
the box a little bit
and see what's
going on in there.
If you do X transpose X, which
we know gives you a d by d,
you'll see that X is
going to have vectors that
are of the form,
Xi, and X transpose
is going to have vectors that
are of the form, Xi transpose,
right?
And so, this is actually
probably the right way to go.
So let's look at what's X
transpose X is giving us.
So I claim that it's actually
going to give us what we want,
but rather than actually
going there, let's--
to actually-- I mean, we
could check it entry by entry,
but there's actually a
nice thing we can do.
Before we go there,
let's write X transpose
as the following sum of
variables, X1 and then
just a bunch of 0's
everywhere else.
So it's still d by n.
So n minus 1 of the columns
are equal to 0 here.
Then I'm going to put
a 0 and then put X2.
And then just a
bunch of 0's, right?
So that's just 0, 0 plus 0,
0, all the way to Xn, OK?
Everybody agrees with it?
See what I'm doing here?
I'm just splitting it into
a sum of matrices that
only have one nonzero columns.
But clearly, that's true.
Now let's look at the product
of this guy with itself.
So, let's call these
matrices M1, M2, Mn.
So when I do X
transpose X, what I
do is the sum of the
Mi's for i equal 1 to n,
times the sum of the
Mi transpose, right?
Now, the sum of
the Mi's transpose
is just the sum of each
of the Mi's transpose, OK?
So now I just have this
product of two sums,
so I'm just going to
re-index the second one by j.
So this is sum for i equal
1 to n, j equal 1 to n of Mi
Mj transpose.
OK?
And now what we
want to notice is
that if i is different
from j, what's happening?
Well if i is different from j,
let's look at say, M1 times XM2
transpose.
So what is the product
between those two matrices?
AUDIENCE: It's a new
entry and [INAUDIBLE]
PHILIPPE RIGOLLET:
There's an entry?
AUDIENCE: Well, it's an entry.
It's like a dot product in that
form next to [? transpose. ?]
PHILIPPE RIGOLLET: You mean
a dot product is just getting
[INAUDIBLE] number, right?
So I want-- this is
going to be a matrix.
It's the product of
two matrices, right?
This is a matrix times a matrix.
So this should be a matrix,
right, of size d by d.
Yeah, I should
see a lot of hands
that look like this, right?
Because look at this.
So let's multiply the first--
let's look at what's going
on in the first column here.
I'm multiplying this column
with each of those rows.
The only nonzero
coefficient is here,
and it only hits
this column of 0's.
So every time, this is going
to give you 0, 0, 0, 0.
And it's going to be the same
for every single one of them.
So this matrix is just
full of 0's, right?
They never hit each
other when I do
the matrix-matrix
multiplication.
There's no-- every
non-zero hits a 0.
So what it means is--
and this, of course,
you can check for every
i different from j.
So this means that Mi times
Mj transpose is actually
equal to 0 when i is
different from j, Right?
Everybody is OK with this?
So what that means is that when
I do this double sum, really,
it's a simple sum.
There's only just the
sum from i equal 1
to n of Mi Mi transpose.
Because this is the only
terms in this double sum
that are not going to be 0 when
[INAUDIBLE] [? M1 ?] with M1
itself.
Now, let's see
what's going on when
I do M1 times M1 transpose.
Well, now, if I do Mi
times and Mi transpose,
now this guy becomes [? X1 ?]
[INAUDIBLE] it's here.
And so now, I really have
X1 times X1 transpose.
So this is really
just the sum from i
equal 1 to n of Xi Xi transpose,
just because Mi Mi transpose
is Xi Xi transpose.
There's nothing else there.
So that's the good news, right?
This term here is really just
X transpose X divided by n.
OK, I can use that
guy again, I guess.
Well, no.
Let's just-- OK, so
let me rewrite S.
All right, that's the
definition we have.
And we know that this guy
already is equal to 1 over n X
transpose X. x bar
x bar transpose--
we know that x bar-- we
just proved that x bar--
sorry, little x
bar was equal to 1
over n X bar transpose
times the all-ones vector.
So I'm just going to do that.
So that's just
going to be minus.
I'm going to pull
my two 1 over n's--
one from this guy,
one from this guy.
So I'm going to get
1 over n squared.
And then I'm going
to get X bar--
sorry, there's no X bar here.
It's just X. Yeah.
X transpose all ones times X
transpose all ones transpose,
right?
And X transpose all
ones transpose--
right, the rule-- if I
have A times B transpose,
it's B transpose times
A transpose, right?
That's just the rule
of transposition.
So this is 1
transpose X transpose.
And so when I put all
these guys together,
this is actually equal to 1
over n X transpose X minus one
over n squared X transpose
1, 1 transpose X. Because X
transpose transposes X, OK?
So now, I can actually--
I have something which is
of the form, X transpose X--
[INAUDIBLE] to the left, X
transpose; to the right, X.
Here, I have X transpose to
the left, X to the right.
So it can factor out
whatever's in there.
So I can write S as 1 over n--
sorry, X transpose times 1 over
n times the identity of Rd.
And then I have minus 1
over n, 1, 1 transpose X.
OK, because if you--
I mean, you can
distribute it back, right?
So here, I'm going to get what?
X transpose identity times X,
the whole thing divided by n.
That's this term.
And then the second one is
going to be-- sorry, 1 over n
squared.
And then I'm going to get 1 over
n squared times X transpose 1,
1 transpose which is
this guy, times X,
and that's the [? right ?]
[? thing, ?] OK?
So, the way it's written, I
factored out one of the 1 over
n's.
So I'm just going to do the
same thing as on this slide.
So I'm just factoring
out this 1 over n here.
So it's 1 over n times
X transpose identity
of our d divided by n
divided by 1 this time,
minus 1 over n 1, 1
transpose times X, OK?
So that's just
what's on the slides.
What does the matrix, 1,
1 transpose, look like?
AUDIENCE: All 1's.
PHILIPPE RIGOLLET: It's
just all 1's, right?
Because the entries are the
products of the all-ones--
of the coordinates of
the all-ones vectors with
the coordinates of the all-ones
vectors, so I only get 1's.
So it's a d by d
matrix with only 1's.
So this matrix, I can
actually write exactly, right?
H, this matrix that
I called H which
is what's sandwiched in-between
this X transpose and X.
By definition, I said this
is the definition of H. Then
this thing, I can write
its coordinates exactly.
We know it's identity
divided by n minus--
sorry, I don't know
why I keep [INAUDIBLE]..
Minus 1 over n 1, 1 transpose--
so it's this matrix
with the only 1's
on the diagonals and 0's and
elsewhere-- minus a matrix that
only has 1 over n everywhere.
OK, so the whole thing is 1
minus 1 over n on the diagonals
and then minus 1
over n here, OK?
And now I claim that this matrix
is an orthogonal projector.
Now, I'm writing this, but
it's completely useless.
This is just a way for you to
see that it's actually very
convenient now to think
about this problem
as being a matrix
problem, because things
are much nicer when you
think about the actual form
of your matrices, right?
They could tell you,
here is the matrix.
I mean, imagine you're
sitting at a midterm,
and I say, here's the
matrix that has 1 minus 1
over n on the diagonals
and minus 1 over n
on the [INAUDIBLE] diagonal.
Prove to me that it's
a projector matrix.
You're going to
have to basically
take this guy times itself.
It's going to be really
complicated, right?
So we know it's symmetric.
That's for sure.
But the fact that it
has this particular way
of writing it is
going to make my life
super easy to check this.
That's the definition
of a projector.
It has to be
symmetric and it has
to square to itself
because we just
said in the chapter
on linear regression
that once you project, if you
apply the projection again,
you're not moving because
you're already there.
OK, so why is H
squared equal to H?
Well let's just write H square.
It's the identity
minus 1 over n 1, 1
transpose times the
identity minus 1 over n 1, 1
transpose, right?
Let's just expand this now.
This is equal to
the identity minus--
well, the identity times 1, 1
transpose is just the identity.
So it's 1, 1 transpose, sorry.
So 1 over n 1, 1 transpose
minus 1 over n 1, 1 transpose.
And then there's
going to be what
makes the deal is that
I get this 1 over n
squared this time.
And then I get the product
of 1 over n trans--
oh, let's write it completely.
I get 1, 1 transpose
times 1, 1 transpose, OK?
But this thing here--
what is this?
n, right, is the end product
of the all-ones vector
with the all-ones vector.
So I'm just summing n times
1 squared, which is n.
So this is equal to n.
So I pull it out,
cancel one of the ends,
and I'm back to
what I had before.
So I had identity minus 2
over n 1, 1 transpose plus 1
over n 1, 1 transpose
which is equal to H.
Because one of the 1
over n's cancel, OK?
So it's a projection matrix.
It's projecting onto
some linear space, right?
It's taking a matrix.
Sorry, it's taking
a vector and it's
projecting onto a
certain space of vectors.
What is this space?
Right, so, how do
you-- so I'm only
asking the answer to this
question in words, right?
So how would you
describe the vectors
onto which this
matrix is projecting?
Well, if you want to
answer this question,
the way you would tackle
it is first by saying, OK,
what does a vector which is of
the form, H times something,
look like, right?
What can I say about
this vector that's
going to be definitely
giving me something
about the space on
which it projects?
I need to know a little more to
know that it projects exactly
onto this.
But one way we can
do this is just
see how it acts on a vector.
What does it do to a
vector to apply H, right?
So I take v. And let's see what
taking v and applying H to it
looks like.
Well, it's the identity
minus something.
So it takes v and
it removes something
from v. What does it remove?
Well, it's 1 over n
times v transpose 1 times
the all-ones vector, right?
Agreed?
I just wrote v transpose 1
instead of 1 transpose v,
which are the same thing.
What is this thing?
What should I call it in
mathematical notation?
v bar, right?
I should all it v bar because
this is exactly the average
of the entries of v, agreed?
This is summing the entries
of v's, and this is dividing
by the number of those v's.
Sorry, now v is in our--
sorry, why do I divide by--
I'm just-- OK, I need to check
what my dimensions are now.
No, it's in Rd, right?
So why do I divide by n?
So it's not really v bar.
It's the sum of the
v's divided by--
right, so it's v bar.
AUDIENCE: [INAUDIBLE]
[INTERPOSING VOICES]
AUDIENCE: Yeah, v
has to be [INAUDIBLE]
PHILIPPE RIGOLLET: Oh, yeah.
OK, thank you.
So everywhere I wrote
Hd, that was actually Hn.
Oh, man.
I wish I had a computer now.
All right.
So-- yeah, because the--
yeah, right?
So why it's not--
well, why I thought it was
this is because I was thinking
about the outer
dimension of X, really
of X transpose, which is
really the inner dimension,
didn't matter to me, right?
So the thing that I can
sandwich between X transpose
and X has to be n by n.
So this was actually n by n.
And so that's actually n by n.
Everything is n by n.
Sorry about that.
So this is n.
This is n.
This is-- well, I
didn't really tell you
what the all-ones vector
was, but it's also in our n.
Yeah, OK.
Thank you.
And n-- actually, I used the
fact that this was of size n
here already.
OK, and so that's indeed v bar.
So what is this projection
doing to a vector?
It's removing its average
on each coordinate, right?
And the effect of this
is that v is a vector.
What is the average of Hv?
AUDIENCE: 0.
PHILIPPE RIGOLLET:
Right, so it's 0.
It's the average of v, which
is v bar, minus the average
of something that only has v
bar's entry, which is v bar.
So this thing is actually 0.
So let me repeat my question.
Onto what subspace
does H project?
Onto the subspace of
vectors that have mean 0.
A vector that has
mean 0 is a vector.
So if you want to talk more
linear algebra, v bar--
for a vector you
have mean 0, it means
that v is orthogonal to the
span of the all-ones vector.
That's it.
It projects to this space.
So in words, it
projects onto the space
of vectors that have 0 mean.
In linear algebra,
it says it projects
onto the hyperplane
which is orthogonal
to the all-ones vector, OK?
So that's all.
Can you guys still
see the screen?
Are you good over there?
OK.
All right, so now, what it
means is that, well, I'm
doing this weird thing, right?
I'm taking the inner product--
so S is taking X. And then
it's removing its mean of each
of the columns of X, right?
When I take H times X, I'm
basically applying this
projection which consists
in removing the mean of all
the X's.
And then I multiply
by H transpose.
But what's actually
nice is that, remember,
H is a projector.
Sorry, I don't
want to keep that.
Which means that when I
look at X transpose HX,
it's the same as looking
at X transpose H squared X.
But since H is equal
to its transpose,
this is actually the same
as looking at X transpose H
transpose HX, which is the
same as looking at HX transpose
HX, OK?
So what it's doing, it's
first applying this projection
matrix, H, which removes the
mean of each of your columns,
and then looks at the inner
products between those guys,
right?
Each entry of this guy
is just the covariance
between those centered things.
That's all it's doing.
All right, so those are actually
going to be the key statements.
So everything we've
done so far is really
mainly linear algebra, right?
I mean, looking at expectations
and covariances was just--
we just used the fact that
the expectation was linear.
We didn't do much.
But now there's a nice
thing that's happening.
And that's why we're
going to switch
from the language
of linear algebra
to more statistical,
because what's happening
is that if I look at this
quadratic form, right?
So I take sigma.
So I take a vector, u.
And I'm going to look at
u-- so let's say, in Rd.
And I'm going to look
at u transpose sigma u.
OK?
What is this doing?
Well, we know that u transpose
sigma u is equal to what?
Well, sigma is the
expectation of XX transpose
minus the expectation of X
expectation of X transpose,
right?
So I just substitute in there.
Now, u is deterministic.
So in particular, I can push
it inside the expectation
here, agreed?
And I can do the
same from the right.
So here, when I push u
transpose here, and u here,
what I'm left with is the
expectation of u transpose X
times X transpose u.
OK?
And now, I can do the
same thing for this guy.
And this tells me that this is
the expectation of u transpose
X times the expectation
of X transpose u.
Of course, u transpose X
is equal to X transpose u.
And u-- yeah.
So what it means is
that this is actually
equal to the expectation
of u transpose X squared
minus the expectation
of u transpose X,
the whole thing squared.
But this is something
that should look familiar.
This is really just the variance
of this particular random
variable which is of
the form, u transpose X,
right? u transpose
X is a number.
It involves a random vector,
so it's a random variable.
And so it has a variance.
And this variance is exactly
given by this formula.
So this is just the
variance of u transpose X.
So what we've proved is
that if I look at this guy,
this is really just the
variance of u transpose X, OK?
I can do the same thing
for the sample variance.
So let's do this.
And as you can
see, spoiler alert,
this is going to be
the sample variance.
OK, so remember, S is 1 over n,
sum of Xi Xi transpose minus X
bar X bar transpose.
So when I do u
transpose, Su, what
it gives me is 1 over
n sum from i equal 1
to n of u transpose Xi times
Xi transpose u, all right?
So those are two numbers
that multiply each other
and that happen to be
equal to each other,
minus u transpose X
bar X bar transpose u,
which is also the product
of two numbers that happen
to be equal to each other.
So I can rewrite
this with squares.
So we're almost there.
All I need to know to check
is that this thing is actually
the average of
those guys, right?
So u transpose X bar.
What is it?
It's 1 over n sum from i equal
1 to n of u transpose Xi.
So it's really something that I
can write as u transpose X bar,
right?
That's the average of
those random variables
of the form, u transpose Xi.
So what it means is that u
transpose Su, I can write as 1
over n sum from i equal 1 to
n of u transpose Xi squared
minus u transpose X
bar squared, which
is the empirical variance
that we need noted by small
s squared, right?
So that's the empirical variance
of u transpose X1 all the way
to u transpose Xn.
OK, and here, same thing.
I use exactly the same thing.
I just use the fact that here,
the only thing I use is really
the linearity of this
guy, of 1 over n sum
or the linearity of expectation,
that I can push things
in there, OK?
AUDIENCE: So what
you have written
at the end of that
sum for uT Su?
PHILIPPE RIGOLLET: This one?
AUDIENCE: Yeah.
PHILIPPE RIGOLLET: Yeah, I
said it's equal to small s,
and I want to make a
difference between the big S
that I'm using here.
So this is equal to small--
I don't know, I'm
trying to make it look
like a calligraphic s squared.
OK, so this is nice, right?
This covariance matrix-- so
let's look at capital sigma
itself right now.
This covariance matrix,
we know that if we
read its entries, what
we get is the covariance
between the coordinates
of the X's, right,
of the random vector, X.
And the coordinates, well,
by definition, are attached
to a coordinate system.
So I only know
what the covariance
of X in of those two things are,
or the covariance of those two
things are.
But what if I want to find
coordinates between linear
combination of the X's?
Sorry, if I want to find
covariances between linear
combination of those X's.
And that's exactly what
this allows me to do.
It says, well, if I pre-
and post-multiply by u,
this is actually telling
me what the variance
of X along direction u is, OK?
So there's a lot of
information in there,
and it's just really
exploiting the fact
that there is some linearity
going on in the covariance.
So, why variance?
Why is variance
interesting for us, right?
Why?
I started by saying,
here, we're going
to be interested
in having something
to do dimension reduction.
We have-- think of your points
as [? being in a ?] dimension
larger than 4, and we're going
to try to reduce the dimension.
So let's just think
for one second,
what do we want about a
dimension reduction procedure?
If I have all my points that
live in, say, three dimensions,
and I have one point
here and one point here
and one point here and one
point here and one point here,
and I decide to project
them onto some plane--
that I take a plane that's
just like this, what's
going to happen is that those
points are all going to project
to the same point, right?
I'm just going to
not see anything.
However, if I take a
plane which is like this,
they're all going to
project into some nice line.
Maybe I can even
project them onto a line
and they will still be
far apart from each other.
So that's what you want.
You want to be able to
say, when I take my points
and I say I project them
onto lower dimensions,
I do not want them to collapse
into one single point.
I want them to be spread as
possible in the direction
on which I project.
And this is what we're
going to try to do.
And of course, measuring
spread between points
can be done in many ways, right?
I mean, you could
look at, I don't know,
sum of pairwise distances
between those guys.
You could look at
some sort of energy.
You can look at
many ways to measure
of spread in a direction.
But variance is a
good way to measure
of spread between points.
If you have a lot of
variance between your points,
then chances are they're
going to be spread.
Now, this is not
always the case, right?
If I have a direction in which
all my points are clumped
onto one big point and
one other big point,
it's going to choose
this because that's
the direction that
has a lot of variance.
But hopefully, the
variance is going
to spread things out nicely.
So the idea of principal
component analysis
is going to try to
identify those variances--
those directions along which
we have a lot of variance.
Reciprocally, we're
going to try to eliminate
the directions along which we do
not have a lot of variance, OK?
And let's see why.
Well, if-- so here's
the first claim.
If you transpose Su is equal
to 0, what's happening?
Well, I know that an empirical
variance is equal to 0.
What does it mean for
an empirical variance
to be equal to 0?
So I give you a bunch
of points, right?
So those points are those
points-- u transpose
X1, u transpose-- those
are a bunch of numbers.
What does it mean to have
the empirical variance
of those points
being equal to 0?
AUDIENCE: They're all the same.
PHILIPPE RIGOLLET:
They're all the same.
So what it means is that
when I have my points, right?
So, can you find a direction
for those points in which they
project to all the same point?
No, right?
There's no such thing.
For this to happen, you have
to have your points which
are perfectly aligned.
And then when you're
going to project
onto the orthogonal
of this guy, they're
going to all project
to the same point
here, which means that
the empirical variance is
going to be 0.
Now, this is an extreme case.
This will never
happen in practice,
because if that
happens, well, I mean,
you can basically figure
that out very quickly.
So in the same way,
it's very unlikely
that you're going to have
u transpose sigma u, which
is equal to 0, which means
that, essentially, all
your points are [INAUDIBLE]
or let's say all of them
are orthogonal to u, right?
So it's exactly the same thing.
It just says that in
the population case,
there's no probability that your
points deviate from this guy
here.
This happens with
zero probability, OK?
And that's just
because if you look
at the variance of this
guy, it's going to be 0.
And then that means that
there's no deviation.
By the way, I'm using
the name projection
when I talk about u
transpose X, right?
So let's just be
clear about this.
If you-- so let's say I
have a bunch of points,
and u is a vector
in this direction.
And let's say that u has the--
so this is 0.
This is u.
And let's say that
u has norm, 1, OK?
When I look, what is the
coordinate of the projection?
So what is the length
of this guy here?
Let's call this guy X1.
What is the length of this guy?
In terms of inner products?
This is exactly u transpose X1.
This length here,
if this is X2, this
is exactly u transpose X2, OK?
So those-- u transpose X
measure exactly the distance
to the origin of those--
I mean, it's really--
think of it as being
just an x-axis thing.
You just have a bunch of points.
You have an origin.
And it's really just
telling you what
the coordinate on this
axis is going to be, right?
So in particular, if the
empirical variance is 0,
it means that all
these points project
to the same point, which
means that they have
to be orthogonal to this guy.
And you can think of it as
being also maybe an entire plane
that's orthogonal
to this line, OK?
So that's why I talk
about projection,
because the inner
products, u transpose X,
is really measuring
the coordinates of X
when u becomes the x-axis.
Now, if u does not have
norm 1, then you just
have a change of scale here.
You just have a
change of unit, right?
So this is really u times X1.
The coordinates should really
be divided by the norm of u.
OK, so now, just in
the same way-- so
we're never going
to have exactly 0.
But if we [INAUDIBLE]
the other end,
if u transpose Su is
large, what does it mean?
It means that when
I look at my points
as projected onto the
axis generated by u,
they're going to have
a lot of variance.
They're going to be far away
from each other in average,
right?
That's what large variance
means, or at least
large empirical variance means.
And same thing for u.
So what we're going
to try to find
is a u that maximizes this.
If I can find a u
that maximizes this
so I can look in
every direction,
and suddenly I find a direction
in which the spread is massive,
then that's a point
on which I'm basically
the less likely
to have my points
project onto each other
and collide, right?
At least I know they're
going to project
at least onto two points.
So the idea now is
to say, OK, let's try
to maximize this spread, right?
So we're going to try to
find the maximum over all u's
of u transpose Su.
And that's going to be the
direction that maximizes
the empirical variance.
Now of course, if I read it
like that for all u's in Rd,
what is the value
of this maximum?
It's infinity, right?
Because I can always
multiply u by 10,
and this entire thing is
going to multiplied by 100.
So I'm just going to take
u as large as I want,
and this thing is going
to be as large as I want,
and so I need to constrain u.
And as I said, I need
to have u of size 1
to talk about coordinates
in the system generated
by u like this.
So I'm just going to
constrain u to have
Euclidean norm equal to 1, OK?
So that's going to
be my goal-- trying
to find the largest
possible u transpose Su,
or in other words, empirical
variance of the points
projected onto the direction
u when u is of norm 1,
which justifies to use
the word, "direction,"
and because there's no
magnitude to this u.
OK, so how am I
going to do this?
I could just fold and
say, let's just optimize
this thing, right?
Let's just take this problem.
It says maximize a function
onto some constraints.
Immediately, the constraint
is sort of nasty.
I'm on a sphere, and I'm trying
to move points on the sphere.
And I'm maximizing
this thing which
actually happens to be convex.
And we know we know how to
minimize convex functions,
but maximize them is
a different question.
And so this problem
might be super hard.
So I can just say,
OK, here's what
I want to do, and let me
give that to an optimizer
and just hope that the optimizer
can solve this problem for me.
That's one thing we can do.
Now as you can imagine, PCA
is so well spread, right?
Principal component
analysis is something
that people do constantly.
And so that means that we
know how to do this fast.
So that's one thing.
The other thing that you should
probably question about why--
if this thing is actually
difficult, why in the world
would you even choose the
variance as a measure of spread
if there's so many
measures of spread, right?
The variance is one
measure of spread.
It's not guaranteed
that everything
is going to project nicely
far apart from each other.
So we could choose
the variance, but we
could choose something else.
If the variance does
not help, why choose it?
Turns out the variance helps.
So this is indeed a
non-convex problem.
I'm maximizing, so
it's actually the same.
I can make this
constraint convex
because I'm maximizing
a convex function,
so it's clear that
the maximum is going
to be attained at the boundary.
So I can actually just fill
this ball into some convex ball.
However, I'm still
maximizing, so this
is a non-convex problem.
And this turns out to be the
fanciest non-convex problem
we know how to solve.
And the reason why we
know how to solve it
is not because of optimization
or using gradient-type things
or anything of the
algorithms that I mentioned
during the maximum likelihood.
It's because of linear algebra.
Linear algebra guarantees that
we know how to solve this.
And to understand this, we
need to go a little deeper
in linear algebra, and we
need to understand the concept
of diagonalization of a matrix.
So who has ever seen the
concept of an eigenvalue?
Oh, that's beautiful.
And if you're not
raising your hand,
you're just playing
"Candy Crush," right?
All right, so, OK.
This is great.
Everybody's seen it.
For my live audience of
millions, maybe you have not,
so I will still go through it.
All right, so one
of the basic facts--
and I remember when
I learned this in--
I mean, when I was
an undergrad, I
learned about the
spectral decomposition
and this diagonalization
of matrices.
And for me, it was just
a structural property
of matrices, but it turns out
that it's extremely useful,
and it's useful for
algorithmic purposes.
And so what this
theorem tells you
is that if you take
a symmetric matrix--
well, with real
entries, but that
really does not matter so much.
And here, I'm
going to actually--
so I take a symmetric matrix,
and actually S and sigma
are two such symmetric
matrices, right?
Then there exists P
and D, which are both--
so let's say d by d.
Which are both d by d
such that P is orthogonal.
That means that P transpose
P is equal to PP transpose
is equal to the identity.
And D is diagonal.
And sigma, let's say, is
equal to PDP transpose, OK?
So it's a diagonalization
because it's
finding a nice transformation.
P has some nice properties.
It's really just the change
of coordinates in which
your matrix is diagonal, right?
And the way you
want to see this--
and I think it sort of helps
to think about this problem
as being--
sigma being a covariance matrix.
What does a covariance
matrix tell you?
Think of a
multivariate Gaussian.
Can everybody visualize a
three-dimensional Gaussian
density?
Right, so it's going to be some
sort of a bell-shaped curve,
but it might be more elongated
in one direction than another.
And then going to chop
it like that, all right?
So I'm going to chop it off.
And I'm going to look at
how it bleeds, all right?
So I'm just going to look
at where the blood is.
And what it's going to look at--
it's going to look like some
sort of ellipsoid, right?
In high dimension, it's
just going to be an olive.
And that is just going
to be bigger and bigger.
And then I chop it
off a little lower,
and I get something a
little bigger like this.
And so it turns out that sigma
is capturing exactly this,
right?
The matrix sigma-- so the
center of your covariance matrix
of your Gaussian is
going to be this thing.
And sigma is going to tell you
which direction it's elongated.
And so in particular, if you
look, if you knew an ellipse,
you know there's something
called principal axis, right?
So you could actually
define something
that looks like this, which is
this axis, the one along which
it's the most elongated.
Then the axis along which
is orthogonal to it,
along which it's
slightly less elongated,
and you go again and again
along the orthogonal ones.
It turns out that
those things here
is the new coordinate system
in which this transformation, P
and P transpose, is
putting you into.
And D has entries
on the diagonal
which are exactly this length
and this length, right?
So that's just what it's doing.
It's just telling
you, well, if you
think of having this Gaussian
or this high-dimensional
ellipsoid, it's elongated
along certain directions.
And these directions are
actually maybe not well aligned
with your original coordinate
system, which might just
be the usual one, right--
north, south, and east, west.
Maybe I need to turn it.
And that's exactly what this
orthogonal transformation is
doing for you, all right?
So, in a way, this is actually
telling you even more.
It's telling you that any
matrix that's symmetric,
you can actually
turn it somewhere.
And that'll start to dilate
things in the directions
that you have, and
then turn it back
to what you originally had.
And that's actually
exactly the effect
of applying a symmetric matrix
through a vector, right?
And it's pretty impressive.
It says if I take sigma
times v. Any sigma that's
of this form, what I'm
doing is-- that's symmetric.
What I'm really
doing to v is I'm
changing its coordinate
system, so I'm rotating it.
Then I'm changing-- I'm
multiplying its coordinates,
and then I'm rotating it back.
That's all it's
doing, and that's
what all symmetric
matrices do, which
means that this is doing a lot.
All right, so OK.
So, what do I know?
So I'm not going to
prove that this is
the so-called spectral theorem.
And the diagonal entries of
D is of the form, lambda 1,
lambda 2, lambda d, 0, 0.
And the lambda j's are
called eigenvalues of D.
Now in general, those numbers
can be positive, negative,
or equal to 0.
But here, I know that
sigma and S are--
well, they're
symmetric for sure,
but they are positive
semidefinite.
What does it mean?
It means that when I take u
transpose sigma u for example,
this number is
always non-negative.
Why is this true?
What is this number?
It's the variance of--
and actually, I don't even
need to finish this sentence.
As soon as I say that
this is a variance, well,
it has to be non-negative.
We know that a variance
is not negative.
And so, that's also a
nice way you can use that.
So it's just to say,
well, OK, this thing
is positive semidefinite because
it's a covariance matrix.
So I know it's a variance, OK?
So I get this.
Now, if I had some
negative numbers--
so the effect of that is that
when I draw this picture,
those axes are always positive,
which is kind of a weird thing
to say.
But what it means is that when
I take a vector, v, I rotate it,
and then I stretch it in the
directions of the coordinate,
I cannot flip it.
I can only stretch or shrink,
but I cannot flip its sign,
all right?
But in general, for
any symmetric matrices,
I could do this.
But when it's positive
symmetric definite,
actually what turns out
is that all the lambda
j's are non-negative.
I cannot flip it, OK?
So all the eigenvalues
are non-negative.
That's a property
of positive semidef.
So when it's symmetric,
you have the eigenvalues.
They can be any number.
And when it's positive
semidefinite, in particular
that's the case of
the covariance matrix
and the empirical
covariance matrix, right?
Because the empirical
covariance matrix
is an empirical variance,
which itself is non-negative.
And so I get that the
eigenvalues are non-negative.
All right, so principal
component analysis is saying,
OK, I want to find
the direction, u,
that maximizes u
transpose Su, all right?
I've just introduced
in one slide
something about eigenvalues.
So hopefully, they should help.
So what is it that I'm
going to be getting?
Well, let's just
see what happens.
Oh, I forgot to mention
that-- and I will use this.
So the lambda j's are
called eigenvectors.
And then the matrix, P,
has columns v1 to vd, OK?
The fact that it's orthogonal--
that P transpose P is equal
to the identity--
means that those guys
satisfied that vi transpose
vj is equal to 0 if i
is different from j.
And vi transpose vi is
actually equal to 1,
right, because the
entries of PP transpose
are exactly going to be of
the form, vi transpose vj, OK?
So those v's are
called eigenvectors.
And v1 is attached to lambda 1,
and v2 is attached to lambda 2,
OK?
So let's see what's
happening with those things.
What happens if I take sigma--
so if you know eigenvalues,
you know exactly what's
going to happen.
If I look at, say, sigma
times v1, well, what is sigma?
We know that sigma
is PDP transpose v1.
What is P transpose times v1?
Well, P transpose has
rows v1 transpose,
v2 transpose, all the
way to vd transpose.
So when I multiply
this by v1, what
I'm left with is
the first coordinate
is going to be equal to 1
and the second coordinate is
going to be equal to 0, right?
Because they're
orthogonal to each other--
0 all the way to the end.
So that's when I
do P transpose v1.
Now I multiply by
D. Well, I'm just
multiplying this guy by lambda
1, this guy by lambda 2,
and this guy by lambda d, so
this is really just lambda 1.
And now I need to
post-multiply by P.
So what is P times this guy?
Well, P is v1 all the way to vd.
And now I multiply
by a vector that
only has 0's except
lambda 1 on the first guy.
So this is just
lambda 1 times v1.
So what we've proved is that
sigma times v1 is lambda 1 v1,
and that's probably the
notion of eigenvalue you're
most comfortable with, right?
So just when I
multiply by v1, I get
v1 back multiplied by something,
which is the eigenvalue.
So in particular, if I look
at v1, transpose sigma v1,
what do I get?
Well, I get lambda
1 v1 transpose v1,
which is 1, right?
So this is actually
lambda 1 v1 transpose v1,
which is lambda 1, OK?
And if I do the same
with v2, clearly I'm
going to get v2 transpose sigma.
v2 is equal to lambda 2.
So for each of the
vj's, I know that if I
look at the variance
along the vj,
it's actually exactly given by
those eigenvalues, all right?
Which proves this, because the
variance along the eigenvectors
is actually equal
to the eigenvalues.
So since they're variances,
they have to be non-negative.
So now, I'm looking for
the one direction that
has the most variance, right?
But that's not only
among the eigenvectors.
That's also among
the other directions
that are in-between
the eigenvectors.
If I were to look only
at the eigenvectors,
it would just tell me, well,
just pick the eigenvector, vj,
that's associated to the
largest of the lambda j's.
But it turns out that that's
also true for any vector--
that the maximum direction is
actually one direction which
is among the eigenvectors.
And among the eigenvectors,
we know that the one that's
the largest--
that carries the
largest variance is
the one that's associated to the
largest eigenvalue, all right?
And so this is what PCA is
going to try to do for me.
So in practice, that's what
I mentioned already, right?
We're trying to
project the point cloud
onto a low-dimensional
space, D prime,
by keeping as much
information as possible.
And by "as much information,"
I mean we do not
want points to collide.
And so what PCA is
going to do is just
going to try to project
[? on two ?] directions.
So there's going
to be a u, and then
there's going to be something
orthogonal to u, and then
the third one, et cetera, so
that once we project on those,
we're keeping as much of the
covariance as possible, OK?
And in particular,
those directions
that we're going to
pick are actually
a subset of the vj's that
are associated to the largest
eigenvalues.
So I'm going to
stop here for today.
We'll finish this on Tuesday.
But basically, the idea is
it's just the following.
You're just going to--
well, let me skip one more.
Yeah, this is the idea.
You're first going to pick
the eigenvector associated
to the largest eigenvalue.
Then you're going to pick
the direction that orthogonal
to the vector that
you've picked,
and that's carrying
the most variance.
And that's actually
the second largest--
the eigenvector associated to
the second largest eigenvalue.
And you're going to go all
the way to the number of them
that you actually want to pick,
which is in this case, d, OK?
And wherever you choose
to chop this process,
not going all the way to d,
is going to actually give you
a lower-dimensional
representation
in the coordinate system
that's given by v1, v2, v3, et
cetera, OK?
So we'll see that in
more details on Tuesday.
But I don't want
to get into it now.
We don't have enough time.
Are there any questions?
