video_id
stringclasses 7
values | text
stringlengths 2
29.3k
|
|---|---|
rLlZpnT02ZU
|
So now I'm going to split
my integral into two parts,
|
rLlZpnT02ZU
|
in A, A star, so on A
star, f is larger than g,
|
rLlZpnT02ZU
|
so the absolute value is
just the difference itself.
|
rLlZpnT02ZU
|
So here I put parenthesis
rather than absolute value.
|
rLlZpnT02ZU
|
And then I have plus 1/2 of
the integral on the complement.
|
rLlZpnT02ZU
|
What are you guys used to to
write the complement, to the C
|
rLlZpnT02ZU
|
or the bar?
|
rLlZpnT02ZU
|
To the C?
|
rLlZpnT02ZU
|
And so here on the complement,
then f is less than g,
|
rLlZpnT02ZU
|
so this is actually really
g of X minus f of X, dx.
|
rLlZpnT02ZU
|
Everybody's with me here?
|
rLlZpnT02ZU
|
So I just said--
|
rLlZpnT02ZU
|
I mean, those are just
rewriting what the definition
|
rLlZpnT02ZU
|
of the absolute value is.
|
rLlZpnT02ZU
|
OK.
|
rLlZpnT02ZU
|
So now there's nice things
that I know about f and g.
|
rLlZpnT02ZU
|
And the two nice things is that
the integral of f is equal to 1
|
rLlZpnT02ZU
|
and the integral
of g is equal to 1.
|
rLlZpnT02ZU
|
This implies that the integral
of f minus g is equal to what?
|
rLlZpnT02ZU
|
AUDIENCE: 0.
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET: 0.
|
rLlZpnT02ZU
|
And so now that
means that if I want
|
rLlZpnT02ZU
|
to just go from the integral
here on A complement
|
rLlZpnT02ZU
|
to the integral on A--
|
rLlZpnT02ZU
|
or on A star, complement
to the integral of A star,
|
rLlZpnT02ZU
|
I just have to flip the sign.
|
rLlZpnT02ZU
|
So that implies that
an integral on A star
|
rLlZpnT02ZU
|
complement of g
of X minus f of X,
|
rLlZpnT02ZU
|
dx, this is simply equal
to the integral on A star
|
rLlZpnT02ZU
|
of f of X minus g of X, dx.
|
rLlZpnT02ZU
|
All right.
|
rLlZpnT02ZU
|
So now this guy becomes
this guy over there.
|
rLlZpnT02ZU
|
So I have 1/2 of this
plus 1/2 of the same guy,
|
rLlZpnT02ZU
|
so that means that 1/2 half
of the integral between of f
|
rLlZpnT02ZU
|
minus g absolute value--
|
rLlZpnT02ZU
|
so that was my
original definition,
|
rLlZpnT02ZU
|
this thing is actually equal
to the integral on A star
|
rLlZpnT02ZU
|
of f of X minus g of X, dx.
|
rLlZpnT02ZU
|
And this is simply
equal to P of A star--
|
rLlZpnT02ZU
|
so say Pf of A start
minus Pg of A star.
|
rLlZpnT02ZU
|
Which one is larger
than the other one?
|
rLlZpnT02ZU
|
AUDIENCE: [INAUDIBLE]
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET: It is.
|
rLlZpnT02ZU
|
Just look at this board.
|
rLlZpnT02ZU
|
AUDIENCE: [INAUDIBLE]
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET: What?
|
rLlZpnT02ZU
|
AUDIENCE: [INAUDIBLE]
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET:
The first one has
|
rLlZpnT02ZU
|
to be larger, because
this thing is actually
|
rLlZpnT02ZU
|
equal to a non-negative number.
|
rLlZpnT02ZU
|
So now I have this absolute
value of two things,
|
rLlZpnT02ZU
|
and so I'm closer to
the actual definition.
|
rLlZpnT02ZU
|
But I still need to show
you that this thing is
|
rLlZpnT02ZU
|
the maximum value.
|
rLlZpnT02ZU
|
So this is definitely at
most the maximum over A of Pf
|
rLlZpnT02ZU
|
of A minus Pg of A.
|
rLlZpnT02ZU
|
That's certainly true.
|
rLlZpnT02ZU
|
Right?
|
rLlZpnT02ZU
|
We agree with this?
|
rLlZpnT02ZU
|
Because this is just
for one specific A,
|
rLlZpnT02ZU
|
and I'm bounding it by the
maximum over all possible A.
|
rLlZpnT02ZU
|
So that's clearly true.
|
rLlZpnT02ZU
|
So now I have to go
the other way around.
|
rLlZpnT02ZU
|
I have to show you that the max
is actually this guy, A star.
|
rLlZpnT02ZU
|
So why would that be true?
|
rLlZpnT02ZU
|
Well, let's just inspect
this thing over there.
|
rLlZpnT02ZU
|
So we want to show
that if I take
|
rLlZpnT02ZU
|
any other A in this integral
than this guy A star,
|
rLlZpnT02ZU
|
it's actually got to
decrease its value.
|
rLlZpnT02ZU
|
So we have this function.
|
rLlZpnT02ZU
|
I'm going to call
this function delta.
|
rLlZpnT02ZU
|
And what we have
is-- so let's say
|
rLlZpnT02ZU
|
this function looks like this.
|
rLlZpnT02ZU
|
Now it's the difference
between two densities.
|
rLlZpnT02ZU
|
It doesn't have to
integrate-- it doesn't
|
rLlZpnT02ZU
|
have to be non-negative.
|
rLlZpnT02ZU
|
But it certainly has
to integrate to 0.
|
rLlZpnT02ZU
|
And so now I take this thing.
|
rLlZpnT02ZU
|
And the A star, what
is the set A star here?
|
rLlZpnT02ZU
|
The set A star is the set
over which the function
|
rLlZpnT02ZU
|
delta is non-negative.
|
rLlZpnT02ZU
|
So that's just the definition.
|
rLlZpnT02ZU
|
A star was the set over
which f minus g was positive,
|
rLlZpnT02ZU
|
and f minus g was
just called delta.
|
rLlZpnT02ZU
|
So what it means is that
what I'm really integrating
|
rLlZpnT02ZU
|
is delta on this set.
|
rLlZpnT02ZU
|
So it's this area
under the curve,
|
rLlZpnT02ZU
|
just on the positive things.
|
rLlZpnT02ZU
|
Agreed?
|
rLlZpnT02ZU
|
So now let's just make some
tiny variations around this guy.
|
rLlZpnT02ZU
|
If I take A to be
larger than A star--
|
rLlZpnT02ZU
|
so let me add, for
example, this part here.
|
rLlZpnT02ZU
|
That means that when
I compute my integral,
|
rLlZpnT02ZU
|
I'm removing this
area under the curve.
|
rLlZpnT02ZU
|
It's negative.
|
rLlZpnT02ZU
|
The integral here is negative.
|
rLlZpnT02ZU
|
So if I start adding something
to A, the value goes lower.
|
rLlZpnT02ZU
|
If I start removing something
from A, like say this guy,
|
rLlZpnT02ZU
|
I'm actually removing this
value from the integral.
|
rLlZpnT02ZU
|
So there's no way.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.