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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. |
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