video_id
stringclasses 7
values | text
stringlengths 2
29.3k
|
|---|---|
rLlZpnT02ZU
|
TV of P theta and
P theta prime is
|
rLlZpnT02ZU
|
the same as the TV between
P theta prime and P theta.
|
rLlZpnT02ZU
|
Well, that's pretty obvious
from this definition.
|
rLlZpnT02ZU
|
I just flip those two,
I get the same number.
|
rLlZpnT02ZU
|
It's actually also true
if I take the maximum.
|
rLlZpnT02ZU
|
Those things are completely
symmetric in theta and theta
|
rLlZpnT02ZU
|
prime.
|
rLlZpnT02ZU
|
You can just flip them.
|
rLlZpnT02ZU
|
It's non-negative.
|
rLlZpnT02ZU
|
Is that clear to everyone that
this thing is non-negative?
|
rLlZpnT02ZU
|
I integrate an absolute
value, so this thing
|
rLlZpnT02ZU
|
is going to give me some
non-negative number.
|
rLlZpnT02ZU
|
And so if I integrate
this non-negative number,
|
rLlZpnT02ZU
|
it's going to be a
non-negative number.
|
rLlZpnT02ZU
|
The fact also that
it's an area tells me
|
rLlZpnT02ZU
|
that it's going to
be non-negative.
|
rLlZpnT02ZU
|
The nice thing is that if
TV is equal to zero, then
|
rLlZpnT02ZU
|
the two distributions, the two
probabilities are the same.
|
rLlZpnT02ZU
|
That means that for
every A, P theta of A
|
rLlZpnT02ZU
|
is equal to P theta
prime of A. Now,
|
rLlZpnT02ZU
|
there's two ways to see that.
|
rLlZpnT02ZU
|
The first one is to say
that if this integral is
|
rLlZpnT02ZU
|
equal to 0, that means
that for almost all X,
|
rLlZpnT02ZU
|
f theta is equal
to f theta prime.
|
rLlZpnT02ZU
|
The only way I can integrate
a non-negative and get 0
|
rLlZpnT02ZU
|
is that it's 0 pretty
much everywhere.
|
rLlZpnT02ZU
|
And so what it means is
that the two densities
|
rLlZpnT02ZU
|
have to be the same
pretty much everywhere,
|
rLlZpnT02ZU
|
which means that the
distributions are the same.
|
rLlZpnT02ZU
|
But this is not really the
way you want to do this,
|
rLlZpnT02ZU
|
because you have
to understand what
|
rLlZpnT02ZU
|
pretty much everywhere means--
|
rLlZpnT02ZU
|
which I should really
say almost everywhere.
|
rLlZpnT02ZU
|
That's the formal
way of saying it.
|
rLlZpnT02ZU
|
But let's go to
this definition--
|
rLlZpnT02ZU
|
which is gone.
|
rLlZpnT02ZU
|
Yeah.
|
rLlZpnT02ZU
|
That's the one here.
|
rLlZpnT02ZU
|
The max of those two guys, if
this maximum is equal to 0--
|
rLlZpnT02ZU
|
I have a maximum of non-negative
numbers, their absolute values.
|
rLlZpnT02ZU
|
Their maximum is
equal to 0, well,
|
rLlZpnT02ZU
|
they better be all equal
to 0, because if one is not
|
rLlZpnT02ZU
|
equal to 0, then the
maximum is not equal to 0.
|
rLlZpnT02ZU
|
So those two guys,
for those two things
|
rLlZpnT02ZU
|
to be-- for the maximum
to be equal to 0,
|
rLlZpnT02ZU
|
then each of the
individual absolute values
|
rLlZpnT02ZU
|
have to be equal to 0, which
means that the probability here
|
rLlZpnT02ZU
|
is equal to this probability
here for every event A.
|
rLlZpnT02ZU
|
So those two things--
|
rLlZpnT02ZU
|
this is nice, right?
|
rLlZpnT02ZU
|
That's called definiteness.
|
rLlZpnT02ZU
|
The total variation equal
to 0 implies that P theta
|
rLlZpnT02ZU
|
is equal to P theta prime.
|
rLlZpnT02ZU
|
So that's really some
notion of distance, right?
|
rLlZpnT02ZU
|
That's what we want.
|
rLlZpnT02ZU
|
If this thing
being small implied
|
rLlZpnT02ZU
|
that P theta could be all
over the place compared
|
rLlZpnT02ZU
|
to P theta prime, that
would not help very much.
|
rLlZpnT02ZU
|
Now, there's also the
triangle inequality
|
rLlZpnT02ZU
|
that follows immediately
from the triangle
|
rLlZpnT02ZU
|
inequality inside this guy.
|
rLlZpnT02ZU
|
If I squeeze in some f
theta prime prime in there,
|
rLlZpnT02ZU
|
I'm going to use the
triangle inequality
|
rLlZpnT02ZU
|
and get the triangle
inequality for the whole thing.
|
rLlZpnT02ZU
|
Yeah?
|
rLlZpnT02ZU
|
AUDIENCE: The fact that
you need two definitions
|
rLlZpnT02ZU
|
of the [INAUDIBLE],,
is it something
|
rLlZpnT02ZU
|
obvious or is it complete?
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET:
I'll do it for you now.
|
rLlZpnT02ZU
|
So let's just prove that
those two things are actually
|
rLlZpnT02ZU
|
giving me the same definition.
|
rLlZpnT02ZU
|
So what I'm going to do
is I'm actually going
|
rLlZpnT02ZU
|
to start with the second one.
|
rLlZpnT02ZU
|
And I'm going to write--
|
rLlZpnT02ZU
|
I'm going to start with
the density version.
|
rLlZpnT02ZU
|
But as an exercise, you can
do it for the PMF version
|
rLlZpnT02ZU
|
if you prefer.
|
rLlZpnT02ZU
|
So I'm going to start
with the fact that f--
|
rLlZpnT02ZU
|
so I'm going to write f of g so
I don't have to write f and g.
|
rLlZpnT02ZU
|
So think of this as being f sub
theta, and think of this guy
|
rLlZpnT02ZU
|
as being f sub theta prime.
|
rLlZpnT02ZU
|
I just don't want to have to
write indices all the time.
|
rLlZpnT02ZU
|
So I'm going to start with
this thing, the integral of f
|
rLlZpnT02ZU
|
of X minus g of X dx.
|
rLlZpnT02ZU
|
The first thing I'm going to do
is this is an absolute value,
|
rLlZpnT02ZU
|
so either the number in the
absolute value is positive
|
rLlZpnT02ZU
|
and I actually kept it
like that, or it's negative
|
rLlZpnT02ZU
|
and I flipped its sign.
|
rLlZpnT02ZU
|
So let's just split
between those two cases.
|
rLlZpnT02ZU
|
So this thing is equal
to 1/2 the integral of--
|
rLlZpnT02ZU
|
so let me actually
write the set A star as
|
rLlZpnT02ZU
|
being the set of X's such that
f of X is larger than g of X.
|
rLlZpnT02ZU
|
So that's the set on
which the difference is
|
rLlZpnT02ZU
|
going to be positive
or the difference is
|
rLlZpnT02ZU
|
going to be negative.
|
rLlZpnT02ZU
|
So this, again,
is equivalent to f
|
rLlZpnT02ZU
|
of X minus g of X is positive.
|
rLlZpnT02ZU
|
OK.
|
rLlZpnT02ZU
|
Everybody agrees?
|
rLlZpnT02ZU
|
So this is the set
I'm interested in.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.