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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. |
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