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rLlZpnT02ZU | I'm actually stuck. |
rLlZpnT02ZU | This A star is the one
that actually maximizes |
rLlZpnT02ZU | the integral of this function. |
rLlZpnT02ZU | So we used the fact
that for any function, |
rLlZpnT02ZU | say delta, the integral
over A of delta |
rLlZpnT02ZU | is less than the integral
over the set of X's |
rLlZpnT02ZU | such that delta of X is
non-negative of delta of X, dx. |
rLlZpnT02ZU | And that's an obvious
fact, just by picture, say. |
rLlZpnT02ZU | And that's true for all A. Yeah? |
rLlZpnT02ZU | AUDIENCE: [INAUDIBLE]
could you use |
rLlZpnT02ZU | like a portion under the
axis as like less than |
rLlZpnT02ZU | or equal to the
portion above the axis? |
rLlZpnT02ZU | PHILIPPE RIGOLLET:
It's actually equal. |
rLlZpnT02ZU | We know that the
integral of f minus g-- |
rLlZpnT02ZU | the integral of delta is 0. |
rLlZpnT02ZU | So there's actually exactly
the same area above and below. |
rLlZpnT02ZU | But yeah, you're right. |
rLlZpnT02ZU | You could go to
the extreme cases. |
rLlZpnT02ZU | You're right. |
rLlZpnT02ZU | No. |
rLlZpnT02ZU | It's actually still be
true, even if there was-- |
rLlZpnT02ZU | if this was a constant,
that would still be true. |
rLlZpnT02ZU | Here, I never use the fact that
the integral is equal to 0. |
rLlZpnT02ZU | I could shift this function by
1 so that the integral of delta |
rLlZpnT02ZU | is equal to 1,
and it would still |
rLlZpnT02ZU | be true that it's maximized
when I take A to be |
rLlZpnT02ZU | the set where it's positive. |
rLlZpnT02ZU | Just need to make sure that
there is someplace where it is, |
rLlZpnT02ZU | but that's about it. |
rLlZpnT02ZU | Of course, we used this before,
when we made this thing. |
rLlZpnT02ZU | But just the last
argument, this last fact |
rLlZpnT02ZU | does not require that. |
rLlZpnT02ZU | All right. |
rLlZpnT02ZU | So now we have this notion of-- |
rLlZpnT02ZU | I need the-- |
rLlZpnT02ZU | OK. |
rLlZpnT02ZU | So we have this
notion of distance |
rLlZpnT02ZU | between probability measures. |
rLlZpnT02ZU | I mean, these things
are exactly what-- |
rLlZpnT02ZU | if I were to be in a formal
math class and I said, |
rLlZpnT02ZU | here are the axioms that
a distance should satisfy, |
rLlZpnT02ZU | those are exactly those things. |
rLlZpnT02ZU | If it's not
satisfying this thing, |
rLlZpnT02ZU | it's called pseudo-distance or
quasi-distance or just metric |
rLlZpnT02ZU | or nothing at all, honestly. |
rLlZpnT02ZU | So it's a distance. |
rLlZpnT02ZU | It's symmetric,
non-negative, equal to 0, |
rLlZpnT02ZU | if and only if the two
arguments are equal, then |
rLlZpnT02ZU | it satisfies the
triangle inequality. |
rLlZpnT02ZU | And so that means that we have
this actual total variation |
rLlZpnT02ZU | distance between
probability distributions. |
rLlZpnT02ZU | And here is now a statistical
strategy to implement our goal. |
rLlZpnT02ZU | Remember, our goal
was to spit out |
rLlZpnT02ZU | a theta hat, which was
close such that P theta |
rLlZpnT02ZU | hat was close to P theta star. |
rLlZpnT02ZU | So hopefully, we were trying
to minimize the total variation |
rLlZpnT02ZU | distance between P theta
hat and P theta star. |
rLlZpnT02ZU | Now, we cannot do that, because
just by this fact, this slide, |
rLlZpnT02ZU | if we wanted to do that
directly, we would just take-- |
rLlZpnT02ZU | well, let's take theta hat
equals theta star and that will |
rLlZpnT02ZU | give me the value 0. |
rLlZpnT02ZU | And that's the minimum
possible value we can take. |
rLlZpnT02ZU | The problem is
that we don't know |
rLlZpnT02ZU | what the total variation is to
something that we don't know. |
rLlZpnT02ZU | We know how to compute total
variations if I give you |
rLlZpnT02ZU | the two arguments. |
rLlZpnT02ZU | But here, one of the
arguments is not known. |
rLlZpnT02ZU | P theta star is not known to
us, so we need to estimate it. |
rLlZpnT02ZU | And so here is the strategy. |
rLlZpnT02ZU | Just build an estimator
of the total variation |
rLlZpnT02ZU | distance between P
theta and P theta star |
rLlZpnT02ZU | for all candidate theta,
all possible theta |
rLlZpnT02ZU | in capital theta. |
rLlZpnT02ZU | Now, if this is a good estimate,
then when I minimize it, |
rLlZpnT02ZU | I should get something
that's close to P theta star. |
rLlZpnT02ZU | So here's the strategy. |
rLlZpnT02ZU | This is my function
that maps theta |
rLlZpnT02ZU | to the total variation between
P theta and P theta star. |
rLlZpnT02ZU | I know it's minimized
at theta star. |
rLlZpnT02ZU | That's definitely TV of P--
and the value here, the y-axis |
rLlZpnT02ZU | should say 0. |
rLlZpnT02ZU | And so I don't know
this guy, so I'm |
rLlZpnT02ZU | going to estimate it
by some estimator that |
rLlZpnT02ZU | comes from my data. |
rLlZpnT02ZU | Hopefully, the more data I have,
the better this estimator is. |
rLlZpnT02ZU | And I'm going to try to
minimize this estimator now. |
rLlZpnT02ZU | And if the two things are
close, then the minima |
rLlZpnT02ZU | should be close. |
rLlZpnT02ZU | That's a pretty good
estimation strategy. |
rLlZpnT02ZU | The problem is that
it's very unclear |
rLlZpnT02ZU | how you would build
this estimator of TV, |
rLlZpnT02ZU | of the Total Variation. |
rLlZpnT02ZU | So building
estimators, as I said, |
rLlZpnT02ZU | typically consists in replacing
expectations by averages. |
rLlZpnT02ZU | But there's no simple way of
expressing the total variation |
rLlZpnT02ZU | distance as the
expectations with respect |
rLlZpnT02ZU | to theta star of anything. |
rLlZpnT02ZU | So what we're going
to do is we're |
rLlZpnT02ZU | going to move from
total variation distance |
rLlZpnT02ZU | to another notion of
distance that sort of has |
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