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ALBERT MEYER: We've looked at
independence for two events.

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What about when we
have a bunch of events?

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Well, in that case
we want to look

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at the idea of
mutual independence.

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So let's check that out.

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I'll say that if I have
n different events,

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I'll say that they're mutually
independent, intuitively.

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If the probability
that one of them occurs

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is unchanged by which other
ones happen to have occurred.

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So expressed in conditional
probability, which

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is the way to make it precise,
what we're really saying

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is that events A1 through An
are mutually independent when

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the probability of Ai is equal
to the probability of Ai,

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given the intersection
of any of the other

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As as long as i is
not one of them.

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So take A1, A2, or
A1, A2, A3, and so on.

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And A5 is going to
be independent of all

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of those other intersections.

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If we shift over to the other
definition of independence

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that we used for two sets,
in terms of products,

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you could say that n sets
are mutually independent when

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the probability of the
intersection of any bunch

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of them is equal to the product
of the individual probabilities

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of the events in
the intersection.

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Let's look at an example
of mutual independence.

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Maybe the simplest
one is the one

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of independent coin flips, which
by definition are independent.

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So the idea is that I will
flip a coin a bunch of times.

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And I will let Hi be
the event that the ith

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time I flip I get a heads.

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So if you think about
what's going on,

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what happens on
the fifth flip has

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nothing to do with what happens
on the first, fourth or seventh

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flip.

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There's no causal
relationship between the flips

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before or after flip five.

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Flip five is an isolated
event by itself.

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And the fact that there
were a bunch of heads before

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or there will be
a bunch of heads

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afterward doesn't have any
impact on the probability

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that the fifth flip
comes up with a head.

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At least that's what we
believe and that's the way

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that we would model them.

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So what that means,
for example, is

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that the probability of
a head on the fifth toss

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is equal to the probability of
a head on the fifth toss given

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that the first toss was a head
and the fourth toss was a head

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and the seventh
toss was not a head.

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This is the complement of H7.

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So that would just
be an example of one

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of the many different
conditional equations that

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hold when you have
mutual independence.

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Let's look at an example.

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Suppose that I flip
a fair coin twice.

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Now, the previous definition
didn't require fairness at all

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in the coin flipping.

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But now I'm going to need it.

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So that means that heads and
tails are equally likely.

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And suppose I flip
the coin twice.

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Well, let H1 be as before,
the event that a head comes up

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on the first flip.

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And H2, the event that a head
comes up on the second flip.

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And let O be the event that
there were an odd number

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of heads in the two flips.

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Now, I claim that O is
independent of whether or not

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there's a head on
the first flip.

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That may seem a
little weird because O

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depends on both the first
flip and the second flip.

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It's whether or not there are
an odd number of heads there,

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but nevertheless, I
claim that whether or not

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there are an odd
number of heads is

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independent of whether or not
the first toss was a head.

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Let's just check it using
the official definition.

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First of all, O is
the event HT TH.

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If I write out Hs and
Ts, a pair of them

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for what the results of the
first and second flips were,

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you get an odd number of heads
exactly when there's first

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a head and then a tail or first
a tail and then a head, which

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means that the probability
of O is exactly a half.

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Because the other
two outcomes are

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TT and HH, which is when you
have an even number of heads.

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Now, O into section
H1 is saying that you

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have an odd number of heads
and the first toss is a head.

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The only outcome that
fits that description

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is HT, which means that--
and the probability of HT

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is a quarter-- so the
probability of O intersection

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H1 is a quarter.

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O into section H1 is
just a peculiar way

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of saying you got a head
and then you got a tail.

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So that means that the
probability of O intersection

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H1 is a quarter.

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And of course, that's equal
to the probability of O,

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which we decided was a half,
and the probability of H1,

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which of course is
a half, because we

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said the coin was fair.

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So I've verified the condition
for the independence of O

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and H1, and therefore, I'm done.

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But the weird
thing to notice now

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is that if you look at O, H1,
and H2, the three of them,

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they are not
mutually independent.

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Because in fact, if you
know any two of them

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you can figure out
what the third one was.

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But just explicitly in terms
of conditional probabilities,

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the probability of there
being an odd number of heads,

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given that the first
toss was a head

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and the second toss
was a head, is 0,

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because once you
know H1 and H2 you

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know exactly how many
heads there were.

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There were two.

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And that's not odd.

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So the probability of odd
given H1 intersection H2

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is 0, which is not equal to the
probability of odd by itself,

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which was a half.

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So the three of them
are not independent.

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They're not mutually
independent,

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even though any two of them
are because O and H1 are.

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And obviously O and
H2 are by symmetry.

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And H1 and H2 to
are coin tosses,

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and they're independent.

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So that leads us to the general
idea of k-way independence.

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And an example would be if
you flip a fair coin k times,

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let Hi be whether or not
there's a head on the ith flip.

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And you let O, again,
be whether or not

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there are an odd
number of heads.

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And by the same
argument, you can

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verify that any set
of k of these events

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are mutually independent.

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But if you give me all k plus 1,
then they are not independent.

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In fact, any k of them will
determine the k plus first one.

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But any k among themselves
will be mutually independent.

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So that's why this notion of how
independent a bunch of sets are

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comes up, and this
is how to count it.

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So in general, events A1 through
an arbitrary set of events

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are k-way independent
if any k of them

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are mutually independent.

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Pairwise, then, is just the
case of two way independence.

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And what we saw was the
example that with k coin

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flips the events
odd and the outcomes

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of head or not on H1 through
Hk are k-way independent,

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but not k plus one
way independent.

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By the way, now
that we understand

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what k-way independence is,
mutual independence of n sets

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is simply n-way independence.

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But I just wanted to
close with the remark

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that checking with n events
are mutually independent

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means that you
actually have to check

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all the intersections
equaling the products

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of the individual events
in the intersection.

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So that there are two to
the n possible collections

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of subsets of A1
through An and you

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have to check for each of
them, that the intersection

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of those ones that you chose
is equal to the product

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of their probabilities.

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But of course, you don't need
to check the empty selection.

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And you don't need to
check the single [? set ?],

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so you just have to
check the 2 to the n

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equations corresponding
to all the subsets of size

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more than one.

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So it's 2 to the n minus n
plus 1 equations to check.

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So in general, it's
not going to be

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easy to verify mutual
independence by doing

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this kind of a calculation.

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And you usually
arrive at it really

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by assumption most of the time.