ALBERT MEYER: We've looked at independence for two events.
What about when we have a bunch of events?
Well, in that case we want to look
at the idea of mutual independence.
So let's check that out.
I'll say that if I have n different events,
I'll say that they're mutually independent, intuitively.
If the probability that one of them occurs
is unchanged by which other ones happen to have occurred.
So expressed in conditional probability, which
is the way to make it precise, what we're really saying
is that events A1 through An are mutually independent when
the probability of Ai is equal to the probability of Ai,
given the intersection of any of the other
As as long as i is not one of them.
So take A1, A2, or A1, A2, A3, and so on.
And A5 is going to be independent of all
of those other intersections.
If we shift over to the other definition of independence
that we used for two sets, in terms of products,
you could say that n sets are mutually independent when
the probability of the intersection of any bunch
of them is equal to the product of the individual probabilities
of the events in the intersection.
Let's look at an example of mutual independence.
Maybe the simplest one is the one
of independent coin flips, which by definition are independent.
So the idea is that I will flip a coin a bunch of times.
And I will let Hi be the event that the ith
time I flip I get a heads.
So if you think about what's going on,
what happens on the fifth flip has
nothing to do with what happens on the first, fourth or seventh
flip.
There's no causal relationship between the flips
before or after flip five.
Flip five is an isolated event by itself.
And the fact that there were a bunch of heads before
or there will be a bunch of heads
afterward doesn't have any impact on the probability
that the fifth flip comes up with a head.
At least that's what we believe and that's the way
that we would model them.
So what that means, for example, is
that the probability of a head on the fifth toss
is equal to the probability of a head on the fifth toss given
that the first toss was a head and the fourth toss was a head
and the seventh toss was not a head.
This is the complement of H7.
So that would just be an example of one
of the many different conditional equations that
hold when you have mutual independence.
Let's look at an example.
Suppose that I flip a fair coin twice.
Now, the previous definition didn't require fairness at all
in the coin flipping.
But now I'm going to need it.
So that means that heads and tails are equally likely.
And suppose I flip the coin twice.
Well, let H1 be as before, the event that a head comes up
on the first flip.
And H2, the event that a head comes up on the second flip.
And let O be the event that there were an odd number
of heads in the two flips.
Now, I claim that O is independent of whether or not
there's a head on the first flip.
That may seem a little weird because O
depends on both the first flip and the second flip.
It's whether or not there are an odd number of heads there,
but nevertheless, I claim that whether or not
there are an odd number of heads is
independent of whether or not the first toss was a head.
Let's just check it using the official definition.
First of all, O is the event HT TH.
If I write out Hs and Ts, a pair of them
for what the results of the first and second flips were,
you get an odd number of heads exactly when there's first
a head and then a tail or first a tail and then a head, which
means that the probability of O is exactly a half.
Because the other two outcomes are
TT and HH, which is when you have an even number of heads.
Now, O into section H1 is saying that you
have an odd number of heads and the first toss is a head.
The only outcome that fits that description
is HT, which means that-- and the probability of HT
is a quarter-- so the probability of O intersection
H1 is a quarter.
O into section H1 is just a peculiar way
of saying you got a head and then you got a tail.
So that means that the probability of O intersection
H1 is a quarter.
And of course, that's equal to the probability of O,
which we decided was a half, and the probability of H1,
which of course is a half, because we
said the coin was fair.
So I've verified the condition for the independence of O
and H1, and therefore, I'm done.
But the weird thing to notice now
is that if you look at O, H1, and H2, the three of them,
they are not mutually independent.
Because in fact, if you know any two of them
you can figure out what the third one was.
But just explicitly in terms of conditional probabilities,
the probability of there being an odd number of heads,
given that the first toss was a head
and the second toss was a head, is 0,
because once you know H1 and H2 you
know exactly how many heads there were.
There were two.
And that's not odd.
So the probability of odd given H1 intersection H2
is 0, which is not equal to the probability of odd by itself,
which was a half.
So the three of them are not independent.
They're not mutually independent,
even though any two of them are because O and H1 are.
And obviously O and H2 are by symmetry.
And H1 and H2 to are coin tosses,
and they're independent.
So that leads us to the general idea of k-way independence.
And an example would be if you flip a fair coin k times,
let Hi be whether or not there's a head on the ith flip.
And you let O, again, be whether or not
there are an odd number of heads.
And by the same argument, you can
verify that any set of k of these events
are mutually independent.
But if you give me all k plus 1, then they are not independent.
In fact, any k of them will determine the k plus first one.
But any k among themselves will be mutually independent.
So that's why this notion of how independent a bunch of sets are
comes up, and this is how to count it.
So in general, events A1 through an arbitrary set of events
are k-way independent if any k of them
are mutually independent.
Pairwise, then, is just the case of two way independence.
And what we saw was the example that with k coin
flips the events odd and the outcomes
of head or not on H1 through Hk are k-way independent,
but not k plus one way independent.
By the way, now that we understand
what k-way independence is, mutual independence of n sets
is simply n-way independence.
But I just wanted to close with the remark
that checking with n events are mutually independent
means that you actually have to check
all the intersections equaling the products
of the individual events in the intersection.
So that there are two to the n possible collections
of subsets of A1 through An and you
have to check for each of them, that the intersection
of those ones that you chose is equal to the product
of their probabilities.
But of course, you don't need to check the empty selection.
And you don't need to check the single [? set ?],
so you just have to check the 2 to the n
equations corresponding to all the subsets of size
more than one.
So it's 2 to the n minus n plus 1 equations to check.
So in general, it's not going to be
easy to verify mutual independence by doing
this kind of a calculation.
And you usually arrive at it really
by assumption most of the time.