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As an introduction to the main
topic of this lecture

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sequence, let us go through a
simple example and on the way,

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review what we have
learned so far.

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The example that we're going
to consider involves three

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tosses of a biased coin.

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It's a coin that results in
heads with probability p.

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We're going to make this
a little more precise.

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And the coin is biased in the
sense that this number p is

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not necessarily the
same as one half.

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We represent this particular
probabilistic experiment in

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terms a tree that shows us the
different stages of the

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experiment.

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Each particular branch
corresponds to a sequence of

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possible results in the
different stages, and the

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leaves of this tree correspond
to the possible outcomes.

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The branches of this tree are
annotated by certain numbers,

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and these numbers are to be
interpreted appropriately as

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probabilities or conditional
probabilities.

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So for example, this number
here is interpreted as the

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probability of heads in the
first toss, an event that we

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denote as H1.

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This number here is to be
interpreted as a conditional

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probability.

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It's the conditional probability
of obtaining heads

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in the second toss given
that the first

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toss resulted in heads.

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And finally, this number here
is to be interpreted as the

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conditional probability of heads
in the third toss, given

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that the first toss resulted in
heads and the second toss

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also resulted in heads.

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Let us now continue with
some calculations.

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First, we're going to practice
the multiplication rule, which

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allows us to calculate
the probability

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of a certain outcome.

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In this case, the outcome of
interest is tails followed by

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heads followed by tails.

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So we're talking about this
particular outcome here.

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According to the multiplication
rule, to find

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the probability of a particular
final outcome, we

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multiply probabilities and
conditional probabilities

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along the path that leads to
this particular outcome.

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So in this case, it's (1 minus
p) times p times (1 minus p).

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Let us now calculate
the probability

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of a certain event.

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The event of interest is the
event that we obtain exactly

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one head in the three tosses.

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This is an event that can
happen in multiple ways.

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Here is one possibility where
we have a single head.

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Here's another possibility.

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And here's a third one.

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These are the three possible
ways that we can have exactly

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one head, depending
on where exactly

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that single head appears.

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Is it in the first toss, in the
second, or in the third.

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To find the total probability of
this event, we need to add

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the probability of the different
outcomes that

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correspond to this event.

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The probability of this outcome
is p times (1 minus p)

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squared, the probability of
this outcome is what we

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calculated.

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It's, again, p times (1
minus p) squared.

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And the probability of the third
one is also p times 1

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minus p squared.

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So the answer is 3p times
(1 minus p) squared.

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Notice that each one of the 3
different ways that this event

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can happen have the
same probability.

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So these 3 outcomes are
equally likely.

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Finally, let us calculate a
conditional probability.

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And this is essentially
the Bayes rule.

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Suppose that we were told that
there was exactly one head.

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So in particular, the blue
event has occurred.

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And we're interested in the
probability that the first

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toss is heads, which
corresponds

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to this event here.

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These are all the outcomes
in which the first

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toss is equal to heads.

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So given that the blue event
happened, what is the

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probability that the green
event happens?

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You can guess the answer,
that it should be 1/3.

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Why is that?

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Each one of these blue
outcomes has the same

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probability.

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So when you condition on the
blue outcome having happened,

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the conditional probability
of each one of

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these should be 1/3.

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So given that the blue outcome
happened, there's probability

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1/3 that this particular
one has happened.

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And this is the only one that
makes the green event happen.

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But let us see if we
can derive this

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answer in a formal manner.

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Let's see if we're
going to get 1/3.

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We use the definition of
conditional probabilities.

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The conditional probability is
the ratio, first, of the

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probability that both events
happen, divided by the

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probability of the conditioning
event, which is

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the probability of 1 head.

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Now, the probability of
both events happening.

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That we have exactly one head
and the first toss is heads.

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This is the intersection of the
blue event and the green

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event which can happen only in
this particular outcome,

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namely the sequence heads,
tails, tails.

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And has probability p times
(1 minus p) squared.

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The denominator is something
that we have already

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calculated.

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It's 3p times (1 minus
p) squared.

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And so the final answer is
1/3 as we had guessed.

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Let me now make a few comments
about this particular example.

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This particular example is
pretty special in the

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following respect.

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We have that of the probability
of H2, heads in

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the second toss, given that the
first one was heads, is

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equal to p.

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And the same is true for the
conditional probability of

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heads in the second toss given
that the first one was tails.

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In other words, our beliefs
about what may happen in the

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second toss remain the same.

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There's a probability, p, of
obtaining heads no matter what

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happened in the first toss.

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Telling you the result of the
first toss doesn't change your

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beliefs about what may happen,
and with what probability, in

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the second toss.

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And if you were to calculate the
unconditional probability

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of heads in the second toss,
what you would get using the

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total probability theorem
would be the following.

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It's probability of heads in
the first toss times the

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probability of heads in the
second, given heads in the

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first, plus the probability of
tails in the first toss times

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the probability of heads in the
second toss, given tails

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in the first.

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And if you do the algebra,
this turns out to

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be equal to p again.

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So the unconditional probability
of heads in the

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second toss turns out to be the
same as the conditional

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probabilities.

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Again, knowing what happened
in the first toss doesn't

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change your beliefs about the
second toss, which were

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associated with this particular
probability, p.

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So what we're going to do next
is to generalize this special

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situation by giving a definition
of independence of

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events, and then discuss
various properties and

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concepts associated
with independence.