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PROFESSOR: So some of
the standard questions

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that we've examined
already about random graphs

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are the probability of getting
from one place to another,

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or the expected time to get
from one place to another.

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But a different kind
of question that

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comes up in a fundamental
way is the probability

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of being someplace.

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So let's examine that.

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Here is the graph with states
blue, orange, and green

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that we've seen before.

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And suppose that I
start at state B.

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And I ask, what's the
probability of being at each

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of these states after one step?

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So to start with, I'm
interested in p B, p O,

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and p G-- which is the
probability of being at state

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B, the probability
of being at state O,

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and the probability
of being at state

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G. The sum of the
probabilities is going to be 1.

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And initially when I tell
you that I'm at state B,

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it means the probability
of being at B is 1,

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and the other two is 0.

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And I'm interested in the
way that these probabilities

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update after one step.

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If p prime B is the
probability of being in state B

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after one step, and p
prime O is the probability

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of being in the orange state
one step later-- and likewise

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for green-- what are
these probabilities?

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Well it's easy to see just
reading off this graph

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that the only place you're
at is B. So the only way

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to get probability
of being somewhere

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is by following
an edge out of B.

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So the probability of being at
one step at the orange vertex

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is 1/4.

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And it's likewise 1/4 for
being at the green state.

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And it's 1/2 for staying
at the blue state.

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So what we can say is that
the updated probabilities

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of being at these
different states

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is 1/2, 1/4, and 1/4,
as we've just reasoned.

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OK.

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Let's keep going.

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Given that the probability
that I'm at the states blue,

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orange, and green are given by
this vector of probabilities,

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what's the distribution
after two steps?

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So let p double-prime B be the
probability of being at state B

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after two steps,
starting from B. Well,

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the way we can
figure that out is

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by using conditional
probabilities.

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Let's look at the
example of calculating

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the probability of being
in the orange state two

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steps after you've
started at the blue state.

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So here was the probabilities
of being at the different states

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after one step.

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How do I get to
the orange state?

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Well I can get to the orange
state from the blue state.

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And so the probability of
being in the orange state

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after two steps
is the probability

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of being at the blue state after
one step times the probability

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that I take this edge
to the orange state.

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That is, it's the probability
of going from B to O-- given

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that I'm at B-- times
the probability of being

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in B after one step.

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This then is the probability
of being in O after two steps.

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And likewise, another
component of the probability

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of being at O is that if
you're at O, and what's

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the probability of
going from O to O?

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And that is this 1/3
times the probability

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of being at O at
all, which is 1/4.

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And the final case, using again
the law of total probability,

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breaking it up into cases,
the third way that I

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can get to the orange
state on step two

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is by being at the green
state on step one following

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the green to O edge-- of
which there isn't any,

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so that's going
to be probability

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0-- times the probability of
being at the green state, which

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is 1/4, but it won't matter.

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So let's just fill
in these amounts

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looking at the first term.

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The probability of going
from B to O when you're at B

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is simply the
probability of this edge.

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It's 1/4.

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And likewise, the
probability of going from O

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to O given that you're
at O is the probability

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of this edge-- namely 1/3.

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So we can fill that term in.

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And finally the probability
of going from G to O

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is 0, given that you're
at G, because there

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isn't any vertex there.

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And then you fill in
those probabilities

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and do the arithmetic.

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You come out with
5/24 probability

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of being in the orange
state after two steps.

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Well the same calculation,
you can figure out

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what's the probability of
being at the blue state

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or the green step
after two steps.

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And there's the answer.

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There's a 50/50 chance of
being at the blue state

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after two steps, 5/24 as
we saw at the orange state,

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and the rest of it is
7/24 is the probability

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of being at the green state.

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OK.

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So what's going on in general?

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And we can explain how to do
these calculations by using

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a little bit of linear algebra.

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So let's define the
edge probability

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matrix of a random
walk graph is just

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the adjacency
matrix of the graph,

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except that instead of using
0's and 1's to indicate

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whether an edge is not
present or present,

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I'll use in the I, J
position of the matrix--

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the probability of the edge
that goes from I to J--

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if there is an edge-- and
0 if there isn't any edge.

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Let's look at an example.

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So here is the way we'd
fill it in abstractly

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for our three-state graph.

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It'll be a 3 by 3 matrix
with the probabilities

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of the successive edges in
the corresponding position.

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So this is the
position in the B,

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B coordinate is the probability
of the edge from B to B. The O,

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B coordinate, if you
think of the columns

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as labeled blue, orange,
green; and the rows as labeled

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blue, orange, green.

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And this is the orange,
blue coordinate.

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And it's the probability
of the edge from 0 to B.

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Let's fill in the first
row, which was-- this

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is just read directly
off the graph.

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It was the edges out of
B that went from B to B,

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from B to O, and
from B to green-- G.

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And it had those weights.

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And if I fill in
the rest of it, I

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get the edge
probability matrix for

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our simple three-state graph.

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And there it is.

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So this last one shows the fact
that there is a certain edge

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from green to blue.

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The only place you can
go from green is to blue,

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and you can't go to either
orange or green in one step.

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OK.

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So why are we bringing
up the matrix?

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Well if you looked at the
way we updated the state

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to go from the
one-step distribution

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to the two-step distribution,
it was really a matrix multiply.

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And in general, to
do an update, you're

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just going to do a vector
matrix multiplication.

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If you have the probabilities of
being in the successive states

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B, O, and G, and you do a vector
matrix multiplication using

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the probability
matrix of the graph,

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you get the updated
vector of distributions.

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And that's easy to check
just from the definitions,

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and from the definition
of vector times matrix,

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which I assume
you're familiar with.

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So now we can ask what's the
distribution after t steps,

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starting from some particular
given distribution-- say,

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starting at state B, or starting
at any possible distribution

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of probabilities to
the different states.

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And the way that we can
figure that out-- so I'm

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interested in other
words is the probability

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of being in O after t
steps G after t steps in B

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after t steps, say,
starting from state B.

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And what happens also as
t approaches infinity?

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And these are sort of
two basic questions

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that we're going to be asking.

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So first of all,
how do you calculate

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starting at a given
distribution p

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B, p O, p G where you're
going to be after t steps?

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Well, you're just
continually updating, which

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means multiplying by M t times.

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So the distribution
after t steps

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is gotten by taking the
initial distribution times

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the t-th power of M.

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Now this is actually already
useful computationally,

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because it means that since
you can compute a matrix

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power by successive
squarings, you actually

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only need about log of
t matrix multiplications

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in order to be
able to figure out

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what's the distribution
of probabilities

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after t steps of the graph.

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Then the crucial concept
that we want to examine--

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and we'll make a lot of use
of in the next video when

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we talk about a page
rank-- is the idea

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of a stationary distribution.

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So a stationary
distribution means

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that once you're in the
stationary distribution,

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it's stable.

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You're going to stay
in that distribution.

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You're not going to be
in any particular state,

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but you'll have a vector
of probabilities of being

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in the different states.

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And one step later, that
vector's not going to change.

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So what it means is that
the next-step distribution

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is the same as the
current distribution.

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What's the stationary
distribution here?

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Well, the way we're going
to have to calculate

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that is here's how you update.

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This is the result of the
vector matrix multiplication.

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But let's just spell
it out in terms

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of the conditional
probabilities.

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After one step, if the
original distribution

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is p B, p O, p G, then
the new probability

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of being in state
B, the only way

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you can get there is by
following the edge from B

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to B with probability 1/2.

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And that's times the
probability of being at B.

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And the other way
you can get to B

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is by being at the green state.

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And then one step later
you're certain to be at B. So

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that adds a contribution
of 1 times p G.

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And likewise for p-- the
updated probability of being

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at the orange state
and the green state.

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And what we want is that
these updated probabilities

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are the same as the ones
that I'm starting with.

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That's the definition
of stability.

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You update the
vector p B, p O, p G,

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and you get the same vector.

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That's what makes it stable.

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And of course, a
side constraint.

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Since you can always solve a
system of equations like this

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by letting all the p's be
0, which is degenerate,

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we add the constraint
that the sum

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of the probabilities of being
in the states has to be 1.

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Well if we solve that simple
3 by 3 system of equations,

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then it turns out that the
stable distribution is there's

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an 8/15 chance of
being in state B,

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a 3/15 chance of
being in state orange,

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and a 4/15 chance of
being in state green.

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And you should
check that yourself

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by asking what's the probability
of being in p B after one step

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given these probabilities?

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And I'm not going to
talk you through that.

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But just to verify and
imprint the idea of stability,

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that's one that's worth
stopping the video for a moment

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to check and do a little
arithmetic with a pencil

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and paper.

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OK.

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So in general, what
we're going to do

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is we're trying to find
the stationary distribution

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vector-- call it
s bar, for vector.

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And we get this by solving
the vector matrix equation--

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that the distribution vector
times the edge probability

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matrix is equal to that
same distribution vector.

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We want to solve this
system of equations.

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If there are n
states, then this is

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an n by n system of equations,
with an additional constraint

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that we want the norm of
the stable vector to be 1,

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because that's to avoid
the degenerate 0 solution.

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Well there are some problems
with stationary distributions

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that we want to think about.

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First of all, what happens in
this example where you have

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just two states, and
the probability of being

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in the first state at 1
and the second state is 0?

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Well if you update that
state, what happens

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is you just go to the second
state with probability 1.

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And you can keep doing that.

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And there may be a
stable distribution here,

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but this particular pattern
doesn't converge to it.

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As you go through time,
at every other step

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you're at state 1, and every
other step you're at state 0.

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But you never get to
a stable distribution

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where step after step you
are at equal probability

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of being at these two places.

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I'm assuming here that
this is a certain edge,

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and that's a certain edge.

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It has to be.

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There's only one edge out.

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So a stable distribution
would be 1/2, 1/2.

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But this thing doesn't
converge to it.

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OK.

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Here's a slightly more
complicated example,

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where again assume that all
the edges are equally likely.

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There's exactly two edges
out of each of these vertices

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so that each edge
has weight 1/2.

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And the problem
with this graph is

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that when you ask what's
the stable distribution

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and, well, if you look
at it, if you assume

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that the probability of
being in the middle is 0,

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and the two places that you
get stuck at have probability p

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and 1 minus p,
then that's stable,

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because once you're at this
state with probability p

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you're following the one certain
edge that goes back around

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to this vertex.

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And therefore
there's probability

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of p of being there
one step later,

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and likewise probability
q of one step later.

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So the split between p and
q is a stable distribution

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for this thing, with
probability 0 and 0 there.

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And of course p and q can be any
real numbers between 0 and 1.

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So there's actually
an uncountable number

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of stable distributions
for this graph.

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Problem here is it's
not strongly connected.

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00:14:16,310 --> 00:14:20,540
And that turns out to be
a sufficient condition

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that it's got a single
stable distribution whenever

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it's strongly connected.

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So in general we can
ask the question,

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is there always a
stationary distribution

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for any random graph? well,
if the graph is finite, yes,

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there's guaranteed to be
a stationary distribution.

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00:14:42,230 --> 00:14:43,530
But is it unique?

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00:14:43,530 --> 00:14:45,840
Well sometimes, sometimes not.

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If the graph is strongly
connected, it will be unique.

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But we've seen examples
in the previous slide

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where it's not unique.

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In fact, it could
be uncountably many.

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00:14:55,670 --> 00:14:59,550
And another crucial question
is, does a random walk

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approach the stable distribution
no matter how you start?

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00:15:04,260 --> 00:15:06,130
And that first
example was one where

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00:15:06,130 --> 00:15:08,960
you went between the first
state and the second state

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00:15:08,960 --> 00:15:09,990
and oscillated.

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00:15:09,990 --> 00:15:12,690
And it never converged on
the stable distribution

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00:15:12,690 --> 00:15:14,160
of 1/2 and 1/2.

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00:15:14,160 --> 00:15:17,000
In general, it's nice when
you can say that no matter

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00:15:17,000 --> 00:15:20,310
how you start, after a
while things stabilize,

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00:15:20,310 --> 00:15:23,580
and you wind up at the
unique stable distribution.

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00:15:23,580 --> 00:15:30,040
So sometimes it'll be the case
that every initial distribution

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00:15:30,040 --> 00:15:33,040
will eventually converge
on the stable one

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00:15:33,040 --> 00:15:34,420
or the stationary one.

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00:15:34,420 --> 00:15:35,810
Sometimes not.

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00:15:35,810 --> 00:15:37,470
And then another
crucial question

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00:15:37,470 --> 00:15:40,920
will be, how quickly does
this convergence happen?

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00:15:40,920 --> 00:15:44,780
If we start off at some
arbitrary probability

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00:15:44,780 --> 00:15:46,520
distribution, or some
particular state,

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00:15:46,520 --> 00:15:49,470
how long does it take
before by and large

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00:15:49,470 --> 00:15:52,130
the probabilities that were
in the different states

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00:15:52,130 --> 00:15:55,060
has become pretty stationary?

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00:15:55,060 --> 00:15:57,620
And the rate at which
that happens again

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varies depending on the graph.