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In this problem, we'll be
working with a object called
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random walk, where we have a
person on the line-- or a
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tight rope, according
to the problem.
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Let's start from the origin, and
each time step, it would
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randomly either go forward
or backward with certain
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probability.
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In our case, with probability
P, the person would go
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forward, and 1 minus
P going backwards.
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Now, the walk is random in the
following sense-- that the
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choice going forward or backward
in each step is
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random, and it's completely
independent
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from all past history.
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So let's look at the problem.
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It has three parts.
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In the first part, we'd like to
know what's the probability
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that after two steps the person
returns to the starting
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point, which in this
case is 0?
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Now, throughout this problem,
I'm going to be using the
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following notation.
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F indicates the action of going
forward and B indicates
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the action of going backwards.
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A sequence says F and B implies
the sample that the
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person first goes forward,
and then backwards.
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If I add another F, it will
mean, forward, backward,
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forward again.
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OK?
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So in order for the person to
go somewhere after two steps
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and return to the origin, the
following must happen.
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Either the person went forward
followed by backward, or
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backward followed by forward.
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And indeed, this event--
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namely, the union of these
two possibilities--
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defines the event of interest
in our case.
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And we'd like to know what's
the probability of A, which
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we'll break down into the
probability of forward,
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backward, backward, forward.
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Now, forward, backward and
backward, forward-- they are
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two completely different
outcomes.
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And we know that because they're
disjoint, this would
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just be the sum of the
two probabilities--
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plus probability of
backward/forward.
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Here's where the independence
will come in.
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When we try to compute the
probability of going forward
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and backward, because
the action--
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each step is completely
independent from the past, we
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know this is the same as saying,
in the first step, we
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have probability P of going
forward, in the next step,
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probability 1 minus P
of going backwards.
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We can do so-- namely, writing
the probability of forward,
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backward as a product of going
forward times the probability
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of going backwards, because
these actions are independent.
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And similarly, for the second
one, we have going backwards
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first, times going forward
the second time.
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Adding these two up, we have 2
times P times 1 minus P. And
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that will be the answer to the
first part of the problem.
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In the second part of the
problem, we're interested in
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the probability that after three
steps, the person ends
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up in position 1, or
one step forward
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compared to where he started.
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Now, the only possibilities here
are that among the three
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steps, exactly two steps are
forward, and one step is
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backwards, because otherwise
there's no way the person will
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end up in position 1.
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To do so, there, again, are
three possibilities in which
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we go forward, forward,
backward, or forward,
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backward, forward, or backward,
forward, forward.
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And that exhausts all the
possibilities that the person
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can end up in position
1 after three steps.
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And we'll define the collection
of all these
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outcomes as event C. The
probability of event C--
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same as before--
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is simply the sum of
the probability of
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each individual outcome.
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Now, based on the independence
assumption that we used
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before, each outcome here has
the same probability, which is
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equal to P squared times 1 minus
P. The P squared comes
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from the fact that two forward
steps are taken, and 1 minus
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P, the probability of that
one backwards step.
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And since there are three of
them, we multiply 3 in front,
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and that will give us
the probability.
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In the last part of the problem,
we're asked to
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compute that, conditional on
event C already took place,
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what is the probability that the
first step he took was a
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forward step?
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Without going into the details,
let's take a look at
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the C, in which we have three
elements, and only the first
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two elements correspond
to a forward step
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in the first step.
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So we can define event
D as simply
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the first two outcomes--
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forward, forward, backward, and
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forward, backward, forward.
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Now, the probability we're
interested in is simply
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probability of D conditional on
C. We'd write it out using
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the law of conditional
probability--
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D intersection C conditional
on C. Now, because D is a
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subset of C, we have probability
of D divided by
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the probability of C.
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Again, because all samples
here have the same
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probability, all we need to do
is to count the number of
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samples here, which is 2, and
divide by the number of
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samples here, which is 3.
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So we end up with 2 over 3.
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And that concludes
the problem.
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See you next time.
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