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