WEBVTT
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In this segment, we
go through an example
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to get some practice with
Poisson process calculations.
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The example is as follows.
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You go fishing and fish
are caught according
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to a Poisson process with
an arrival rate-- that
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is the rate at which fish are
caught-- of 0.6 fish per hour.
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You will fish for two
hours no matter what.
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And if during
those two hours you
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have caught at least
one fish, then you stop.
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As in this scenario, in which
you have caught three fish
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during the first two
hours, and then you stop.
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Otherwise, you will continue
until you catch one fish.
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And at that time, you will stop.
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We will answer a few questions
and we will write down
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the answers to these questions
in terms of the notation
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that we have introduced.
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And here, for reference,
we have all of the formulas
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that we have developed so far.
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The first question is,
what is the probability
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that you get to fish
for more than two hours?
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Well, you get to fish for more
than two hours if and only
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if you didn't catch any fish
during the first two hours.
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So this is the
probability of catching
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zero fish in the
first two hours.
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And we can use this formula.
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Substitute k equal
to 0, tau equal to 2,
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lambda equal to 0.6.
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Plug in the numbers and
obtain a numerical answer.
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We will not bother [with]
the numerical answers,
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we will just be writing
down the expressions that
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give us the answers.
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Now, in this question we could
have a different approach.
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You will fish for
more than two hours
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if and only if there are no
arrivals during the first two
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hours, which means
that the first arrival
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in the Poisson process of
fishing happens after time 2.
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So we're talking about the event
that the first arrival, T1,
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is bigger than 2.
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And we're looking
into the probability
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of this event, which is
the integral of the density
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of the first arrival time.
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And the integral is
taken over the range
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of values of interest--
larger than 2.
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That is, from 2 to infinity.
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We know that this
density is exponential,
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so we can write down this
integral and evaluate it.
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So notice the difference
between these two approaches.
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Here we think in terms of
the random variables that
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correspond to the
number of arrivals
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during a certain time interval.
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Here we're reasoning in
terms of inter-arrival times.
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And more generally, for
Poisson process problems,
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an event of interest
sometimes can
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be expressed in terms
of number of arrivals.
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Or sometimes it can
be expressed in terms
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of arrival and
inter-arrival times.
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Or sometimes both
approaches are possible.
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But usually one of
the two approaches
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will be simpler than the other.
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For example, here we
already have a formula,
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whereas here we would need
to evaluate an integral.
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Here is now our next question,
which is a little bit more
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complicated.
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What is the probability
that you get
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to fish for more than
two hours, and also you
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get to fish for less
than five hours?
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This is the scenario,
again, that you
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fish for more than two hours,
which means that no fish were
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caught during the
first two hours.
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And then you continue fishing.
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And by time 5 you have
already caught your fish
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and you have left.
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Now, it is convenient of
thinking about the Poisson
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process as follows.
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Think about the Poisson
process of catching fish
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at the rate of 0.6 per
hour as going on forever.
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So there's a fisherman
who fishes forever,
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except that this fisherman
at either this time,
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under this scenario, or at
that time, under this scenario,
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raises a flag and
says, OK, this is
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the time I should be stopping.
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So even though the fisherman
will stop fishing at this time,
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we can imagine the Poisson
process keeps going on.
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With this way of
thinking, the fact
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that you stopped fishing
before time 5 is the event
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and that the number
of fish caught,
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or the number of Poisson
arrivals during the interval
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from 2 to 5 is at least equal
to 1, larger than or equal to 1.
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So the event of interest
consists of the intersection
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of two events, zero fish caught
during the first two hours--
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which has this probability--
and at least one
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arrival in the Poisson process
between times 2 and 5--
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this is a time
interval of length 3.
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And having at least
one arrival is
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1 minus the probability
of 0 arrivals
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in a time interval of length 3.
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Notice that I'm multiplying
those two probabilities here.
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Why am I doing this?
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In a Poisson process,
whatever happens
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in disjoint time intervals
are independent events.
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So an event having to
do with this interval--
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the event that no
fish were caught--
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and the event that has to do
with this interval-- at least
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one fish caught, at least
one arrival during this time
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interval-- these two
events are independent.
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And this is why we can
multiply their probabilities.
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Now, let us think of the
alternative approach,
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a different way of
describing this event using
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arrival and inter-arrival times.
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What is this event?
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This is the event that
no arrival happened
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until this time, but an
arrival happened before time 5.
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So this is the event
that the first arrival
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happens after time
2, but before time 5.
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And we're looking
at the probability
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of this event, which we
can find by integrating
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the PDF of the first
arrival time from 2 until 5.
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The next question is,
what is the probability
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that we catch at least two fish?
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Under this scenario, we
catch one fish and we stop.
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Therefore, this scenario
must have materialized.
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The event of catching
at least two fish
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is the scenario that
from time from 0 until 2,
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we caught at least two fish.
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So we're talking
about the probability
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of catching at least
two fish, which
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is the probability of
catching k fish during a time
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interval of length 2, where
k can be anything from 2
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to infinity.
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So this is the probability
that the number
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of fish caught during
these two time units
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is at least equal to 2.
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And an alternative way of
writing this expression
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so that we do not have to
evaluate an infinite sum,
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it is 1 minus the probability
of catching 0 fish,
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and minus the probability
of catching 1 fish.
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If we were to write
this in terms of arrival
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and inter-arrival times, we
will catch at least two fish
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if and only if by the time we
stop-- which will be time 2--
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we've had 2 arrivals, which
means that the second arrival
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time happened before time 2.
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So we're looking
into the probability
00:08:02.870 --> 00:08:08.010
that the second arrival time
happened before time 2, which
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is the integral from
0 to 2 of the density
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of the second arrival time.
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We have a formula
for this density
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given by the Erlang PDF.
00:08:20.510 --> 00:08:24.090
So we could take this
expression, plug it in here,
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evaluate the integral,
and obtain the same answer
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as we would have obtained here.
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Clearly, in this case as
well, this first approach
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would be a simpler one because
these probabilities are already
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available to us.
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The next question
is the following.
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Suppose that you already
fished for three hours.
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This is something
that can only happen
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under the second scenario.
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So no fish have being
caught until time 2.
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You continue.
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No fish were caught
until time 3.
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And given that this
event has occurred,
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what is the expected value
of the future fishing time?
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The expected value until a fish
gets caught for the first time?
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Well, the Poisson process
starts fresh at time 3,
00:09:13.670 --> 00:09:16.090
no matter what
happened in the past,
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no matter what information
we're given in the past.
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Now you're sitting at time 3
and looking into the future.
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You have a fresh
starting Poisson process,
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as if this was the initial time.
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Starting from this time, the
time until the first arrival
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is going to have an exponential
distribution with parameter
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lambda, and the expected
value of this distribution
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is equal to 1 over lambda.
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Finally, let's look at a
different type of question.
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What is the expected
value of the total time
00:09:51.010 --> 00:09:53.510
that you get to fish?
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Let us introduce
here some notation.
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Let us call the total
fishing time capital F.
00:10:01.070 --> 00:10:04.900
And the total fishing time
consists of two pieces.
00:10:04.900 --> 00:10:09.500
There's a time until
time 2 that you
00:10:09.500 --> 00:10:13.300
will be fishing no matter what.
00:10:13.300 --> 00:10:16.660
And then there's going to
be the remaining fishing
00:10:16.660 --> 00:10:18.610
time after time 2.
00:10:22.210 --> 00:10:24.880
So now let's look
at the expectation
00:10:24.880 --> 00:10:27.870
of the remaining fishing
time after time 2.
00:10:27.870 --> 00:10:29.480
Here there are two scenarios.
00:10:29.480 --> 00:10:31.880
In the first scenario, you stop.
00:10:31.880 --> 00:10:34.440
In the second
scenario, you continue.
00:10:34.440 --> 00:10:36.830
And we will take into
account these two scenarios
00:10:36.830 --> 00:10:40.780
by using the total
expectation theorem.
00:10:40.780 --> 00:10:45.600
The first scenario happens
with some probability.
00:10:45.600 --> 00:10:49.580
This is the probability that
you stop fishing at time 2.
00:10:49.580 --> 00:10:53.620
And in that case, your
remaining fishing time
00:10:53.620 --> 00:10:56.980
is going to be equal
to 0 because you do not
00:10:56.980 --> 00:10:59.510
continue under that scenario.
00:10:59.510 --> 00:11:01.710
But there's the other
scenario that you
00:11:01.710 --> 00:11:06.870
get to fish for more
than 2 time units.
00:11:06.870 --> 00:11:10.840
And then we multiply
this term with
00:11:10.840 --> 00:11:14.670
the conditional expectation
of the remaining fishing time,
00:11:14.670 --> 00:11:18.060
given that you got to fish
for more than 2 times units.
00:11:21.540 --> 00:11:22.620
So what is this term?
00:11:22.620 --> 00:11:25.510
The probability that you get
to fish for more than 2 time
00:11:25.510 --> 00:11:28.610
units, this is the
probability that no fish
00:11:28.610 --> 00:11:31.630
were caught during
the first time units.
00:11:31.630 --> 00:11:34.700
This is the probability
of the second scenario.
00:11:34.700 --> 00:11:38.660
And then we have this
conditional expectation.
00:11:38.660 --> 00:11:42.100
Given that you
didn't catch anything
00:11:42.100 --> 00:11:45.750
and, therefore, you will
be continuing fishing,
00:11:45.750 --> 00:11:51.090
what is the expected amount
of time-- F minus 2--
00:11:51.090 --> 00:11:53.520
that you will be fishing?
00:11:53.520 --> 00:11:56.890
Now the Poisson process
starts fresh at time 2.
00:11:56.890 --> 00:11:58.700
So looking into
the future, we're
00:11:58.700 --> 00:12:00.650
faced with a Poisson
process and we're
00:12:00.650 --> 00:12:04.550
asking for the expected time
until the first arrival.
00:12:04.550 --> 00:12:11.510
And this time has an expected
value equal to 1 over lambda.
00:12:11.510 --> 00:12:14.710
Our last question is
of a similar type.
00:12:14.710 --> 00:12:18.650
What is the expected number
of fish you're going to catch?
00:12:18.650 --> 00:12:23.260
Once more, we will break
this into two pieces.
00:12:23.260 --> 00:12:26.510
Number of fish caught
during the first two hours,
00:12:26.510 --> 00:12:31.780
and number of fish caught
during the remaining hours.
00:12:31.780 --> 00:12:35.480
During the first two
hours, the expected number
00:12:35.480 --> 00:12:39.650
of fish that you catch
is given by this formula.
00:12:39.650 --> 00:12:46.650
It is equal to lambda tau--
and in our case lambda is 0.6
00:12:46.650 --> 00:12:52.870
and tau is equal to 2-- plus
the expected number of fish
00:12:52.870 --> 00:12:56.150
that are caught afterwards.
00:12:56.150 --> 00:12:58.060
What is this expected value?
00:12:58.060 --> 00:13:01.520
Again, we think in terms of
the total expectation theorem.
00:13:01.520 --> 00:13:05.000
There's one scenario that
has a certain probability,
00:13:05.000 --> 00:13:08.390
and under that scenario
you do not catch any fish.
00:13:08.390 --> 00:13:11.070
So that doesn't give
us any contribution.
00:13:11.070 --> 00:13:13.840
Then there is this
scenario, the second one,
00:13:13.840 --> 00:13:17.220
that has this
probability of occurring.
00:13:17.220 --> 00:13:19.470
The probability of
catching no fish here,
00:13:19.470 --> 00:13:22.280
so that the second
scenario materializes.
00:13:22.280 --> 00:13:24.720
And if that second
scenario materializes,
00:13:24.720 --> 00:13:28.480
how many fish are you going
to catch after time 2?
00:13:28.480 --> 00:13:31.045
It's going to be one
fish with certainty.
00:13:33.820 --> 00:13:36.780
And this is the final
answer to this question.
00:13:36.780 --> 00:13:40.110
Notice that in answering
both of these questions
00:13:40.110 --> 00:13:44.440
we used the divide and
conquer strategy twice.
00:13:44.440 --> 00:13:48.830
We first divided the time
horizon into two pieces
00:13:48.830 --> 00:13:52.850
and dealt separately with
each one of the pieces.
00:13:52.850 --> 00:13:56.660
And then in order to deal
with this second piece--
00:13:56.660 --> 00:14:00.740
the time after time
2-- we used divide
00:14:00.740 --> 00:14:03.820
and conquer into the
two different scenarios
00:14:03.820 --> 00:14:07.400
and using the total
expectation theorem.