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00:00:00,650 --> 00:00:03,320
In this segment, we
go through an example

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00:00:03,320 --> 00:00:06,690
to get some practice with
Poisson process calculations.

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00:00:06,690 --> 00:00:08,420
The example is as follows.

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00:00:08,420 --> 00:00:11,610
You go fishing and fish
are caught according

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00:00:11,610 --> 00:00:15,020
to a Poisson process with
an arrival rate-- that

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00:00:15,020 --> 00:00:20,970
is the rate at which fish are
caught-- of 0.6 fish per hour.

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00:00:20,970 --> 00:00:24,530
You will fish for two
hours no matter what.

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00:00:24,530 --> 00:00:26,140
And if during
those two hours you

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00:00:26,140 --> 00:00:28,940
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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00:00:32,280 --> 00:00:35,380
during the first two
hours, and then you stop.

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00:00:35,380 --> 00:00:40,640
Otherwise, you will continue
until you catch one fish.

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00:00:40,640 --> 00:00:44,280
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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00:00:47,580 --> 00:00:51,100
the answers to these questions
in terms of the notation

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00:00:51,100 --> 00:00:52,850
that we have introduced.

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00:00:52,850 --> 00:00:55,580
And here, for reference,
we have all of the formulas

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00:00:55,580 --> 00:00:58,570
that we have developed so far.

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00:00:58,570 --> 00:01:01,240
The first question is,
what is the probability

20
00:01:01,240 --> 00:01:04,269
that you get to fish
for more than two hours?

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00:01:04,269 --> 00:01:07,710
Well, you get to fish for more
than two hours if and only

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00:01:07,710 --> 00:01:12,490
if you didn't catch any fish
during the first two hours.

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00:01:12,490 --> 00:01:15,289
So this is the
probability of catching

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00:01:15,289 --> 00:01:19,039
zero fish in the
first two hours.

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00:01:19,039 --> 00:01:20,740
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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00:01:32,229 --> 00:01:35,320
we will just be writing
down the expressions that

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give us the answers.

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00:01:37,860 --> 00:01:43,470
Now, in this question we could
have a different approach.

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00:01:43,470 --> 00:01:46,020
You will fish for
more than two hours

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00:01:46,020 --> 00:01:49,350
if and only if there are no
arrivals during the first two

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00:01:49,350 --> 00:01:52,190
hours, which means
that the first arrival

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00:01:52,190 --> 00:01:57,229
in the Poisson process of
fishing happens after time 2.

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00:01:57,229 --> 00:02:02,080
So we're talking about the event
that the first arrival, T1,

38
00:02:02,080 --> 00:02:04,440
is bigger than 2.

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00:02:04,440 --> 00:02:06,320
And we're looking
into the probability

40
00:02:06,320 --> 00:02:12,170
of this event, which is
the integral of the density

41
00:02:12,170 --> 00:02:15,040
of the first arrival time.

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00:02:15,040 --> 00:02:17,740
And the integral is
taken over the range

43
00:02:17,740 --> 00:02:20,030
of values of interest--
larger than 2.

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00:02:20,030 --> 00:02:23,329
That is, from 2 to infinity.

45
00:02:23,329 --> 00:02:25,900
We know that this
density is exponential,

46
00:02:25,900 --> 00:02:29,840
so we can write down this
integral and evaluate it.

47
00:02:29,840 --> 00:02:33,090
So notice the difference
between these two approaches.

48
00:02:33,090 --> 00:02:36,490
Here we think in terms of
the random variables that

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00:02:36,490 --> 00:02:39,360
correspond to the
number of arrivals

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00:02:39,360 --> 00:02:41,670
during a certain time interval.

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00:02:41,670 --> 00:02:45,560
Here we're reasoning in
terms of inter-arrival times.

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00:02:45,560 --> 00:02:48,700
And more generally, for
Poisson process problems,

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00:02:48,700 --> 00:02:50,760
an event of interest
sometimes can

54
00:02:50,760 --> 00:02:54,720
be expressed in terms
of number of arrivals.

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00:02:54,720 --> 00:02:57,710
Or sometimes it can
be expressed in terms

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00:02:57,710 --> 00:03:00,590
of arrival and
inter-arrival times.

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00:03:00,590 --> 00:03:03,520
Or sometimes both
approaches are possible.

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00:03:03,520 --> 00:03:05,850
But usually one of
the two approaches

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00:03:05,850 --> 00:03:07,380
will be simpler than the other.

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00:03:07,380 --> 00:03:10,130
For example, here we
already have a formula,

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00:03:10,130 --> 00:03:13,520
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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00:03:18,570 --> 00:03:20,079
What is the probability
that you get

65
00:03:20,079 --> 00:03:24,270
to fish for more than
two hours, and also you

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00:03:24,270 --> 00:03:29,329
get to fish for less
than five hours?

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00:03:29,329 --> 00:03:32,360
This is the scenario,
again, that you

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00:03:32,360 --> 00:03:35,460
fish for more than two hours,
which means that no fish were

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00:03:35,460 --> 00:03:37,500
caught during the
first two hours.

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00:03:37,500 --> 00:03:39,470
And then you continue fishing.

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00:03:39,470 --> 00:03:44,310
And by time 5 you have
already caught your fish

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00:03:44,310 --> 00:03:46,180
and you have left.

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00:03:46,180 --> 00:03:49,440
Now, it is convenient of
thinking about the Poisson

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00:03:49,440 --> 00:03:51,460
process as follows.

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00:03:51,460 --> 00:03:54,130
Think about the Poisson
process of catching fish

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00:03:54,130 --> 00:03:57,950
at the rate of 0.6 per
hour as going on forever.

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00:03:57,950 --> 00:04:02,490
So there's a fisherman
who fishes forever,

78
00:04:02,490 --> 00:04:07,020
except that this fisherman
at either this time,

79
00:04:07,020 --> 00:04:11,490
under this scenario, or at
that time, under this scenario,

80
00:04:11,490 --> 00:04:14,520
raises a flag and
says, OK, this is

81
00:04:14,520 --> 00:04:17,600
the time I should be stopping.

82
00:04:17,600 --> 00:04:21,600
So even though the fisherman
will stop fishing at this time,

83
00:04:21,600 --> 00:04:27,860
we can imagine the Poisson
process keeps going on.

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00:04:27,860 --> 00:04:30,490
With this way of
thinking, the fact

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00:04:30,490 --> 00:04:35,460
that you stopped fishing
before time 5 is the event

86
00:04:35,460 --> 00:04:38,570
and that the number
of fish caught,

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00:04:38,570 --> 00:04:42,100
or the number of Poisson
arrivals during the interval

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00:04:42,100 --> 00:04:48,560
from 2 to 5 is at least equal
to 1, larger than or equal to 1.

89
00:04:48,560 --> 00:04:51,310
So the event of interest
consists of the intersection

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00:04:51,310 --> 00:04:56,440
of two events, zero fish caught
during the first two hours--

91
00:04:56,440 --> 00:05:02,880
which has this probability--
and at least one

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00:05:02,880 --> 00:05:07,020
arrival in the Poisson process
between times 2 and 5--

93
00:05:07,020 --> 00:05:10,870
this is a time
interval of length 3.

94
00:05:10,870 --> 00:05:13,280
And having at least
one arrival is

95
00:05:13,280 --> 00:05:17,450
1 minus the probability
of 0 arrivals

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00:05:17,450 --> 00:05:20,720
in a time interval of length 3.

97
00:05:20,720 --> 00:05:24,450
Notice that I'm multiplying
those two probabilities here.

98
00:05:24,450 --> 00:05:26,340
Why am I doing this?

99
00:05:26,340 --> 00:05:28,740
In a Poisson process,
whatever happens

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00:05:28,740 --> 00:05:33,230
in disjoint time intervals
are independent events.

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00:05:33,230 --> 00:05:36,010
So an event having to
do with this interval--

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00:05:36,010 --> 00:05:38,290
the event that no
fish were caught--

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00:05:38,290 --> 00:05:41,200
and the event that has to do
with this interval-- at least

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00:05:41,200 --> 00:05:44,540
one fish caught, at least
one arrival during this time

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00:05:44,540 --> 00:05:47,020
interval-- these two
events are independent.

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00:05:47,020 --> 00:05:51,040
And this is why we can
multiply their probabilities.

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00:05:51,040 --> 00:05:53,370
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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00:05:59,710 --> 00:06:02,140
This is the event that
no arrival happened

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00:06:02,140 --> 00:06:06,920
until this time, but an
arrival happened before time 5.

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00:06:06,920 --> 00:06:11,580
So this is the event
that the first arrival

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00:06:11,580 --> 00:06:17,020
happens after time
2, but before time 5.

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00:06:17,020 --> 00:06:18,810
And we're looking
at the probability

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00:06:18,810 --> 00:06:22,440
of this event, which we
can find by integrating

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00:06:22,440 --> 00:06:29,910
the PDF of the first
arrival time from 2 until 5.

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00:06:29,910 --> 00:06:32,040
The next question is,
what is the probability

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00:06:32,040 --> 00:06:34,425
that we catch at least two fish?

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00:06:34,425 --> 00:06:38,770
Under this scenario, we
catch one fish and we stop.

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00:06:38,770 --> 00:06:43,490
Therefore, this scenario
must have materialized.

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00:06:43,490 --> 00:06:46,430
The event of catching
at least two fish

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00:06:46,430 --> 00:06:51,170
is the scenario that
from time from 0 until 2,

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we caught at least two fish.

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00:06:54,170 --> 00:06:56,020
So we're talking
about the probability

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00:06:56,020 --> 00:06:59,090
of catching at least
two fish, which

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00:06:59,090 --> 00:07:04,540
is the probability of
catching k fish during a time

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00:07:04,540 --> 00:07:11,441
interval of length 2, where
k can be anything from 2

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00:07:11,441 --> 00:07:11,940
to infinity.

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00:07:15,030 --> 00:07:17,100
So this is the probability
that the number

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00:07:17,100 --> 00:07:20,800
of fish caught during
these two time units

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00:07:20,800 --> 00:07:22,830
is at least equal to 2.

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00:07:22,830 --> 00:07:25,900
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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00:07:33,060 --> 00:07:36,385
and minus the probability
of catching 1 fish.

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00:07:39,060 --> 00:07:41,790
If we were to write
this in terms of arrival

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00:07:41,790 --> 00:07:47,150
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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00:07:56,760 --> 00:07:59,990
time happened before time 2.

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00:07:59,990 --> 00:08:02,870
So we're looking
into the probability

143
00:08:02,870 --> 00:08:08,010
that the second arrival time
happened before time 2, which

144
00:08:08,010 --> 00:08:11,660
is the integral from
0 to 2 of the density

145
00:08:11,660 --> 00:08:15,130
of the second arrival time.

146
00:08:15,130 --> 00:08:17,890
We have a formula
for this density

147
00:08:17,890 --> 00:08:20,510
given by the Erlang PDF.

148
00:08:20,510 --> 00:08:24,090
So we could take this
expression, plug it in here,

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00:08:24,090 --> 00:08:27,720
evaluate the integral,
and obtain the same answer

150
00:08:27,720 --> 00:08:29,700
as we would have obtained here.

151
00:08:29,700 --> 00:08:32,750
Clearly, in this case as
well, this first approach

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00:08:32,750 --> 00:08:36,100
would be a simpler one because
these probabilities are already

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00:08:36,100 --> 00:08:39,280
available to us.

154
00:08:39,280 --> 00:08:41,010
The next question
is the following.

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00:08:41,010 --> 00:08:44,080
Suppose that you already
fished for three hours.

156
00:08:44,080 --> 00:08:46,620
This is something
that can only happen

157
00:08:46,620 --> 00:08:49,150
under the second scenario.

158
00:08:49,150 --> 00:08:51,860
So no fish have being
caught until time 2.

159
00:08:51,860 --> 00:08:52,650
You continue.

160
00:08:52,650 --> 00:08:55,380
No fish were caught
until time 3.

161
00:08:55,380 --> 00:08:58,460
And given that this
event has occurred,

162
00:08:58,460 --> 00:09:03,070
what is the expected value
of the future fishing time?

163
00:09:03,070 --> 00:09:08,710
The expected value until a fish
gets caught for the first time?

164
00:09:08,710 --> 00:09:13,670
Well, the Poisson process
starts fresh at time 3,

165
00:09:13,670 --> 00:09:16,090
no matter what
happened in the past,

166
00:09:16,090 --> 00:09:18,660
no matter what information
we're given in the past.

167
00:09:18,660 --> 00:09:21,870
Now you're sitting at time 3
and looking into the future.

168
00:09:21,870 --> 00:09:25,520
You have a fresh
starting Poisson process,

169
00:09:25,520 --> 00:09:28,730
as if this was the initial time.

170
00:09:28,730 --> 00:09:32,870
Starting from this time, the
time until the first arrival

171
00:09:32,870 --> 00:09:36,550
is going to have an exponential
distribution with parameter

172
00:09:36,550 --> 00:09:39,770
lambda, and the expected
value of this distribution

173
00:09:39,770 --> 00:09:44,270
is equal to 1 over lambda.

174
00:09:44,270 --> 00:09:47,780
Finally, let's look at a
different type of question.

175
00:09:47,780 --> 00:09:51,010
What is the expected
value of the total time

176
00:09:51,010 --> 00:09:53,510
that you get to fish?

177
00:09:53,510 --> 00:09:56,470
Let us introduce
here some notation.

178
00:09:56,470 --> 00:10:01,070
Let us call the total
fishing time capital F.

179
00:10:01,070 --> 00:10:04,900
And the total fishing time
consists of two pieces.

180
00:10:04,900 --> 00:10:09,500
There's a time until
time 2 that you

181
00:10:09,500 --> 00:10:13,300
will be fishing no matter what.

182
00:10:13,300 --> 00:10:16,660
And then there's going to
be the remaining fishing

183
00:10:16,660 --> 00:10:18,610
time after time 2.

184
00:10:22,210 --> 00:10:24,880
So now let's look
at the expectation

185
00:10:24,880 --> 00:10:27,870
of the remaining fishing
time after time 2.

186
00:10:27,870 --> 00:10:29,480
Here there are two scenarios.

187
00:10:29,480 --> 00:10:31,880
In the first scenario, you stop.

188
00:10:31,880 --> 00:10:34,440
In the second
scenario, you continue.

189
00:10:34,440 --> 00:10:36,830
And we will take into
account these two scenarios

190
00:10:36,830 --> 00:10:40,780
by using the total
expectation theorem.

191
00:10:40,780 --> 00:10:45,600
The first scenario happens
with some probability.

192
00:10:45,600 --> 00:10:49,580
This is the probability that
you stop fishing at time 2.

193
00:10:49,580 --> 00:10:53,620
And in that case, your
remaining fishing time

194
00:10:53,620 --> 00:10:56,980
is going to be equal
to 0 because you do not

195
00:10:56,980 --> 00:10:59,510
continue under that scenario.

196
00:10:59,510 --> 00:11:01,710
But there's the other
scenario that you

197
00:11:01,710 --> 00:11:06,870
get to fish for more
than 2 time units.

198
00:11:06,870 --> 00:11:10,840
And then we multiply
this term with

199
00:11:10,840 --> 00:11:14,670
the conditional expectation
of the remaining fishing time,

200
00:11:14,670 --> 00:11:18,060
given that you got to fish
for more than 2 times units.

201
00:11:21,540 --> 00:11:22,620
So what is this term?

202
00:11:22,620 --> 00:11:25,510
The probability that you get
to fish for more than 2 time

203
00:11:25,510 --> 00:11:28,610
units, this is the
probability that no fish

204
00:11:28,610 --> 00:11:31,630
were caught during
the first time units.

205
00:11:31,630 --> 00:11:34,700
This is the probability
of the second scenario.

206
00:11:34,700 --> 00:11:38,660
And then we have this
conditional expectation.

207
00:11:38,660 --> 00:11:42,100
Given that you
didn't catch anything

208
00:11:42,100 --> 00:11:45,750
and, therefore, you will
be continuing fishing,

209
00:11:45,750 --> 00:11:51,090
what is the expected amount
of time-- F minus 2--

210
00:11:51,090 --> 00:11:53,520
that you will be fishing?

211
00:11:53,520 --> 00:11:56,890
Now the Poisson process
starts fresh at time 2.

212
00:11:56,890 --> 00:11:58,700
So looking into
the future, we're

213
00:11:58,700 --> 00:12:00,650
faced with a Poisson
process and we're

214
00:12:00,650 --> 00:12:04,550
asking for the expected time
until the first arrival.

215
00:12:04,550 --> 00:12:11,510
And this time has an expected
value equal to 1 over lambda.

216
00:12:11,510 --> 00:12:14,710
Our last question is
of a similar type.

217
00:12:14,710 --> 00:12:18,650
What is the expected number
of fish you're going to catch?

218
00:12:18,650 --> 00:12:23,260
Once more, we will break
this into two pieces.

219
00:12:23,260 --> 00:12:26,510
Number of fish caught
during the first two hours,

220
00:12:26,510 --> 00:12:31,780
and number of fish caught
during the remaining hours.

221
00:12:31,780 --> 00:12:35,480
During the first two
hours, the expected number

222
00:12:35,480 --> 00:12:39,650
of fish that you catch
is given by this formula.

223
00:12:39,650 --> 00:12:46,650
It is equal to lambda tau--
and in our case lambda is 0.6

224
00:12:46,650 --> 00:12:52,870
and tau is equal to 2-- plus
the expected number of fish

225
00:12:52,870 --> 00:12:56,150
that are caught afterwards.

226
00:12:56,150 --> 00:12:58,060
What is this expected value?

227
00:12:58,060 --> 00:13:01,520
Again, we think in terms of
the total expectation theorem.

228
00:13:01,520 --> 00:13:05,000
There's one scenario that
has a certain probability,

229
00:13:05,000 --> 00:13:08,390
and under that scenario
you do not catch any fish.

230
00:13:08,390 --> 00:13:11,070
So that doesn't give
us any contribution.

231
00:13:11,070 --> 00:13:13,840
Then there is this
scenario, the second one,

232
00:13:13,840 --> 00:13:17,220
that has this
probability of occurring.

233
00:13:17,220 --> 00:13:19,470
The probability of
catching no fish here,

234
00:13:19,470 --> 00:13:22,280
so that the second
scenario materializes.

235
00:13:22,280 --> 00:13:24,720
And if that second
scenario materializes,

236
00:13:24,720 --> 00:13:28,480
how many fish are you going
to catch after time 2?

237
00:13:28,480 --> 00:13:31,045
It's going to be one
fish with certainty.

238
00:13:33,820 --> 00:13:36,780
And this is the final
answer to this question.

239
00:13:36,780 --> 00:13:40,110
Notice that in answering
both of these questions

240
00:13:40,110 --> 00:13:44,440
we used the divide and
conquer strategy twice.

241
00:13:44,440 --> 00:13:48,830
We first divided the time
horizon into two pieces

242
00:13:48,830 --> 00:13:52,850
and dealt separately with
each one of the pieces.

243
00:13:52,850 --> 00:13:56,660
And then in order to deal
with this second piece--

244
00:13:56,660 --> 00:14:00,740
the time after time
2-- we used divide

245
00:14:00,740 --> 00:14:03,820
and conquer into the
two different scenarios

246
00:14:03,820 --> 00:14:07,400
and using the total
expectation theorem.