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We now follow a program that
parallels our development

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for the case of the
Bernoulli process.

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We will study the time
until the first arrival,

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a random variable
that we denote by T1.

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We're interested in
finding the probability

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distribution of this
random variable.

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And later on, we
will continue and try

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to study the time
until the kth arrival.

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Now T1 is a continuous
random variable,

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because the Poisson process
runs in continuous time.

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And therefore, it has a PDF.

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But instead of finding
the PDF directly,

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we will first find the CDF
of this random variable.

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So we fix a certain
time, T. And we're

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asking for the probability
that the first arrival happens

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during this interval.

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Now this is 1 minus
the probability

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that the first arrival
happens outside this interval.

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So we can write this probability
as 1 minus the probability

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that T1 is bigger than t.

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But what is this event?

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The first arrival occurring
after time, little t,

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is the same as saying that there
were no arrivals in the time

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interval from 0 to little t.

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And this probability
of 0 arrivals

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in a time interval
of length t is

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something for which we
already have a formula.

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Take this formula and
replace k by 0, tau by t.

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When k is equal
to 0, this term is

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something to the 0-th
power equal to 1.

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Using our convention, that
0 factorial is equal to 1,

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we're left just with e
to the minus lambda t.

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And this is the answer
for the CDF of the time

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until the first arrival.

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We then take the derivative.

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And we find that the PDF of the
time until the first arrival

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has this form, which is the
PDF of an exponential random

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variable.

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Of course, this
calculation here is

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only valid for t's
that are non-negative.

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For negative t's, the PDF
of T1 is, of course, 0.

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For the exponential
random variable,

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we have seen that it has certain
memorylessness properties.

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Namely, if I
condition on an event

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that nothing has occurred
until a certain time,

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t, and I am interested
in the time from now

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until the first arrival
occurs, this remaining

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until the first arrival is again
an exponential distribution.

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That is, looking
ahead from this time,

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I will still wait an
exponentially distributed

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amount of time until I
see the first arrival.

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Whatever happened in
the past and how long

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I have been waiting
doesn't matter.

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Starting from this
time, I will still

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wait an exponentially
distributed amount of time.

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This is essentially
an expression

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of a fresh start property of
the Poisson process, which

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is analogous to the fresh start
properties for the Bernoulli

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process.

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And we will be discussing
this fresh start

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property a lot more.

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Having figured out
the distribution

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of the time of
the first arrival,

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let us now study the time of the
k-th arrival, a random variable

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that we denote by Y sub
k, similar to the case

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of the Bernoulli process.

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This random variable
is a continuous one,

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because arrivals happen
in continuous time,

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so it takes continuous values.

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And therefore, it will
be described by a PDF.

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And this is what
we want to find.

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In order to find it, we will
make use of the Poisson PMF

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that we have already derived
for the number of arrivals

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during an interval
of a fixed length.

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One approach to
finding the PDF of Yk

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is the usual program,
similar again

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to what we did for the case
of the first arrival time.

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We can first find CDF, and then
differentiate to find the PDF.

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So what is the CDF?

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We want to calculate
the probability

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that Yk is less than or equal
to some number, little y.

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Now what is this event?

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The k-th arrival
occurs by time y.

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This means that by time y,
we've had at least k arrivals.

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We've had k arrivals, or maybe
k plus 1, or maybe k plus 2.

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We've had some number
of arrivals, n,

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in an interval of length, y.

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And this is an event that
happens with this probability.

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But we need to take into account
all of the possible values of n

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that are at least as large as k.

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Now we have a formula for this
probability, the probability

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of n arrivals in an
interval of given length.

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This is the Poisson PMF with
appropriate changes of symbols.

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So we can take this expression,
substitute it here, and then

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differentiate to do some
algebra and find the answer.

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Instead of carrying
out this algebra,

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however, we will proceed in
a more intuitive way that

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will get us there
perhaps faster.

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And the derivation that
we would follow actually

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parallels the one
that we went through

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in the case of the
Bernoulli process.

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The intuitive argument
that we will use

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will rest on the
interpretation of a PDF

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in terms of probabilities
of small intervals.

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So the PDF evaluated at
some particular point,

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y, times delta, is
approximately the probability

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that our random variable
falls within a delta interval

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from this number, little
y, that we're considering.

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So here's time 0, here's time
y, and here's time y plus delta.

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We want to find or
to say something

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about the probability of falling
inside this small interval.

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Now what does it mean
for the k-th arrival

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to fall inside this interval?

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This is an event that
can happen as follows.

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The k-th arrival falls
in this interval,

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and we've had k minus 1 arrivals
during the previous interval.

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What is the probability
of this event?

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A basic assumption about
the Poisson process

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is the independence assumption.

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Therefore, having k minus
1 arrivals in this interval

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and having one arrival
in this interval

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are independent events.

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Therefore, the probability
of this scenario

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is the product of
the probabilities

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that we've had k minus 1
arrivals in an interval

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of length, y, times
the probability

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that we've had one arrival in
an interval of length delta.

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And that latter probability
is approximately

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equal to lambda times delta.

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So I should write here an
approximate equality instead

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of an exact
equality, to indicate

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that there are other terms,
order of delta squared,

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for example, but which are much
smaller compared to the delta.

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However, this is
not the only way

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that we can get the k-th
arrival in this interval.

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There's an alternative scenario.

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We might have had k minus 2
arrivals during this interval,

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and then two arrivals
during that little interval.

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In this case, the
k-th arrival again

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occurs within that
little interval.

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So we need to also calculate the
probability of this scenario.

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The probability of
that scenario is

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the probability of k minus
2 arrivals in an interval

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of length, y, times the
probability of two arrivals.

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But the probability
of two arrivals

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is something that's
order of delta squared.

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And order of delta
squared is much smaller

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than this term, which
is linear in delta.

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And so this term can be
ignored as long as we're just

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keeping track of
the dominant terms,

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those are linear in delta.

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And then, they would
be similar expressions.

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For example, the scenario
that we have three arrivals up

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to time y, and then
three more arrivals

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during that little
interval, which is again

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an event of probability,
order of delta

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squared, that we get
three arrivals there.

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And all of those terms
are insignificant,

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and we can ignore them.

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And we end up with an
approximate equality

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between this term and
this expression here.

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Delta shows up on both sides,
so we can cancel delta.

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And therefore, we have ended
up with a formula for the PDF.

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In particular, the PDF is
equal to this probability times

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lambda.

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What is this probability?

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We have a formula for it.

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But we just need to substitute.

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Put k minus 1 in the place of k,
and put y in the place of tau.

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00:09:04,500 --> 00:09:10,050
This gives us lambda y
to the power k minus 1, e

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00:09:10,050 --> 00:09:16,850
to the minus lambda y, divided
by k minus 1 factorial.

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00:09:16,850 --> 00:09:18,510
And then we have
the extra factor

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of lambda, which can be put
together with this lambda

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00:09:22,890 --> 00:09:24,860
to the k minus 1 here.

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And we end up with this final
formula for the PDF of Yk.

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00:09:30,100 --> 00:09:31,720
The distribution
that we have here

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is called an Erlang
distribution.

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But actually, it's not
just one distribution.

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We have different
distributions depending

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on what k we're considering.

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The distribution of the
time of the third arrival

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is different from
the distribution

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00:09:45,510 --> 00:09:47,120
of the 10th arrival.

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00:09:47,120 --> 00:09:49,580
So if we fix a
particular k, then

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we say that we have an Erlang
distribution of order k.

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00:09:56,300 --> 00:09:59,520
For the case where
k is equal to 1,

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this term here disappears,
k minus 1 is equal to 0.

188
00:10:03,840 --> 00:10:06,060
And the denominator
term disappears,

189
00:10:06,060 --> 00:10:10,260
and we end up with lambda
times e to the minus lambda y.

190
00:10:10,260 --> 00:10:12,480
But this is the
exponential distribution

191
00:10:12,480 --> 00:10:16,730
that we had already derived
with a different method earlier.

192
00:10:16,730 --> 00:10:18,900
As you increase
k, of course, you

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00:10:18,900 --> 00:10:20,580
get different distributions.

194
00:10:20,580 --> 00:10:23,820
And these tend to shift
towards the right.

195
00:10:23,820 --> 00:10:25,610
This makes sense.

196
00:10:25,610 --> 00:10:28,430
The time of the
second arrival is

197
00:10:28,430 --> 00:10:30,280
likely to take certain values.

198
00:10:30,280 --> 00:10:32,890
The time of the third
arrival is likely to take

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00:10:32,890 --> 00:10:34,310
values that are higher.

200
00:10:34,310 --> 00:10:37,640
And the more you increase
k, the more the distribution

201
00:10:37,640 --> 00:10:39,900
will be shifting
towards the right.