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Hi, in this problem, we're going
to look at competing

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exponential.

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So we have three exponential
random variables, X with

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parameter lambda, Y with
parameters mu, and Z with

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parameter nu.

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And we want to calculate
some probability.

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And the probability that we
want to calculate is the

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probability that X is less
than Y is less than Z.

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Now we can reinterpret this as
3 plus Poisson processes.

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Because the link between
exponentials and Poisson

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processes is that the
inter-arrival time of Poisson

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processes are exponentially
distributed.

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So you can think of X being
the time until the first

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arrival in a Poisson process
with parameter lambda.

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And same thing for Y is the
first arrival time of a

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Poisson process with
parameter mu.

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The same thing for Z and nu.

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And so in that interpretation, X
less than Y less than Z, you

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could interpret as a race,
meaning that X finishes first

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followed by Y and then
Z comes in last.

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So with that interpretation,
let's see if we can calculate

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what this probability is.

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We can rewrite this probability
as a combination

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of two things occurring.

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One is that X is less than the
minimum of Y and Z. And then

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the other is that Y
is less than Z.

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So what is this first
event mean?

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This first event means that
X comes in first.

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And it doesn't matter whether
Y comes in second

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or Z comes in second.

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So we first say that X has to
come in first which means it

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has to beat the better of Y and
Z. And then that combined

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with the fact that Y does better
than Z is the same

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thing as saying that
X is first, Y is

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second, and Z is third.

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And now, let's try to argue
using Poisson processes, that

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these two events are actually
independent.

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So this event occurring means
that X is smaller than Y and

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Z. So let's take these Poisson
processes, and because these

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random variables are assumed to
be independent, these are

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independent Poisson processes.

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So we can merge them.

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So let's merge these two.

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And we'll get a Poisson
process that has

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rate mu plus nu.

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And we can also merge this
first one and that one.

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And we'll get another Poisson
process with predator lambda

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plus mu plus nu.

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So in that context, what does
it mean that X is less than

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the minimum Y and Z?

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It just means that in this
merged process, the first

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arrival came from
the X process.

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In that case, if that's true,
then X is less than minimum Y

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and Z. Well, let's say that
event does occur that the

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first arrival is from
the X process.

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Now we're interested in what
the order of the second two

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arrivals are.

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Is it Y first and then Z?

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Or Z first and then Y?

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Well, it doesn't matter
because of

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the fresh start property.

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Because after this arrival
comes, and say it is from the

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X process, the Poisson processes
start anew, and

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they're still independent.

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And so what happens after that
is independent of what

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happened here, when X arrived.

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And so whether Y came first
followed by Z, or Z came first

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followed by Y is independent
of what happened here.

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And so because of that, these
two events are independent,

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and so when we have the
probability of the

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intersection of two independent
events, we can

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write that as the product of
those two probabilities.

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Now, what is the probability
of this first event?

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The probability that X is less
than the minimum Y and Z?

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Well, we just said that that
corresponds to the first

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arrival of this merge process
being from the X process.

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Well, that probability
is lambda over lambda

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plus mu plus nu.

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So it's equal to this ratio
where the process that you're

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interested in, its rate comes
in the numerator.

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And then the merged rate
is on the denominator,

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And what about the second one?

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What's the probability that
Y is less than Z?

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Well, let's go now to this merge
process where we merged

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just the Y and Z processes and
see which one comes first.

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Well, in that case what we want
to know is in this merge

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process, what is the probability
that the first

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arrival came from
the Y process?

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Well, analogously, that
probability is going to mu

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over mu plus nu.

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And that gives us our answer.

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And so we see that what looked
like a pretty complex

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calculation when we
reinterpreted it in terms of

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Poisson processes, it becomes
relatively easy to solve.

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But this still seems like a
complicated expression.

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So let's try to check
to see whether it

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actually makes sense.

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So one way to do that is to look
at a specific example of

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the choice of lambda, mu, and
nu, and see if it actually

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makes sense.

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So one example is suppose that
all three of these parameters

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are the same.

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Well, if they're all the same
then this probability, the

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first part becomes
1 over 3, 1/3.

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And the second one
is 1 over 2.

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And so if all three parameters
are the same, probability

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becomes 1/6.

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And let's see if that
makes sense.

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If all three parameters are
the same, that means that

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these rates, these arrival
rates are all the same.

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And what that means is that any
three ordering of these

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three arrivals is as likely
as any other ordering.

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And what we're interested in
is the probability of one's

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particular ordering happening
which is X first then Y then

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Z.

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But if everything is symmetric
then any of the orderings is

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as likely as any other one.

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And how many orderings
are there?

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Well, there's three choices
for who comes in first.

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Two for who comes in second.

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And one for who comes in last.

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So there's a total of six
possible orders in which these

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three contestants, if you think
of it that way, could

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finish this race.

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And out of none of those, we
want the probability that one

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of those outcomes happens.

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And so the probability
should be 1/6.

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And that's what our
formula tells us.

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So as I said, in this problem,
we saw how to reinterpret

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exponentials in the context of
Poisson processes that helped

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us solve a--

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