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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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