Hi, in this problem, we're going to look at competing
exponential.
So we have three exponential random variables, X with
parameter lambda, Y with parameters mu, and Z with
parameter nu.
And we want to calculate some probability.
And the probability that we want to calculate is the
probability that X is less than Y is less than Z.
Now we can reinterpret this as 3 plus Poisson processes.
Because the link between exponentials and Poisson
processes is that the inter-arrival time of Poisson
processes are exponentially distributed.
So you can think of X being the time until the first
arrival in a Poisson process with parameter lambda.
And same thing for Y is the first arrival time of a
Poisson process with parameter mu.
The same thing for Z and nu.
And so in that interpretation, X less than Y less than Z, you
could interpret as a race, meaning that X finishes first
followed by Y and then Z comes in last.
So with that interpretation, let's see if we can calculate
what this probability is.
We can rewrite this probability as a combination
of two things occurring.
One is that X is less than the minimum of Y and Z. And then
the other is that Y is less than Z.
So what is this first event mean?
This first event means that X comes in first.
And it doesn't matter whether Y comes in second
or Z comes in second.
So we first say that X has to come in first which means it
has to beat the better of Y and Z. And then that combined
with the fact that Y does better than Z is the same
thing as saying that X is first, Y is
second, and Z is third.
And now, let's try to argue using Poisson processes, that
these two events are actually independent.
So this event occurring means that X is smaller than Y and
Z. So let's take these Poisson processes, and because these
random variables are assumed to be independent, these are
independent Poisson processes.
So we can merge them.
So let's merge these two.
And we'll get a Poisson process that has
rate mu plus nu.
And we can also merge this first one and that one.
And we'll get another Poisson process with predator lambda
plus mu plus nu.
So in that context, what does it mean that X is less than
the minimum Y and Z?
It just means that in this merged process, the first
arrival came from the X process.
In that case, if that's true, then X is less than minimum Y
and Z. Well, let's say that event does occur that the
first arrival is from the X process.
Now we're interested in what the order of the second two
arrivals are.
Is it Y first and then Z?
Or Z first and then Y?
Well, it doesn't matter because of
the fresh start property.
Because after this arrival comes, and say it is from the
X process, the Poisson processes start anew, and
they're still independent.
And so what happens after that is independent of what
happened here, when X arrived.
And so whether Y came first followed by Z, or Z came first
followed by Y is independent of what happened here.
And so because of that, these two events are independent,
and so when we have the probability of the
intersection of two independent events, we can
write that as the product of those two probabilities.
Now, what is the probability of this first event?
The probability that X is less than the minimum Y and Z?
Well, we just said that that corresponds to the first
arrival of this merge process being from the X process.
Well, that probability is lambda over lambda
plus mu plus nu.
So it's equal to this ratio where the process that you're
interested in, its rate comes in the numerator.
And then the merged rate is on the denominator,
And what about the second one?
What's the probability that Y is less than Z?
Well, let's go now to this merge process where we merged
just the Y and Z processes and see which one comes first.
Well, in that case what we want to know is in this merge
process, what is the probability that the first
arrival came from the Y process?
Well, analogously, that probability is going to mu
over mu plus nu.
And that gives us our answer.
And so we see that what looked like a pretty complex
calculation when we reinterpreted it in terms of
Poisson processes, it becomes relatively easy to solve.
But this still seems like a complicated expression.
So let's try to check to see whether it
actually makes sense.
So one way to do that is to look at a specific example of
the choice of lambda, mu, and nu, and see if it actually
makes sense.
So one example is suppose that all three of these parameters
are the same.
Well, if they're all the same then this probability, the
first part becomes 1 over 3, 1/3.
And the second one is 1 over 2.
And so if all three parameters are the same, probability
becomes 1/6.
And let's see if that makes sense.
If all three parameters are the same, that means that
these rates, these arrival rates are all the same.
And what that means is that any three ordering of these
three arrivals is as likely as any other ordering.
And what we're interested in is the probability of one's
particular ordering happening which is X first then Y then
Z.
But if everything is symmetric then any of the orderings is
as likely as any other one.
And how many orderings are there?
Well, there's three choices for who comes in first.
Two for who comes in second.
And one for who comes in last.
So there's a total of six possible orders in which these
three contestants, if you think of it that way, could
finish this race.
And out of none of those, we want the probability that one
of those outcomes happens.
And so the probability should be 1/6.
And that's what our formula tells us.
So as I said, in this problem, we saw how to reinterpret
exponentials in the context of Poisson processes that helped
us solve a--