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In this problem, Romeo and
Juliet are to meet up for a

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date, where Romeo arrives at
time x and Juliet at time y,

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where x and y are independent
exponential random variables,

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with parameters lambda.

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And we're interested in knowing
the difference between

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the two times of arrivals,
we'll call it z,

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written as x minus y.

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And we'll like to know what
the distribution of z is,

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expressed by the probability
density function, f of z.

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Now, we'll do so by using the
so-called convolution formula

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that we learn in the lecture.

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Recall that if we have a random
variable w that is the

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sum of two independent random
variables, x plus y, now, if

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that's the case, we can write
the probability [INAUDIBLE]

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function, fw, [INAUDIBLE]

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as the following integration--

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negative infinity to infinity
fx little x times f of y w

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minus x, integrated over x.

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And to use this expression to
calculate f of z, we need to

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do a bit more work.

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Notice w is expressed as a sum
of two random variables,

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whereas z is expressed as the
subtraction of y from x.

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But that's fairly easy to fix.

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Now, we can write z.

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Instead of a subtraction, write
it as addition of x plus

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negative y.

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So in the expression of the
convolution formula, we'll

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simply replace y by negative
y, as it will

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show on the next slide.

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Using the convolution formula,
we can write f of z little z

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as the integration of f of x
little x and f of negative y z

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minus x dx.

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Now, we will use the fact that
f of negative y, evaluated z

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minus x, is simply equal to f
of y evaluated at x minus z.

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To see why this is true, let's
consider, let's say, a

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discreet random variable, y.

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And now, the probability that
negative y is equal to

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negative 1 is simply the same
as probability that

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y is equal to 1.

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And the same is true for
probability density functions.

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With this fact in mind, we can
further write equality as the

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integration of x times
f of y x minus z dx.

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We're now ready to compute.

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We'll first look at the case
where z is less than 0.

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On the right, I'm writing out
the distribution of an

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exponential random variable
with a parameter lambda.

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In this case, using the
integration above, we could

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write it as 0 to infinity,
lambda e to the negative

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lambda x times lambda e to the
negative lambda x minus z dx.

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Now, the reason we chose a
region to integrate from 0 to

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positive infinity is because
anywhere else, as we can

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verify from the expression of
fx right here, that the

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product of fx times
fy here is 0.

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Follow this through.

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We'll pull out the constant.

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Lambda e to the lambda z, the
integral from 0 to infinity,

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lambda e to the negative
2 lambda x dx.

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This will give us lambda e to
the lambda z minus 1/2 e to

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the negative 2 lambda x infinity
minus this expression

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value at 0.

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And this will give us lamdba
over 2 e to the lambda z.

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So now, we have an expression
for f of z evaluated at little

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z when little z is
less than 0.

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Now that have the distribution
of f of z when z is less than

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0, we'd like to know what
happens when z is greater or

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equal to 0.

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In principle, we can go through
the same procedure of

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integration and calculate
that value.

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But it turns out, there's
something much simpler.

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z is the difference between x
and y, at negative z, simply

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the difference between
y and x.

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Now, x and y are independent and
identically distributed.

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And therefore, x minus
y has the same

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distribution as y minus x.

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So that tells us z and negative
z have the same

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distribution.

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What that means is, is the
distribution of z now must be

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symmetric around 0.

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In other words, if we know that
the shape of f of z below

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0 is something like that, then
the shape of it above 0 must

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be symmetric.

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So here's the origin.

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For example, if we were to
evaluate f of z at 1, well,

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this will be equal to the value
of f of z at negative 1.

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So this will equal to f
of z at negative 1.

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Well, with this information in
mind, we know that in general,

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f of z little z is equal to
f of z negative little z.

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So what this allows us to do is
to get all the information

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for z less than 0 and generalize
it to the case

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where z is greater
or equal to 0.

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In particular, by the symmetry
here, we can write, for the

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case z greater or equal to 0,
as lambda over 2 e to the

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negative lambda z.

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So the negative sign comes
from the fact that the

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distribution of f of z is
symmetric around 0.

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And simply, we can go back
to the expression

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here to get the value.

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And all in all, this implies
that f of z little z is equal

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to lambda over 2 e to
the negative lambda

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absolute value of z.

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This completes our problem.

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