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In this problem, we'll be
looking at an ambulance that

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is traveling back and forth
in interval of size l.

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Say from 0 to l.

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At some point in time, there's
an accident occurring, let's

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say at location x.

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And we'll assume the accident
occurs in a random location so

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that x is uniformly distributed
between 0 and l.

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Now, at this point in time,
let's say the ambulance turns

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out to be at location y.

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Again, we'll assume that y is
a uniform random variable

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between 0 and l, and also that
x and y are independently

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distributed.

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The question we're interested
in answering is how long it

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would take an ambulance to
respond to travel from point y

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to point x.

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Let's call this time T. And in
particular, we want to know

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everything about distribution of
T. For example, what is the

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CDF of T given by the
probability of big T, less

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than or equal to little t, or
the PDF, which is done by

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differentiating the CDF
once we have it.

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Now, to start, we'll express T
you as a function of X and Y.

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Since we know that
the ambulance

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travels at a speed V--

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V meters or V units of
distance per second--

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then we can write that big T is
simply equal to Y minus X,

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absolute value the distance
between X and Y, divided by

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the speed at which the ambulance
is traveling at, V.

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So now if we look at the
probability of T less than or

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equal to little t, this is then
equal to the probability

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that Y minus X divided
by V less than or

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equal to little t.

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We now take off the absolute
value by writing the

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expression as negative vt less
equal to Y minus X less equal

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to positive vt.

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Here we multiply v on the other
side of t, and then took

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out the absolute value sign.

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As a final step, we'll also move
X to the other side of

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inequalities by writing this as
X minus vt less equal to y

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less equal to x plus vt.

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To compute this quantity, we'll
define a set A as a set

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of all points that satisfies
this condition right here.

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In particular, it's a pair of
all X and Y such that X minus

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vt less equal to little y less
equal to X plus vt, and also

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that X is within 0 and
l, and so is Y.

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So the set A will be the
set of values we'll

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be integrating over.

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Now that we have A, we can
express the above probability

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as the integral of all X and Y,
this pair within the set A,

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integrating the PDF of f of
X, Y, little x, little y.

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Let's now evaluate this
expression right here in a

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graphical way.

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On the right, we're plotting out
what we just illustrated

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here, where the shaded region is
precisely the set A. As we

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can see, this is a set of values
of X and Y where Y is

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sandwiched between two lines,
the upper one being X plus vt

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right here, and the lower line
being X minus vt, right here.

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So these are the values that
correspond to the set A.

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Now that we have A, let's
look f of x, y.

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We know that both x and y are
uniform random variables

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between 0 and l, and therefore,
since they're

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independent, the probability
density of x and y being at

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any point between 0 and l is
precisely 1 over l squared,

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where l squared is the size of
this square box right here.

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So given this picture, all we
need to do is to multiply by 1

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over l squared the area of the
region A. And depending on the

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value of T, we'll get different

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answers as right here.

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If T is less than 0, obviously,
the area of A

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diminishes to nothing,
so we get 0.

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If T is greater than l over V,
the area of A fills up the

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entire square, and we get 1.

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Now, if T is somewhere in
between 0 and l over v, we

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will have 1 over l squared,
multiply by the area looking

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like something like that right
here-- the shaded region.

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Now, if you wonder how we
arrive at exactly this

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expression right here, here is
a simple way to calculate it.

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What we want is 1 over l squared
times the area A. Now,

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area A can be viewed as the
entire square, l squared,

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minus whatever's not in area
A, which is these two

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triangles right here.

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Now, each triangle has area
1/2, l minus vt squared.

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This multiply 2, and this, after
some algebra, will give

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the answer right here.

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At this point, we have obtained
the probability of

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big T less equal to little t.

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Namely, we have gotten the CDF
for T. And as a final step, we

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can also compute the probability
density function

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for T. We'll call it
little f of t.

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And we do so by simply
differentiating the CDF in

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different regions of T.

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To begin, we'll look at t
between 0 and l over v right

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here at differentiating the
expression right here with

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respect to t.

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And doing so will give
us 2v over l minus 2v

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squared t over L squared.

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And this applies to t greater or
equal to 0, less than l/v.

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Now, any other region, either
t less than 0 or t greater

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than l/v, we have a constant
for the CDF, and hence its

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derivative will be 0.

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So this is for any other t.

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We call it otherwise.

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Now, this completely
characterized the PDF of big

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T, and hence, we've also
finished a problem.

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