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We now introduce a new of
random variable, the

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exponential random variable.

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It has a probability density
function that is determined by

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a single parameter lambda, which
is a positive number.

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And the form of the PDF
is as shown here.

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Note that the PDF is equal to
0 when x is negative, which

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means that negative values
of X will not occur.

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They have zero probability.

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And so our random variable is a

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non-negative random variable.

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The shape of the PDF is as
shown in this diagram.

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It's 0 for negative values, and
then for positive values,

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it starts off, it starts off
at a value equal to lambda.

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This is because if you plug
in x equal to 0 in this

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expression, you get lambda times
e to the 0, which leaves

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you just with lambda.

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So it starts off with lambda,
and then it decays

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at the rate of lambda.

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Notice that when lambda is
small, the initial value of

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the PDF is small.

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But then the decay rate is also
small, so that the PDF

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extends over a large
range of x's.

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On the other hand, when lambda
is large, then the PDF starts

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large, so there's a fair amount
of probability in the

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vicinity of 0.

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But then it decays pretty fast,
so there's much less

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probability for larger
values of x.

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Another observation to make
is that the shape of this

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exponential PDF is quite similar
to the shape of the

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geometric PDF that we have
seen before, the only

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difference being that here we
have a discrete distribution,

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but here we have a continuous
analog of that distribution.

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Let's now carry out
a calculation.

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Let us fix some positive number
a, and let us calculate

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the probability that our random
variable takes a value

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larger than or equal to a.

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So what we're trying to do is
to calculate the probability

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that the random variable
falls inside this

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interval from a to infinity.

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Whenever we have a PDF, we can
calculate the probability of

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falling inside an interval by
integrating over that interval

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the value of the PDF.

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Therefore, we have to calculate

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this particular integral.

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And at this point, we can recall
a fact from calculus,

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namely that the integral of the
function e to the ax is 1

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over a times e to the ax.

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We can use this fact by making
the correspondence between a

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and minus lambda.

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And using this correspondence,
we can now continue the

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calculation of our integral.

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We have a factor of lambda.

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And then a factor of
1 over a, where a

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stands for minus lambda.

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So we get the minus
1 over lambda.

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And then the same exponential
function, e to the

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minus lambda x.

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And because the range of
integration is from a to

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infinity, we need to evaluate
the integral at a and infinity

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and take the difference.

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Now, this lambda cancels
that lambda.

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We're left with a minus sign.

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And from the upper limit, we
get e to the minus lambda

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times infinity.

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And then from the second term,
we have a minus sign that

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cancels with that minus sign and
gives us a plus term, plus

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e to the minus lambda a.

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Now, e to the minus
infinity is 0.

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And so we're left just
with the last term.

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And the answer is e to
the minus lambda a.

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So this gives us the tail
probability for an exponential

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random variable.

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It tells us that the probability
of falling higher

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than a certain number falls off
exponentially with that

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certain number.

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An interesting additional
observation--

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if we let a equal to 0 in this
calculation, we obtain the

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integral of the PDF over the
entire range of x's.

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And in that case, this
probability becomes e to the

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minus lambda 0, which
is equal to 1.

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So we have indeed verified that
the integral of this PDF

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is equal to 1, as
it should be.

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Now, let's move to the
calculation of the expected

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value of this random variable.

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We can use the definition.

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Since the PDF is non-zero only
for positive values of x, we

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only need to integrate
from 0 to infinity.

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We integrate x times the PDF.

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And this is an integral that
you may have encountered at

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some point before.

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It is evaluated by using
integration by parts.

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And the final answer turns
out to be 1 over lambda.

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Regarding the calculation of
the expected value of the

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square of the random variable,
we need to write down a

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similar integral, except
that now we

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will have here x squared.

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This is just another integration
by parts, only a

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little more tedious.

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And the answer turns out to
be 2 over lambda squared.

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Finally, to calculate the
variance, we use the handy

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formula that we have.

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And the expected value of
X squared is this term.

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The expected value of
X is this term.

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When we square it, it becomes
similar to this term, but we

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have here a 2.

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There we have a 1.

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And so we're left with just
1 over lambda squared.

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And this is the variance of the

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exponential random variable.

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Notice that when lambda is
small, the PDF, as we

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discussed before, falls off very
slowly, which means that

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large x's are also
quite possible.

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And so the average of this
random variable will be on the

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higher side.

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The PDF extends over a large
range, and that translates

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into having a large mean.

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And because when that happens,
the PDF actually spreads, the

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variance also increases.

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And this is reflected in this
formula for the variance.

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The exponential random variable
is, in many ways,

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similar to the geometric.

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For example, the expression for
the mean, which is 1 over

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lambda, can be contrasted with
the expression for the mean of

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the geometric, which
is 1 over p.

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And the relationship between
these two distributions, the

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discrete and the continuous
analog, is a theme that we

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will revisit several times.

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At this point, let me just say
that the exponential random

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variable is used to model
many important

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and real world phenomena.

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Generally, it models the time
that we have to wait until

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something happens.

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In the discrete case, the
geometric random variable

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models the time until we see a
success for the first time.

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In the continuous case, an
exponential can be used to

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model the time until a customer
arrives, the time

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until a light bulb burns out,
the time until a machine

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breaks down, the time until you
receive an email, or maybe

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the time until a meteorite
falls on your house.