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We now revisit the exponential
random variable that we

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introduced earlier and develop
some intuition about what it

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represents.

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We do this by establishing a
memorylessness property,

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similar to the one that we
established earlier in the

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discrete case for the
geometric PMF.

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Suppose that it is known that
light bulbs have a lifetime

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until they burn out,
which is an

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exponential random variable.

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You go to a store, and you are
given two choices, to buy a

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new light bulb, or to buy a used
light bulb that has been

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working for some time and
has not yet burned out.

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Which one should you take?

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We want to approach this
question mathematically.

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So let us denote by capital T
the lifetime of the bulb.

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So time starts at time 0, and
then at some random time that

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we denote by capital T, the
light bulb will burn out.

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And we assume that this random
variable is exponential with

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some given parameter lambda.

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In one of our earlier
calculations, we have shown

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that the probability that
capital T is larger than some

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value little x falls
exponentially

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with that value x.

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We are now told that a certain
light bulb has already been

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operating for t time units
without failing.

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So we know that the value of the
random variable capital T

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is larger than little t.

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We are interested in how much
longer the light bulb will be

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operating, and so we look at
capital X, which is the

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remaining lifetime from the
current time until the light

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bulb burns out.

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So capital X is this particular
random variable

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here, and it is equal to capital
T minus little t.

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Let us now calculate the
probability that the light

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bulb lasts for another
little x time units.

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That is, that this random
variable, capital X, is at

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least as large as
some little x.

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That is, that the light
bulb remains alive

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until time t plus x.

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We use the definition of
conditional probabilities to

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write this expression as the
probability that capital X is

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bigger than little x.

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On the other hand, capital
X is T minus t, so we

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write it this way--

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T minus t is bigger than little
x, and also that T is

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bigger than little t, divided
by the probability of the

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conditioning event.

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Just write this event in a
cleaner form, capital T being

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larger than little t plus x and
being larger than little

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t, again divided by the
probability of the

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conditioning event.

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And now notice that capital T
will be greater than little t

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and also greater than little t
plus x, that is, capital T is

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larger than this number and this
number, if and only if it

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is larger than this second
number here.

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So in other words, the
intersection of these two

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events is just this event here,
that capital T is larger

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than little t plus x.

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Now, we can use the formula
for the probability that

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capital T is larger
than something.

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We apply this formula, except
that instead of little x, we

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have t plus x.

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And so here we have e to the
minus lambda t plus x divided

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by the probability that capital
T is bigger than t.

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So we use this formula, but with
little t in the place of

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little x, and we obtain e
to the minus lambda t.

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We have a cancellation, and
we're left with e to the minus

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lambda x, which is a final
answer in this calculation.

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What do we observe here?

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The probability that the used
light bulb will live for

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another x time units is exactly
the same as the

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corresponding probability that
the new light bulb will be

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alive for another
x time units.

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So new and used light bulbs
are described by the same

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probabilities, and they're
probabilistically

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identical, the same.

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Differently said, the used light
bulb does not remember,

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and it is not affected by how
long it has been running.

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And this is the memorylessness
property of

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exponential random variables.

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Let us now build some additional
insights on

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exponential random variables.

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We have a formula for the
density, the PDF.

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And from this, we can calculate
the probability that

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T lies in a small interval.

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For example, for a small delta,
this probability here

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is going to be approximately
equal to the density of T

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evaluated at 0 times delta,
which is lambda times e to the

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0, which is 1, times delta.

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What if we are told that the
light bulb has been alive for

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t time units?

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What is the probability that it
burns out during the next

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delta times units?

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Since a used but still
alive light bulb is

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probabilistically identical to
a new one, this conditional

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probability is the same as this
probability here that a

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new light bulb burns out in the
next delta times units.

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And so this is also
approximately

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equal to lambda delta.

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So we see that independently of
how long a light bulb has

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been alive, during the next
delta time units it will have

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a lambda delta probability
of failing.

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One way of thinking about this
situation is that the time

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interval is split into little
intervals of length delta.

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And as long as the light bulb
is alive, if it is alive at

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this point, it will have
probability lambda delta of

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burning out during the next
interval of length delta.

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This is like flipping a coin.

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Once every delta time steps,
there is a probability lambda

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delta that there is a success
in that coin flip, where

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success corresponds to having
the light bulb actually burn

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down, and the exponential random
variable corresponds to

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the total time elapsed until
the first success.

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In this sense, the exponential
random variable is a close

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analog of the geometric random
variable, which was the time

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until the first success in
a discrete time setting.

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This analogy turns out to be
the foundation behind the

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Poisson process that
we will be studying

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later in this course.