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We will now go through a
beautiful example, in which we

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approach the same question
in a number of different ways

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and see that by reasoning
based on the intuitive

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properties of a
Poisson process, we

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can arrive quickly
to the right answer.

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The problem is as follows.

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We have three lightbulbs,
and each light bulb

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is being lit at time
zero, it starts working,

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and the light bulb lasts
for a certain amount

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of time, which is random.

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00:00:29,630 --> 00:00:33,090
So this light bulb lasts so
long, this one lasts so long,

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this one lasts that long.

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The lifetime of a light bulb,
the time until it burns out,

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will be a random
variable, and we

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make the following assumptions.

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The lifetimes of the
three light bulbs,

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which we denote by
X, Y, and Z, will

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be independent random
variables, each of which

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is an exponential random
variable with the parameter

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lambda.

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We're interested in the
question of calculating

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the expected time until
a light bulb burns out

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for the first time.

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So in this picture, the light
bulb that burned out first

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is the second light
bulb, and this quantity

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is the time until
the first burnout.

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How do we calculate
this quantity?

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The time until a first
light bulb burns out

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is the minimum of the times
at which each one of them

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burns out.

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So we're interested
in the expected

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value of this quantity.

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It's a random variable,
which is a function of three

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random variables.

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How do we calculate
the expected value

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of a function of
random variables?

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We can use the
expected value rule.

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Let us take the
function of interest,

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which is the minimum
of three numbers.

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Then we need to multiply by
the joint PDF or these three

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random variables
X, Y and Z. Now,

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because these three random
variables are independent,

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the joint PDF is the product
of their individual PDFs,

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which are all exponential.

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And so we obtain this
expression for the joint PDF.

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And we need to integrate this
over all x's, y's, and z's.

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So it's going to
be an integral that

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goes for each one of the three
variables from 0 to infinity.

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And here we have dx, dy, dz.

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So this is one approach
that can give us

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the answer if you're able
to manipulate and keep

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track of everything that happens
in this three-dimensional

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integral.

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But this is too tedious and
this is not a good way to go.

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Let us try to think
of an alternative way.

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Can we figure out
the distribution

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of this random variable?

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It's a function of three random
variables, so in some sense,

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it's a derived
distribution problem,

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so we can try to calculate
the cumulative distribution

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function of the
minimum of the three.

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So for the cumulative, we would
be looking at the probability

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that the minimum of these three
random variables is less than

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or equal to a certain number.

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It turns out that
the calculations

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are a little faster if we
look at the probability

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that this is larger than or
equal to a certain number.

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What is this event?

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The minimum is larger
than or equal to t,

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if, and only if,
all three of them

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are larger than or equal to t.

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So the probability of this event
is the same as the probability

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that X is larger
than or equal to t,

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Y larger than or equal to t,
and Z larger than or equal to t.

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But now X, Y, and
Z are independent,

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so we need to multiply
the probability that X

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is larger than or
equal to t, which

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for an exponential
random variable

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is e to the minus lambda
t, then the probability

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of the second event which is
again e to the minus lambda t,

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and, finally, the probability
of the third event, which

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is, again, e to
the minus lambda t,

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00:04:04,010 --> 00:04:06,620
which is e to the
minus 3 lambda t.

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Now, we look at this
expression for the probability

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that the random variable
is larger than or equal

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to t and recognize that this
is the expression that we have

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when we have an exponential
random variable with parameter

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equal to 3 lambda.

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If you want to do it formally,
subtract this quantity from 1

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to find the CDF,
take the derivative,

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and you will find
an exponential PDF.

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So the conclusion is
that this random variable

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is exponential with
parameter 3 lambda.

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And since it's an exponential
with parameter 3 lambda,

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then we know what the
expected value is.

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It is going to be
1 over 3 lambda.

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And this is the
answer to the question

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that we were interested in.

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Now, the fact that this is an
exponential random variable,

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but with a different parameter,
is a pretty clean fact,

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and so it should have
a good explanation.

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Let us now try to think about a
good explanation for this fact.

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Whenever we deal with an
exponential random variable,

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one way of thinking
about it is that

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this exponential random variable
is the first time in a Poisson

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process.

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So imagine that there's
a Poisson process that

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runs forever, and X is
the first arrival time.

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For this light bulb, we
can think the same way,

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and since X and Y are
assumed to be independent,

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we can assume that here we
have an independent Poisson

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process, independent
from the first one,

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it has its own arrival times,
and Y is the first arrival

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time in this Poisson process.

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And finally, we have a third
independent Poisson process,

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and the random variable
Z is the first arrival

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time in that Poisson process.

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So X, Y and Z are
interpreted as first arrivals

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in three independent
Poisson processes.

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Now, let us take these
three Poisson processes

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and merge them.

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If we merge these three
processes, what we obtain

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is a merged process, which
is Poisson with parameter

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equal to the sum of the rates
or parameters of each one

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of the processes, so
it's a Poisson process

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with parameter 3 lambda.

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Now, how can we interpret the
random variable of interest,

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the first burnout time, in
terms of the merged process?

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So the merged process
has an arrival

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whenever one of those three
processes has an arrival.

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The first arrival in
the merged process

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will happen the first time that
one of these three processes

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is going to have an arrival.

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Therefore, we can interpret
the random variable

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of interest, the
first burnout time,

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as the first arrival
time in a merged process.

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But now the merged process
is Poisson with parameter 3

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lambda, therefore,
this random variable

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is going to be an exponential
random variable with parameter

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3 lambda.

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And from this, we also
obtain the expected value

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of that random variable.

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The beauty of this last approach
for coming up with the answer

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by reasoning in terms of
merged Poisson processes

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is that we didn't have to do
any calculations at all, just

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use the intuitive understanding
of Poisson processes

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and, especially,
the interpretation

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of an exponential random
variable as the first arrival

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time in a Poisson process.

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Let us now try to solve a
somewhat harder problem.

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Let us try to calculate
the expected time

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until all the light
bulbs burn out.

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So one light bulb will burn out,
then another one will burn out,

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00:08:05,540 --> 00:08:08,110
and, finally, the third
one will burn out.

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00:08:08,110 --> 00:08:11,700
We want to find the expected
time until this happens.

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Once more, we will be thinking
of these burnout times

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as the first arrival times
in Poisson processes.

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The total time until
the third burnout

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can be split into
different periods.

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There's a time until one
light bulb burns out.

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And the expected
value of this period

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here is going to
be 1 over 3 lambda.

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What happens at this time?

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The second light bulb has burned
out, so we can forget about it,

165
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take it out of the picture.

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00:08:52,960 --> 00:08:55,830
We have two lightbulbs.

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Let us look at the
time it will take

168
00:08:58,390 --> 00:09:02,610
until one of these two
light bulbs burns out.

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00:09:02,610 --> 00:09:07,110
So we're interested in
this period of time.

170
00:09:07,110 --> 00:09:11,370
Now, the Poisson process
starts fresh at this time.

171
00:09:11,370 --> 00:09:13,940
After this time,
whatever happens

172
00:09:13,940 --> 00:09:16,640
is just an ordinary
Poisson process

173
00:09:16,640 --> 00:09:19,350
as if it were
starting at this time.

174
00:09:19,350 --> 00:09:23,630
So this is going to be an
exponential random variable

175
00:09:23,630 --> 00:09:24,930
starting from this time.

176
00:09:24,930 --> 00:09:28,680
And this is going to be another
exponential random variable.

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00:09:28,680 --> 00:09:32,670
So the time until the next light
bulb burns out in this case

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00:09:32,670 --> 00:09:34,410
is going to be
the minimum of two

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00:09:34,410 --> 00:09:36,370
exponential random variables.

180
00:09:36,370 --> 00:09:39,300
We can think again about
merging these two Poisson

181
00:09:39,300 --> 00:09:42,475
processes to obtain a Poisson
process with total rate 2

182
00:09:42,475 --> 00:09:46,450
lambda, and the time until
one of these two turns out

183
00:09:46,450 --> 00:09:50,670
is going to be the first arrival
time in that merged process.

184
00:09:50,670 --> 00:09:54,775
And so the expected time
until the first arrival

185
00:09:54,775 --> 00:09:59,730
of the merged process is
going to be 1 over 2 lambda.

186
00:09:59,730 --> 00:10:03,540
And finally, once this
burnout has happened,

187
00:10:03,540 --> 00:10:06,160
we can forget about
this light bulb.

188
00:10:06,160 --> 00:10:10,200
We're left just with one light
bulb, and starting from here,

189
00:10:10,200 --> 00:10:13,130
we wait until that
light bulb burns out.

190
00:10:13,130 --> 00:10:16,480
Once more, because of the fresh
start property of the Poisson

191
00:10:16,480 --> 00:10:19,960
process, starting from
here until it burns out

192
00:10:19,960 --> 00:10:22,010
is going to be a
random variable, which

193
00:10:22,010 --> 00:10:24,210
is an exponential
random variable.

194
00:10:24,210 --> 00:10:27,150
And in this case, an exponential
random variable with rate

195
00:10:27,150 --> 00:10:28,620
just lambda.

196
00:10:28,620 --> 00:10:31,100
And by adding these
three quantities,

197
00:10:31,100 --> 00:10:38,440
we get the expected time until
all three have burned out.

198
00:10:38,440 --> 00:10:39,970
This is a problem
that would have

199
00:10:39,970 --> 00:10:44,960
been quite hard to solve
in a more analytical way.

200
00:10:44,960 --> 00:10:47,670
We're dealing with a random
variable, which is now

201
00:10:47,670 --> 00:10:52,450
the maximum of X, Y, and
Z. And the distribution

202
00:10:52,450 --> 00:10:56,770
of this random variable is
not so simple to write down.

203
00:10:56,770 --> 00:11:00,130
So that would not be
a very good approach

204
00:11:00,130 --> 00:11:02,430
for going about this problem.

205
00:11:02,430 --> 00:11:04,930
But we managed to find
the expected value

206
00:11:04,930 --> 00:11:08,030
of this random variable
without having to write down

207
00:11:08,030 --> 00:11:12,020
its distribution, by
breaking this random variable

208
00:11:12,020 --> 00:11:17,650
into a sum of three particular
random variables, each of which

209
00:11:17,650 --> 00:11:20,490
had a nice intuitive
interpretation.

210
00:11:20,490 --> 00:11:24,580
And that was the key to the
solution of this problem.