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Hi.

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In this video, we're going to
compute some useful quantities

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for the exponential
random variable.

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So we're given that x is
exponential with rate lambda.

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PDF looks like this, and
the formula is here.

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First question, part
a, what's the CDF?

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So let's go right in.

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The CDF of x is the probability
that X is less

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than or equal to little x.

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Let's look at some cases here.

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What if little x
is less than 0?

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Well, x random variable
only takes on these

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non-negative values.

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And so the probability that X is
less than or equal to some

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negative number is
going to be 0.

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On the other hand, if x is
greater than or equal to 0, we

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do actually have to
integrate here.

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So to do that, we take the
integral from minus infinity

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to x of fx of t--

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the dummy variable
here used is t.

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Notice that again, fx of t is
going to be 0 for negative

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values, so we take the
integral here from 0.

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And now we plug in
for fx of t.

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That's minus lambda t dt.

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And recall that the integral
of u to the a t is 1 over a

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times e to the a t.

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So here in this case,
we'll get lambda,

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which is just a constant.

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And then a here is going
to be negative lambda.

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So we get this, 0 to x.

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Lambdas cancel and we actually
get 1 minus e to the

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minus lambda x.

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So do this.

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And we are done with the CDF.

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Now for the expectation.

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We use the standard formula,
which is minus infinity to

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infinity t times fx of t dt.

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So again, fx of t is going to
be 0 for a negative value.

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So we do the integral from 0.

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We get 0 to infinity t lambda
e to the minus lambda t dt.

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Now, you can try all you want
to get rid of this t.

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It's not going to go even if
you try all kinds of u

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substitution.

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But at the end the day, you're
going to have to pull out your

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calculus textbook and find
the integration by parts

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formula, which is--

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v du.

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So the hope is that this
integral is going to be easier

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than the one on the left.

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Notice that this is the
integral of one

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of the terms here.

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And this is the derivative
of one of the terms.

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So that may help you decide
on how you select u and v.

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In our case actually, I'm
going to use u as--

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t for u.

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Because when you take
the derivative, it's

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going to become 1.

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And the derivative is what's
going to go in that integral.

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So this is going to
be dt for du.

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And then, dv I'm going
to select as

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whatever's left over.

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It's lambda e to minus
lambda t dt.

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So v is going to be--

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we already did the integral--

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minus e to the minus lambda t.

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And so if we do this, it's going
to be negative t times e

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to the minus lambda t.

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So that's uv.

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Minus v, which is negative
e to the minus

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lambda t times dt.

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That goes from 0 to infinity.

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This is evaluated from
0 to infinity.

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Well, what does it mean
for this to be

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evaluated from 0 to infinity?

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A better and easier way to look
at this is to say, well,

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it's going to go from 0 to x.

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But then you take the limit
as x goes to infinity.

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So that's going to
help us here.

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And this negative--

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these negatives cancel.

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And we're left with--

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let's plug in the bounds.

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We're left with negative x
minus lambda x plus the

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integral of this is going to be
1 over negative lambda e to

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the minus lambda t evaluated
from 0 to infinity.

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All right, so now the limit.

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So for the limit, notice
that x increases

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as x goes to infinity.

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And this exponential decays.

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So they're kind of competing
for each other.

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But the exponential is going to
win because it decays way

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faster than x.

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And so this first term
is going to go off--

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the limit is going to go to 0.

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All right.

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For this, if you evaluate
the balance, the

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infinity makes this 0.

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And 0, you're going to
get 1 over lambda.

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So that's 1 over lambda.

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All right.

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And so the expectation
is 1 over lambda.

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OK, so now what's
the variance?

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That's part c, right?

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So we use the standard formula
for variance, which is this.

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We already figured out
the expectation.

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We just need to figure out the
expectation of x squared.

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Well, we're just going to follow
the same set of steps

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from before.

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For x squared, it's just going
to be t squared, t squared, t

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squared, x squared.

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The only thing that's going to
change is what we choose for u

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here, for the u substitution.

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So it's going to be t squared.

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So the derivative is going
to change to 2t dt.

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v is going to be exactly
the same.

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And so here in this term, we
get negative 2t e to the

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minus lambda t.

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But there's a negative sign
out here, so the negatives

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cancel and we're left with
a positive sign here.

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This is going to change.

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All right.

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OK.

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So in order to do this integral,
we can use a trick.

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We can move this--

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so there's a 2t here.

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We move this 2 in here,
leave the t inside.

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And you have to leave
the t inside.

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But multiply by lambda
and divide by lambda.

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Now, look at that integral.

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0 to infinity t times lambda
e to the minus lambda t dt.

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Exactly the expectation
that we computed.

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We already did that.

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That is just 1 over lambda, so
it's 2 over lambda times 1

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over lambda.

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Again, the limit as x
goes to infinity--

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the exponential will
beat x squared.

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No matter what polynomial
we put in there, the

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exponential's going to win.

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So this is going
to be 0 still.

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This one's going to be 2
over lambda squared.

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So we're left with 2 over lambda
squared for expectation

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of x squared.

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And so we have 1 over lambda
squared for the variance.

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OK, so we're done with
the variance.

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Part d.

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We're given that x1, x2, and
x3 are independent and

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identically distributed.

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They're exponentials
with rate lambda.

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We're asked for the PDF
of z, which is the max

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of x1, x2, and x2.

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How do we generally
find a PDF?

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We take the CDF and then take
the derivative, right?

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We first find the CDF, and
then take the derivative.

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So let's do that.

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So first, let's see.

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Part d, find the CDF of z,
which is going to be the

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probability that Z is less than
or equal to little z,

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which is going to be equal to
the probability that the max

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of x1, x2, x3 is less
than or equal to z.

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And this is going to have
the same sort of

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structure as before.

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If z is less than 0, x1,
x2, x3 are positive--

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non-negative.

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And so this is the probability
that if you get little z less

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than 0, you're not going to have
any probability there.

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And so if z is greater than or
equal to 0 is where it gets

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interesting.

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We need to do something
special.

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So the special thing here
is to recognize that the

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probability of the max being
less than or equal to z is

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actually also the probability
of each of these random

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variables individually being
less than or equal to z.

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Why is that true?

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One way to check whether the
events-- these two events are

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the same is to check
the two directions.

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One direction say, if the max of
x1, x2, x3 is less than or

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equal to z, does that mean x1 is
less than or equal to z, x2

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is less than or equal
to z, and x3 is less

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than or equal to z?

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Yes.

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OK.

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And then, if x1, x2, and x3 are
individually less than or

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equal to z, then the max is also
less than or equal to z.

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So these two events are
equivalent and this is true.

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By independence we can
break this up.

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And we get--

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these are all CDFs of the
exponential and they

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all have this form.

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So it's just going to
be 1 minus e to the

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minus lambda z cubed.

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Plug this in here.

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And then, try to take the
derivative to get the PDF.

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Let's see.

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So it's going to be the same,
like this for z less than 0.

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For z greater than or equal
to 0, it's going to be the

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derivative of this thing.

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Derivative of this thing is by
chain rule, 3 times 1 minus e

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to the minus lambda z squared.

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Then the derivative of negative
e to the minus lambda

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z, that's just lambda e
to the minus lambda z.

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There we go.

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This is the PDF we
were looking for.

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So last problem.

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We're looking for the PDF
of w, which is the

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min of x1 and x2.

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So let's try this as
a similar approach.

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Try the same thing, actually.

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See if it works.

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00:12:06,920 --> 00:12:15,260

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So w, w, w, w, min, less
than or equal to w.

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00:12:23,785 --> 00:12:26,700

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OK.

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So let's see if this works.

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Is it true that the min--

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if the min of x1 and x2 is less
than or equal to w, that

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each of them is less
than or equal to w?

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No, right?

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X1 could be less than or
equal to w and x2 could

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be bigger than w.

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And the min could still be
less than or equal to w.

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So that's definitely not true.

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So what do we do here?

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The trick is to flip it and say
we want to compute the min

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of x1 and x2 being
greater than w.

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In that case, let's check
if we can do this trick.

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If the min of x1 and x2 is
greater than w, then clearly

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x1 is bigger than w and
x2 is bigger than w.

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And if x1 and x2 are
individually bigger than w,

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then clearly the min's
also bigger than w.

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So this works.

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And now we can use independence
as before.

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And for this, this is just
1 minus the CDF here.

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So it's just going to be
e to the minus lambda

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w for each of them.

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But that's the same as e
to the minus lambda 2w.

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00:14:03,020 --> 00:14:09,070

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00:14:09,070 --> 00:14:12,140
Or e to the 2 lambda w.

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So it's going to be--

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Notice the similarity between
this and this.

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00:14:21,110 --> 00:14:24,350

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00:14:24,350 --> 00:14:28,120
The only difference is this
has a 2 lambda in there.

261
00:14:28,120 --> 00:14:31,210
That means that w is an
exponential random variable

262
00:14:31,210 --> 00:14:34,260
with rate 2 lambda.

263
00:14:34,260 --> 00:14:42,720
So then the PDF is going to be
an exponential, whatever it is

264
00:14:42,720 --> 00:14:45,260
for an exponential.

265
00:14:45,260 --> 00:14:48,280
Except with rate 2 lambda.

266
00:14:48,280 --> 00:14:55,470

267
00:14:55,470 --> 00:14:58,550
You can also take the derivative
of this and find

268
00:14:58,550 --> 00:15:00,950
that you get this.

269
00:15:00,950 --> 00:15:05,182
OK, so we're done with
the problems.

270
00:15:05,182 --> 00:15:08,590
We computed some interesting
quantities for the exponential

271
00:15:08,590 --> 00:15:10,650
random variable in this--