Hi.
In this video, we're going to compute some useful quantities
for the exponential random variable.
So we're given that x is exponential with rate lambda.
PDF looks like this, and the formula is here.
First question, part a, what's the CDF?
So let's go right in.
The CDF of x is the probability that X is less
than or equal to little x.
Let's look at some cases here.
What if little x is less than 0?
Well, x random variable only takes on these
non-negative values.
And so the probability that X is less than or equal to some
negative number is going to be 0.
On the other hand, if x is greater than or equal to 0, we
do actually have to integrate here.
So to do that, we take the integral from minus infinity
to x of fx of t--
the dummy variable here used is t.
Notice that again, fx of t is going to be 0 for negative
values, so we take the integral here from 0.
And now we plug in for fx of t.
That's minus lambda t dt.
And recall that the integral of u to the a t is 1 over a
times e to the a t.
So here in this case, we'll get lambda,
which is just a constant.
And then a here is going to be negative lambda.
So we get this, 0 to x.
Lambdas cancel and we actually get 1 minus e to the
minus lambda x.
So do this.
And we are done with the CDF.
Now for the expectation.
We use the standard formula, which is minus infinity to
infinity t times fx of t dt.
So again, fx of t is going to be 0 for a negative value.
So we do the integral from 0.
We get 0 to infinity t lambda e to the minus lambda t dt.
Now, you can try all you want to get rid of this t.
It's not going to go even if you try all kinds of u
substitution.
But at the end the day, you're going to have to pull out your
calculus textbook and find the integration by parts
formula, which is--
v du.
So the hope is that this integral is going to be easier
than the one on the left.
Notice that this is the integral of one
of the terms here.
And this is the derivative of one of the terms.
So that may help you decide on how you select u and v.
In our case actually, I'm going to use u as--
t for u.
Because when you take the derivative, it's
going to become 1.
And the derivative is what's going to go in that integral.
So this is going to be dt for du.
And then, dv I'm going to select as
whatever's left over.
It's lambda e to minus lambda t dt.
So v is going to be--
we already did the integral--
minus e to the minus lambda t.
And so if we do this, it's going to be negative t times e
to the minus lambda t.
So that's uv.
Minus v, which is negative e to the minus
lambda t times dt.
That goes from 0 to infinity.
This is evaluated from 0 to infinity.
Well, what does it mean for this to be
evaluated from 0 to infinity?
A better and easier way to look at this is to say, well,
it's going to go from 0 to x.
But then you take the limit as x goes to infinity.
So that's going to help us here.
And this negative--
these negatives cancel.
And we're left with--
let's plug in the bounds.
We're left with negative x minus lambda x plus the
integral of this is going to be 1 over negative lambda e to
the minus lambda t evaluated from 0 to infinity.
All right, so now the limit.
So for the limit, notice that x increases
as x goes to infinity.
And this exponential decays.
So they're kind of competing for each other.
But the exponential is going to win because it decays way
faster than x.
And so this first term is going to go off--
the limit is going to go to 0.
All right.
For this, if you evaluate the balance, the
infinity makes this 0.
And 0, you're going to get 1 over lambda.
So that's 1 over lambda.
All right.
And so the expectation is 1 over lambda.
OK, so now what's the variance?
That's part c, right?
So we use the standard formula for variance, which is this.
We already figured out the expectation.
We just need to figure out the expectation of x squared.
Well, we're just going to follow the same set of steps
from before.
For x squared, it's just going to be t squared, t squared, t
squared, x squared.
The only thing that's going to change is what we choose for u
here, for the u substitution.
So it's going to be t squared.
So the derivative is going to change to 2t dt.
v is going to be exactly the same.
And so here in this term, we get negative 2t e to the
minus lambda t.
But there's a negative sign out here, so the negatives
cancel and we're left with a positive sign here.
This is going to change.
All right.
OK.
So in order to do this integral, we can use a trick.
We can move this--
so there's a 2t here.
We move this 2 in here, leave the t inside.
And you have to leave the t inside.
But multiply by lambda and divide by lambda.
Now, look at that integral.
0 to infinity t times lambda e to the minus lambda t dt.
Exactly the expectation that we computed.
We already did that.
That is just 1 over lambda, so it's 2 over lambda times 1
over lambda.
Again, the limit as x goes to infinity--
the exponential will beat x squared.
No matter what polynomial we put in there, the
exponential's going to win.
So this is going to be 0 still.
This one's going to be 2 over lambda squared.
So we're left with 2 over lambda squared for expectation
of x squared.
And so we have 1 over lambda squared for the variance.
OK, so we're done with the variance.
Part d.
We're given that x1, x2, and x3 are independent and
identically distributed.
They're exponentials with rate lambda.
We're asked for the PDF of z, which is the max
of x1, x2, and x2.
How do we generally find a PDF?
We take the CDF and then take the derivative, right?
We first find the CDF, and then take the derivative.
So let's do that.
So first, let's see.
Part d, find the CDF of z, which is going to be the
probability that Z is less than or equal to little z,
which is going to be equal to the probability that the max
of x1, x2, x3 is less than or equal to z.
And this is going to have the same sort of
structure as before.
If z is less than 0, x1, x2, x3 are positive--
non-negative.
And so this is the probability that if you get little z less
than 0, you're not going to have any probability there.
And so if z is greater than or equal to 0 is where it gets
interesting.
We need to do something special.
So the special thing here is to recognize that the
probability of the max being less than or equal to z is
actually also the probability of each of these random
variables individually being less than or equal to z.
Why is that true?
One way to check whether the events-- these two events are
the same is to check the two directions.
One direction say, if the max of x1, x2, x3 is less than or
equal to z, does that mean x1 is less than or equal to z, x2
is less than or equal to z, and x3 is less
than or equal to z?
Yes.
OK.
And then, if x1, x2, and x3 are individually less than or
equal to z, then the max is also less than or equal to z.
So these two events are equivalent and this is true.
By independence we can break this up.
And we get--
these are all CDFs of the exponential and they
all have this form.
So it's just going to be 1 minus e to the
minus lambda z cubed.
Plug this in here.
And then, try to take the derivative to get the PDF.
Let's see.
So it's going to be the same, like this for z less than 0.
For z greater than or equal to 0, it's going to be the
derivative of this thing.
Derivative of this thing is by chain rule, 3 times 1 minus e
to the minus lambda z squared.
Then the derivative of negative e to the minus lambda
z, that's just lambda e to the minus lambda z.
There we go.
This is the PDF we were looking for.
So last problem.
We're looking for the PDF of w, which is the
min of x1 and x2.
So let's try this as a similar approach.
Try the same thing, actually.
See if it works.
So w, w, w, w, min, less than or equal to w.
OK.
So let's see if this works.
Is it true that the min--
if the min of x1 and x2 is less than or equal to w, that
each of them is less than or equal to w?
No, right?
X1 could be less than or equal to w and x2 could
be bigger than w.
And the min could still be less than or equal to w.
So that's definitely not true.
So what do we do here?
The trick is to flip it and say we want to compute the min
of x1 and x2 being greater than w.
In that case, let's check if we can do this trick.
If the min of x1 and x2 is greater than w, then clearly
x1 is bigger than w and x2 is bigger than w.
And if x1 and x2 are individually bigger than w,
then clearly the min's also bigger than w.
So this works.
And now we can use independence as before.
And for this, this is just 1 minus the CDF here.
So it's just going to be e to the minus lambda
w for each of them.
But that's the same as e to the minus lambda 2w.
Or e to the 2 lambda w.
So it's going to be--
Notice the similarity between this and this.
The only difference is this has a 2 lambda in there.
That means that w is an exponential random variable
with rate 2 lambda.
So then the PDF is going to be an exponential, whatever it is
for an exponential.
Except with rate 2 lambda.
You can also take the derivative of this and find
that you get this.
OK, so we're done with the problems.
We computed some interesting quantities for the exponential
random variable in this--