Statistics - Uniform Distribution

Let X1, X2, ... Xn be independent, uniformly distributed random variables on the interval [0, theta].

a.) Find the c.d.f. of Yn = max(X1, X2, ..., Xn).

b.) Find the p.d.f. of Yn = max(X1, X2, ..., Xn).

c.) Find the mean and variance of Yn = max(X1, X2, ..., Xn).

d.) Suppose that the number of minutes that you need to wait for a bus is uniformly distributed on the interval [0, 15]. If you take the bus five times, what is the probability that your longest wait is less than 10 minutes?

e.) Find the p.d.f. of Yr, the rth-order statistic, where r is an integer between 1 and n.

f.) Find the mean of Yr.

For part a, I got the c.d.f of Yn = [F(y)]^n

For part b, I got the p.d.f pf Yn = g(y) = n*[F(y)]^(n-1) * f(y)

I'm not sure if I am doing this right. And I'm lost for the rest of the problem with what to do.

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** Let X1, X2, ... Xn be independent, uniformly distributed random variables on the interval [0, theta].

a.) Find the c.d.f. of Yn = max(X1, X2, ..., Xn).

b.) Find the p.d.f. of Yn = max(X1, X2, ..., Xn).

For part a, I got the c.d.f of Yn = [F(y)]^n

For part b, I got the p.d.f pf Yn = g(y) = n*[F(y)]^(n-1) * f(y)

I'm not sure if I am doing this right. And I'm lost for the rest of the problem with what to do.

if by $F(y)$ you mean $F_X(y)$ then yes (a) is correct. But you should be able to explicitly determine what $F_X(y)$ is given that $X_i$ is uniform on $[0, \theta]$. The same goes for part (b). You are given $f_X(y)$.

(c) you should know how to do, you found the mean and variance of a distribution in one of the other problems.

(d) just apply the results above.

(e) you can either try and work this out yourself or you can look it up on the web. There are plenty of derivations of this available. Working it out isn't that hard. Just think about what it means to be the ith order statistic in terms of the $X_i$'s.

(f) is like (c)

Re: Statistics - Uniform Distribution

For part (a) and (b) are you saying that I should do:

(a) P[ Yn ≤ y] = P[ X1 ≤ y, X2 ≤ y, ... , Xn ≤ y ] = P[ X1 ≤ y] * P[ X2 ≤ y] * ... * P[ Xn ≤ y] = [F(y)]^n

(b) f(y) = derivative of the cdf = n * [F(y)]^(n-1)

(c) Is the formula for the mean and variance the same as another problem that I did?

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** For part (a) and (b) are you saying that I should do:

(a) P[ Yn ≤ y] = P[ X1 ≤ y, X2 ≤ y, ... , Xn ≤ y ] = P[ X1 ≤ y] * P[ X2 ≤ y] * ... * P[ Xn ≤ y] = [F(y)]^n

(b) f(y) = derivative of the cdf = n * [F(y)]^(n-1)

(c) Is the formula for the mean and variance the same as another problem that I did?

dude...

You're told that the X's are distributed uniform[0,$\theta$]. So what is $F_X(x)$ ? What is $f_X(x)$? Just plug these into what you've got to make it explicit for uniform rvs.

The formula for the mean and the variance is not exactly the same as the other problem but the method is.

You'll have $f_{Y_n}(y)$

$\large E[Y_n]=\displaystyle{\int_{-\infty}^{\infty}} y f_{Y_n}(y)~dy$

$\large Var[Y_n]=\displaystyle{\int_{-\infty}^{\infty}}\left(y-E[Y_n]\right)^2 f_{Y_n}(y)~dy$

Re: Statistics - Uniform Distribution

For part a, then

The cdf is:

{ 0 for x < 0

{ (x-0)/(θ -0) for 0 ≤ x < θ

{ 1 for x ≥ θ

The pdf is:

{ 1/(θ - 0) for 0 ≤ x ≤ θ

{ 0 for x < 0 or x > θ

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** For part a, then

The cdf is:

{ 0 for x < 0

{ (x-0)/(θ -0) for 0 ≤ x < θ

{ 1 for x ≥ θ

what are the -0's ? do you mean $\dfrac{x}{\theta}$ ? if so good, that's right.

Note this is just the individual pdf. you still have to take this the nth power.

Quote:

The pdf is:

{ 1/(θ - 0) for 0 ≤ x ≤ θ

{ 0 for x < 0 or x > θ

do you just mean $\dfrac{1}{\theta}$ ? I hope so because that's correct too.

so use these to find your CDF and pdf of the max which you already found the formula for in terms of

$F_X(y)$ and $f_X(y)$

Re: Statistics - Uniform Distribution

I put the minus zeros in there just because it's part of the formula, I know it wasn't really necessary to do that.

To find the cdf, would I do this:

Fx(y) = the integral from 0 to θ of x/θ dx?

And for the pdf:

fx(y) = the integral from 0 to θ of 1/θ dx?

I'm confused what you mean when you say max

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** I put the minus zeros in there just because it's part of the formula, I know it wasn't really necessary to do that.

To find the cdf, would I do this:

Fx(y) = the integral from 0 to θ of x/θ dx?

And for the pdf:

fx(y) = the integral from 0 to θ of 1/θ dx?

I'm confused what you mean when you say max

remember the problem was originally to find the distribution of the max and min of n uniform rvs.

what you did above is just the cdf and pdf of a single uniform rv. You already calculated the formula for the cdf and pdf of the max and min in terms of these a few posts back. Just plug these into that formula.

Re: Statistics - Uniform Distribution

Oh, so for part (a):

cdf = [F(y)]^n = (x/θ)^n

and for part (b)

pdf = n * [F(y)]^(n-1) = n*(x/θ)^(n - 1)

Re: Statistics - Uniform Distribution

Is part (a) and (b) correct?

If so, then moving onto part (c) I got:

E(Y) = the integral from -∞ to ∞ of y*n*(x/θ)^(n - 1) dy = 1/2 *θ*n*x*(x/θ)^(n-2)

Var(Y) = the integral from -∞ to ∞ of (y - 1/2 *θ*n*x*(x/θ)^(n-2))^2 * n*(x/θ)^(n - 1) dy = something complicated

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** Oh, so for part (a):

cdf = [F(y)]^n = (x/θ)^n

and for part (b)

pdf = n * [F(y)]^(n-1) = n*(x/θ)^(n - 1)

you left out a piece, look at your formula for the pdf again.

Re: Statistics - Uniform Distribution

For the pdf, do I also multiply it by 1/θ ?

So, it would be:

pdf = n*(x/θ)^(n - 1) * 1/θ

Re: Statistics - Uniform Distribution

Re: Statistics - Uniform Distribution

Then

E(Y) = the integral from -∞ to ∞ of y*n*(x/θ)^(n - 1) dy = 1/2 *θ*n*x*(x/θ)^(n-2)

Var(Y) = the integral from -∞ to ∞ of (y - 1/2 *θ*n*x*(x/θ)^(n-2))^2 * n*(x/θ)^(n - 1) * 1/θ dy = I have the answer, but it is extremely complicated, so I'm not sure if it is correct.

Re: Statistics - Uniform Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** Then

E(Y) = the integral from -∞ to ∞ of y*n*(x/θ)^(n - 1) dy = 1/2 *θ*n*x*(x/θ)^(n-2)

Var(Y) = the integral from -∞ to ∞ of (y - 1/2 *θ*n*x*(x/θ)^(n-2))^2 * n*(x/θ)^(n - 1) * 1/θ dy = I have the answer, but it is extremely complicated, so I'm not sure if it is correct.

the integrals range over $[0,\theta]$