# Thread: Statistics - Uniform Distribution

1. ## Re: Statistics - Uniform Distribution

Yes, I did the integrals from [0, θ], but that's the answer that I kept on getting. I don't know what I'm doing wrong.

2. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
Yes, I did the integrals from [0, θ], but that's the answer that I kept on getting. I don't know what I'm doing wrong.
$E[Y]=\displaystyle{\int_0^\theta} \dfrac{n}{\theta}\left(\dfrac{y}{\theta}\right)^{n-1}~dy$

that's the integral you are doing?

3. ## Re: Statistics - Uniform Distribution

I did that integral and now I got one as my answer for the E(Y)

Then Var(Y) = integral from 0 to θ of (y - 1)^2 * (1/θ) dy = ?

4. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
I did that integral and now I got one as my answer for the E(Y)

Then Var(Y) = integral from 0 to θ of (y - 1)^2 * (1/θ) dy = ?
wat?

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

now plug in $f_{Y_n}(y)=\dfrac{n}{ \theta}\left(\dfrac{y}{ \theta}\right)^{n-1}~~0\leq y \leq \theta$

and $E[Y_n]=$ whatever you got for it.

5. ## Re: Statistics - Uniform Distribution

This is what I did:

E(Y) = the integral from 0 to θ of n*x *(x/θ)^(n-1) dy = (n/n+1)*θ

E(Y^2) = integral from 0 to θ of n*x^2*(x/θ)^(n-1) dy = (n/n+2)*θ^2

Var(Y) = E(Y^2) - (E(Y)^2) = n*(θ/n+1)^2 * (n+2)

6. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
This is what I did:

E(Y) = the integral from 0 to θ of n*x *(x/θ)^(n-1) dy = (n/n+1)*θ

E(Y^2) = integral from 0 to θ of n*x^2*(x/θ)^(n-1) dy = (n/n+2)*θ^2
correct

Var(Y) = E(Y^2) - (E(Y)^2) = n*(θ/n+1)^2 * (n+2)
this formula is fine and if you mean

$Var[Y]=\dfrac{n \theta^2}{(n+1)^2(n+2)}$

then this is also correct

7. ## Re: Statistics - Uniform Distribution

Ok, the for part (d) I got:

F(10) = (10/15)^5 = 0.132

And for part (e):

F(Yr) = P(Yr < y) = P( Y < y^(1/r)) = [y^(n/r)]/θ^n

f(yr) = F'(Yr) = (n/r) * [ y^((n-r)/r) / θ^n ]

8. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
Ok, the for part (d) I got:

F(10) = (10/15)^5 = 0.132

And for part (e):

F(Yr) = P(Yr < y) = P( Y < y^(1/r)) = [y^(n/r)]/θ^n

f(yr) = F'(Yr) = (n/r) * [ y^((n-r)/r) / θ^n ]
d) is correct

e) you need to read what an order statistic is.

Order statistic - Wikipedia, the free encyclopedia

and

Order statistic - Wikipedia, the free encyclopedia

9. ## Re: Statistics - Uniform Distribution

I know what an order statistic is, I just don't know how to apply it to the problem.

For part (d) would the pdf be:

g(y) = [ n!/(r-1)! * (n-r)! ] * [F(y)]^(r-1) * f(y) * [1 - F(y)]^(n-r)

and I would just plug in the F(y) and f(y) from above?

10. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
I know what an order statistic is, I just don't know how to apply it to the problem.

For part (d) would the pdf be:

g(y) = [ n!/(r-1)! * (n-r)! ] * [F(y)]^(r-1) * f(y) * [1 - F(y)]^(n-r)

and I would just plug in the F(y) and f(y) from above?
yep that all looks correct

11. ## Re: Statistics - Uniform Distribution

And for part f, to get the mean, you just do the integral from 0 to theta of g(y) ?

12. ## Re: Statistics - Uniform Distribution

Originally Posted by AwesomeHedgehog
And for part f, to get the mean, you just do the integral from 0 to theta of g(y) ?
y g(y)

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