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Math Help - Statistics - Uniform Distribution

  1. #16
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    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.
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  2. #17
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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?
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  3. #18
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    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 = ?
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  4. #19
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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.
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  5. #20
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    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)
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  6. #21
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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
    Thanks from AwesomeHedgehog
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  7. #22
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    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 ]
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  8. #23
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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
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  9. #24
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    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?
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  10. #25
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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
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  11. #26
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    Re: Statistics - Uniform Distribution

    And for part f, to get the mean, you just do the integral from 0 to theta of g(y) ?
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  12. #27
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    Re: Statistics - Uniform Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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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