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Thread: Order statistics help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Order statistics help

    Q: Let $\displaystyle Y_{1},...,Y_{n}$ by independent, uniformly distributed random variables on the interval $\displaystyle [0,\theta]$. Find the

    a) Probability distribution function of $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$.

    b) density function of $\displaystyle Y_{(n)}$.

    c) mean and variance of $\displaystyle Y_{(n)}$.

    A: If I can figure out a) and identify the pdf, the cdf (part b) and mean / variance (part c) should follow. So, here is my attempt

    Let $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$. Then, by Thereom blah, we have

    $\displaystyle f_{Y_{(n)}}=g_{(n)}(y_{n})$$\displaystyle =
    \frac{n!}{(n-1)!(n-k)!}[F(y_{k})]^{k-1}[1-F(y_{k})]^{n-k}f(y_{k})$,$\displaystyle -\infty<y_{k}<\infty$ with $\displaystyle k=n$.

    Thus, $\displaystyle f_{Y_{(n)}}=g_{(n)}(y_{n})=
    \frac{n!}{(n-1)!(n-n)!}[\frac{y_{n}}{\theta}]^{n-1}[1-\frac{y_{n}}{\theta}]^{n-n}
    \frac{1}{\theta}=
    \frac{n}{\theta}[\frac{y_{n}}{\theta}]^{n-1}, 0<y_{k}<\theta$.

    I do not recognize the density function $\displaystyle \frac{n}{\theta}[\frac{y_{n}}{\theta}]^{n-1}, 0<y_{k}<\theta$ and its support.

    Any help would be great. Not sure where I went wrong.

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let $\displaystyle Y_{1},...,Y_{n}$ by independent, uniformly distributed random variables on the interval $\displaystyle [0,\theta]$. Find the

    a) Probability distribution function of $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$.
    Observe:

    $\displaystyle F_{Y_{(n)}}(y)=P(Y_{(n)}<y)=\prod_{i=1}^n P(Y_i<y)=\begin{cases}0&,\ y<0 \\ \left(\frac{y}{\theta}\right)^n & ,\ 0 \le y \le \theta \\1 & , \ y>\theta \end{cases} $


    CB
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  3. #3
    MHF Contributor matheagle's Avatar
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    Just because you can't recognize this, doesn't mean it's wrong.
    AND there is only one y, whether you call it $\displaystyle y_{k}$ or $\displaystyle y_{n}$.....

    $\displaystyle \frac{n}{\theta}[\frac{y}{\theta}]^{n-1}, 0<y<\theta$

    is the derivative of the CDF that CB derived.

    If you are curious....
    If we let $\displaystyle X=Y/\theta$ then X is a Beta and you can obtain the mean and variance that way if you wish.
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by matheagle View Post
    If you are curious....
    If we let $\displaystyle X=Y/\theta$ then X is a Beta and you can obtain the mean and variance that way if you wish.
    I am not seeing how X is a beta random variable. What would my alpha and beta be?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Danneedshelp View Post
    I am not seeing how X is a beta random variable. What would my alpha and beta be?
    $\displaystyle \beta=1$ and $\displaystyle \alpha=N$

    CB
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