Results 1 to 9 of 9

Thread: statistics help needed

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    statistics help needed

    Let $\displaystyle X1 X2,...Xn$ be a random sample from exponential distribution with parameter $\displaystyle q$ that is $\displaystyle f(x,q)=(1/q)e^{(-x/q)}$ for $\displaystyle x>0$ and $\displaystyle q>0$.



    Find the Uniformly Minimum Variance Unbiased Estimator for $\displaystyle P[X<c]$ where $\displaystyle c$ is a known positive constant.



    please help me on this. thank you.
    Last edited by Kat-M; May 1st 2009 at 09:36 AM. Reason: hard to read without typing it in tex
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    The suff stat for q is the sum, $\displaystyle S_n=\sum_{i=1}^nX_i$.
    Next we need to find a function of this sum that's unbiased for $\displaystyle P(X<c)=1-e^{-c/q}$.

    Since each $\displaystyle X_i\sim\Gamma(1,q)$, we have $\displaystyle S_n\sim\Gamma(n,q)$.
    Last edited by matheagle; May 1st 2009 at 08:47 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    154
    Would you help a little more?
    How does knowing that $\displaystyle S$~$\displaystyle Gam(n,q)$help me figure out the Uniformly Minimum Variance Estimator?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    You will need to do the Rao-Blackwell Theorem directly.

    Compute $\displaystyle E(I(X_1<c)|S_n=s)$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    154
    i am sorry to ask you this but what is $\displaystyle I(X1<c)$? Is $\displaystyle I$a funtion?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by Kat-M View Post
    i am sorry to ask you this but what is $\displaystyle I(X1<c)$? Is $\displaystyle I$a funtion?
    I stands for indicator.
    It's 1 when that event happens and zero when it doesn't.
    $\displaystyle EI(X\in A)=P(X\in A)$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2008
    Posts
    154
    i thought about this for a long time but still i am not sure how to compute $\displaystyle E(I(X1<c)|S=s)$.
    X1 is part of S. and i dont know how it affects in computing $\displaystyle \int I(X1<c)f(x1|s)$.
    i got $\displaystyle f(x1|s)=\Gamma(n)/s^{n-1}$ So when i integrate it from 0 to c, i got $\displaystyle c\Gamma(n)/s^{n-1}$. did i do it right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Back to $\displaystyle E(I(X_1<c)|S_n=s)$ where s>c.

    We need the density $\displaystyle f(x_1|s_n) ={f(x_1) f(s_{n-1}) \over f(s_n) }$, where $\displaystyle s_{n-1}=x_2+\cdots +x_n$.

    The density of $\displaystyle X_1$ is $\displaystyle f(x_1)=e^{-x_1/q}/q$.

    While $\displaystyle f(s_{n-1})= {s_{n-1}^{n-2} e^{-s_{n-1}/q}\over \Gamma(n-1)q^{n-1}}$ and $\displaystyle f(s_n) ={s_n^{n-1} e^{-s_n/q}\over \Gamma(n)q^n}$

    Noting that $\displaystyle x_1+s_{n-1}=s_n$ we have

    $\displaystyle f(x_1|s_n) ={(n-1) (x_2+\cdots +x_n)^{n-2} \over (x_1+\cdots +x_n)^{n-1}}$

    Thus $\displaystyle P(X_1<c|S_n=s)=\int_0^cf(x_1|S_n=s)dx_1=(n-1)\int_0^c {(s-x_1)^{n-2}dx_1\over s^{n-1}}$

    which gives me $\displaystyle 1-\biggl(1-{c\over s}\biggr)^{n-1}$, which doesn't make sense when n=1.
    Last edited by matheagle; May 2nd 2009 at 09:39 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by matheagle View Post
    Back to $\displaystyle E(I(X_1<c)|S_n=s)$ where s>c.
    We need the density $\displaystyle f(x_1|s_n) ={f(x_1) f(s_{n-1}) \over f(s_n) }$, where $\displaystyle s_{n-1}=x_2+\cdots +x_n$.

    The density of $\displaystyle X_1$ is $\displaystyle f(x_1)=e^{-x_1/q}/q$.

    While $\displaystyle f(s_{n-1})= {s_{n-1}^{n-2} e^{-s_{n-1}/q}\over \Gamma(n-1)q^{n-1}}$ and $\displaystyle f(s_n) ={s_n^{n-1} e^{-s_n/q}\over \Gamma(n)q^n}$

    Noting that $\displaystyle x_1+s_{n-1}=s_n$ we have

    $\displaystyle f(x_1|s_n) ={(n-1) (x_2+\cdots +x_n)^{n-2} \over (x_1+\cdots +x_n)^{n-1}}$

    Thus $\displaystyle P(X_1<c|S_n=s)=\int_0^cf(x_1|S_n=s)dx_1=(n-1)\int_0^c {(s-x_1)^{n-2}dx_1\over s^{n-1}}$

    which gives me $\displaystyle 1-\biggl(1-{c\over s}\biggr)^{n-1}$, which doesn't make sense when n=1.
    I don't think this is right.
    The ideas are good, but the answer seems lame.

    Don't mess with guard duck, he's one mean sob.
    Last edited by matheagle; May 4th 2009 at 04:57 PM. Reason: Added quote. (Given your current avatar, don't you mean lame duck?)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help on statistics needed please help
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 3rd 2009, 01:10 PM
  2. Statistics help needed
    Posted in the Business Math Forum
    Replies: 5
    Last Post: Jan 9th 2009, 10:51 AM
  3. statistics
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 27th 2008, 07:43 PM
  4. URGENT help needed in Statistics!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Mar 31st 2008, 06:29 AM
  5. statistics help needed
    Posted in the Statistics Forum
    Replies: 0
    Last Post: Mar 17th 2007, 04:44 AM

Search Tags


/mathhelpforum @mathhelpforum