# statistics help needed

• May 1st 2009, 02:07 AM
Kat-M
statistics help needed
Let $X1 X2,...Xn$ be a random sample from exponential distribution with parameter $q$ that is $f(x,q)=(1/q)e^{(-x/q)}$ for $x>0$ and $q>0$.

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

• May 1st 2009, 09:34 PM
matheagle
The suff stat for q is the sum, $S_n=\sum_{i=1}^nX_i$.
Next we need to find a function of this sum that's unbiased for $P(X.

Since each $X_i\sim\Gamma(1,q)$, we have $S_n\sim\Gamma(n,q)$.
• May 1st 2009, 11:11 PM
Kat-M
Would you help a little more?
How does knowing that $S$~ $Gam(n,q)$help me figure out the Uniformly Minimum Variance Estimator?
• May 1st 2009, 11:34 PM
matheagle
You will need to do the Rao-Blackwell Theorem directly.

Compute $E(I(X_1
• May 1st 2009, 11:42 PM
Kat-M
i am sorry to ask you this but what is $I(X1? Is $I$a funtion?
• May 1st 2009, 11:54 PM
matheagle
Quote:

Originally Posted by Kat-M
i am sorry to ask you this but what is $I(X1? Is $I$a funtion?

I stands for indicator.
It's 1 when that event happens and zero when it doesn't.
$EI(X\in A)=P(X\in A)$.
• May 2nd 2009, 02:57 AM
Kat-M
i thought about this for a long time but still i am not sure how to compute $E(I(X1.
X1 is part of S. and i dont know how it affects in computing $\int I(X1.
i got $f(x1|s)=\Gamma(n)/s^{n-1}$ So when i integrate it from 0 to c, i got $c\Gamma(n)/s^{n-1}$. did i do it right?
• May 2nd 2009, 03:03 PM
matheagle
Back to $E(I(X_1 where s>c.

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

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

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

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

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

Thus $P(X_1

which gives me $1-\biggl(1-{c\over s}\biggr)^{n-1}$, which doesn't make sense when n=1.
• May 3rd 2009, 08:45 PM
matheagle
Quote:

Originally Posted by matheagle
Back to $E(I(X_1 where s>c.
We need the density $f(x_1|s_n) ={f(x_1) f(s_{n-1}) \over f(s_n) }$, where $s_{n-1}=x_2+\cdots +x_n$.

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

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

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

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

Thus $P(X_1

which gives me $1-\biggl(1-{c\over s}\biggr)^{n-1}$, 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.