Order statistics

**ankuoli** · February 28th 2010, 06:24 PM

Let Y1<Y2<Y3 be the order statistics of a random sample from a Uniform $(\theta,1)$ distribution.

1) Show that Y1 is an unbiased estimator of $\theta$
2)Find an unbiased estimator of $\theta$ and that it is unbiased.

Thanks for the help, biased and unbiased have always confused me

**matheagle** · February 28th 2010, 07:24 PM

well, without doing any math I'm sure Y1 is BIASED for theta

unbiased is $E(Y_1)=\theta$

But since the density of Y1 will be between theta and 1 there is no way it's expected value can be either endpoint.

**ankuoli** · March 1st 2010, 06:20 AM

How can I show that Y1 is a biased estimator of $\theta$ ?

**matheagle** · March 1st 2010, 09:13 AM

obtain it's density and compute it's expected value

**ankuoli** · March 1st 2010, 10:40 AM

attempt of an answer to show that Y1 is a biased estimator of $\theta$

Y1 being the minimum of the order statistic, i get that the distribution of the minimum is $f(y1)=n*[1-F(y)]^(n-1)*f(y)$

after plugging in $f(y1)=3*[1-\frac{y-\theta}{1-\theta}]^2*\frac{1}{1-\theta}$

and then to get the expected value of Y1 i get the $\int$ from $\theta$ to 1 of Y1 times f(y1)

am i in the right direction or completely off?

**matheagle** · March 1st 2010, 03:46 PM

$f_{Y_1}(y)=3*\left(1-\frac{y-\theta}{1-\theta}\right)^2*\frac{1}{1-\theta}$

is the density of the first order stat. Next integrate

$yf_{Y_1}(y)$ wrt y from theta to 1.

${3\over 1-\theta}\int_{\theta}^1 y\left(1-\frac{y-\theta}{1-\theta}\right)^2 dy$

**Statistik** · April 3rd 2010, 05:37 PM

Hi,
I have gone through the steps on a similar type of problem - the main difference is that the distribution is $U(\theta_1, \theta_2)$ with n > 10 and $\theta_1 < \theta_2$ .

I found the MLE's to be
$ \theta_1 = X_{(1)} $
and
$ \theta_2 = X_{(n)} $

I'm trying to show whether they are biased on unbiased. I get the density and it looks like Beta (1, n) and Beta (n,1) for $\theta_1$ and $\theta_2$ , respectively, except that I have an extra term of
$ \frac {1} {\theta_2 - \theta_1} $
in the function.

I am not sure what to do with that - do I just multiply the mean of the beta function with that fraction? It seems like I would need to "do something" with it in order for the equation to integrate to 1?
I would love a hint... Thanks!

**Moo** · April 4th 2010, 01:40 AM

Hello Statistik,

Please show your working, because it's impossible to see if you've made a mistake somewhere or what you call "looks like a Beta"

**Statistik** · April 4th 2010, 07:41 AM

Moo,

Sorry. Here it is:

$ iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)}) $

$ f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1}$ ~ $ \frac {1} {\theta_2 - \theta_1} * Beta (1,n)$

I am then trying to get $E[X_{(1)}]$ = ? would this be $\frac{1} {\theta_2 - \theta_1} * \frac {1} {n+1}$ ?

For $X_{(n)}$ , I get the same thing except the distribution is Beta (n,1) and gives $E[X_{(n)}] = \frac{1} {\theta_2 - \theta_1} * \frac {n} {n+1}$ ?

Thanks!

**Moo** · April 4th 2010, 08:41 AM

Originally Posted by Statistik

Moo,

Sorry. Here it is:

$ iid: U(\theta_1, \theta_2) => MLE (\theta_1, \theta_2) = (X_{(1)}, X_{(n)}) $

$ f_{X_{(1)}}(x) = n * (1 - \frac {x-\theta_1} {\theta_2 - \theta_1})^{(n-1)} * \frac {1} {\theta_2 - \theta_1}$ ~ $ \frac {1} {\theta_2 - \theta_1} * Beta (1,n)$

I am then trying to get $E[X_{(1)}]$ = ? would this be $\frac{1} {\theta_2 - \theta_1} * \frac {1} {n+1}$ ?

For $X_{(n)}$ , I get the same thing except the distribution is Beta (n,1) and gives $E[X_{(n)}] = \frac{1} {\theta_2 - \theta_1} * \frac {n} {n+1}$ ?

Thanks!

This is not a Beta distribution... Especially with the support ! With a substitution, you can get something like n(1-t)^(n-1) (the 1/(theta2-theta1) simplifies with the du).
But it's not a Beta distribution.

Your pdf is correct though you have to state the support, in which interval it belongs. Namely $[\theta_1,\theta_2]$

In order to find its expectation, just calculate $\int_{\theta_1}^{\theta_2} x ~f_{X_{(1)}}(x) ~dx$ , like you would do for any other random variable.

**Statistik** · April 4th 2010, 12:43 PM

Moo,

THANK YOU! I won't forget to consider the support again, either. ;-)

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