Estimators...

**Moo** · August 20th 2009, 12:28 PM

Hello,

Okay, I don't know at all if I did these questions correctly...

Let $f(x;\theta)=\begin{cases} 1-\theta & \text{ if } -1\leq x\leq 0 \\ 2\theta (1-x) & \text{ if } 0<x\leq 1 \\ 0 & \text{ if } x>1 \end{cases}$

and where $0<\theta<1$

Let X be a rv with pdf f.
We know that $\mathbb{E}(X)=\tfrac 56 \cdot\theta-\tfrac 12$

Let $(X_1,\dots,X_n)$ be a sample of iid rvs following the same distribution as X.

Let the rv $Z_i=\begin{cases} 1 &\text{ if } X_i>0 \\ 0 &\text{ otherwise}\end{cases}$ and let $Y=\sum_{i=1}^n Z_i$

Preliminary questions : we proved that Y/n is an unbiased and convergent estimator for $\theta$ .
We also proved that Y follows a binomial distribution with parameter $\theta$

First question : calculate an estimator T for $\theta$ by the method of moments.
--------------------
So for this one, ... We know that $\theta=\left(\mathbb{E}(X)+\tfrac 12\right) \cdot \tfrac 65$

So we can take $T=\left(\tfrac Yn+\tfrac 12\right) \cdot \tfrac 65$ , right ?

I thought of using observations of $Z_i$ but then there is a table of values for $X_i$ , not $Z_i$ . So I considered it was too easy...

--------------------

Second question : prove that T converges in probability to $\theta$ and calculate its bias
--------------------
So in order to prove that it converges, I used the LLN (the one stating the convergence in probability), then used Slutsky's theorem.

Then for its bias, I find 0... is it normal ?? This is where I am the most doubtful...

--------------------

My own questions :
- is the estimator by the method of moments unique ?
- is it always an unbiased estimator ?

it may sound stupid, but I don't know how to be sure...

Thanks in advance !

**matheagle** · August 20th 2009, 03:29 PM

do you mean consistent estimator for

?

And f(x)= 0 for x<-1 too

Your mean of X is correct.
But I would think that the MOM estimator of theta would be the solution to

${5\over 6}\hat\theta-{1\over 2}={\sum_{i=1}^nXi\over n}=\bar X$

You're using the Z's when you use that Y, those are the truncated X's.
I'm just setting the population mean of the X's to IT's sample mean.

Since

$\bar X \buildrel P\over\to {5\over 6}\theta -{1\over 2}$

$\hat\theta={6\over 5}\biggl(\bar X +{1\over 2}\biggr)\buildrel P\over\to \theta$

**matheagle** · August 20th 2009, 11:12 PM

IF you want to use the Y in the MOM estimation, you should set

$\Hat E(Z_i)={\sum_{i=1}^n Z_i\over n}={Y\over n}$

Now $E(Z_i)=P\{X_i>0\}=2\theta \int_0^1 (1-x)dx=\theta$

In that case you get $\Hat\theta={Y\over n}$

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