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# ISI MStat PSB 2006 Problem 9 | Consistency and MVUE

This is a very simple sample problem from ISI MStat PSB 2006 Problem 9. It’s based on point estimation and finding consistent estimator and a minimum variance unbiased estimator and recognizing the subtle relation between the two types. Go for it!

This is a very simple sample problem from ISI MStat PSB 2006 Problem 9. It’s based on point estimation and finding consistent estimator and a minimum variance unbiased estimator and recognizing the subtle relation between the two types. Go for it!

## Problem– ISI MStat PSB 2006 Problem 9

Let $X_1,X_2,……$ be i.i.d. random variables with density $f_{\theta}(x), \ x \in \mathbb{R}, \ \theta \in (0,1)$, being the unknown parameter. Suppose that there exists an unbiased estimator $T$ of $\theta$ based on sample size 1, i.e. $E_{\theta}(T(X_1))=\theta$. Assume that $Var(T(X_1))< \infty$.

(a) Find an estimator $V_n$ for $\theta$ based on $X_1,X_2,……,X_n$ such that $V_n$ is consistent for $\theta$ .

(b) Let $S_n$ be the MVUE( minimum variance unbiased estimator ) of $\theta$ based on $X_1,X_2,….,X_n$. Show that $\lim_{n\to\infty}Var(S_n)=0$.

### Prerequisites

Consistent estimators

Minimum Variance Unbiased Estimators

Rao-Blackwell Theorem

## Solution :

Often, problems on estimation seems a bit of complicated and we feel directionless, but most cases its always beneficiary do go with the flow.

Here, it is given that $T$ is an unbiased estimator of $\theta$ based on one observation, and we are to find a consistent estimator for $\theta$ based on a sample of size $n$. Now first, we should consider what are the requisition of an estimator to be consistent?

• The required estimator $V_n$ have to be unbiased for $\theta$ as $n \uparrow \infty$ . i.e. $\lim_{n \uparrow \infty} E_{\theta}(V_n)=\theta$.
• The variance of the would be consistent estimator must converge to 0, as n grows large .i.e. $\lim_{n \uparrow \infty}Var_{\theta}(V_n)=0$.

First thing first, let us fulfill the unbiased criteria of $V_n$, so, from each of the observation from the sample , $X_1,X_2,…..,X_n$ , of size n, we can get as set of n unbiased estimator of $\theta$ $T(X_1), T(X_2), ….., T(X_n)$. So, can we write $V_n=\frac{1}{n} \sum_{i=1}^n(T(X_i)+a)$ ? where $a$ is a constant, ( kept for generality). Can you verify that $V_n$ satisfies the first requirement of being a consistent estimator?

Now, proceeding towards fulfilling the final requirement, that is the variance of $V_n$ converges to 0 as $n \uparrow \infty$ . Since we have defined $V_n$ based on $T$, and it is given that $Var(T(X_i))$ exists for $i \in \mathbb{N}$, and $X_1,X_2,…X_n$ are i.i.d. (which is a very important realization here), leads us to

$Var(V_n)= \frac{Var(T(X_1))}{n}$ , (why ??) . So, clearly, $Var(V_n) \downarrow 0$ a $n \uparrow \infty$, fulfilling both required conditions for being a consistent estimator. So, $V_n= \sum_{i=1}^n(T(X_i)+a)$ is a consistent estimator for $\theta$.

(b) For this part one may also use Rao-Blackwell theorem, but I always prefer using as less formulas and theorem as possible, and in this case we can do the required problem from the previous part. Since given $S_n$ is MVUE for $\theta$ and we found that $V_n$ is consistent for $\theta$, so, by the nature of MVUE,

$Var(S_n) \le Var(V_n)$, so as n gets bigger, $\lim_{ n \to \infty} Var(S_n) \le \lim{n \to infty} Var(V_n) \Rightarrow \lim_{n \to \infty}Var(S_n) \le 0$

again, $Var(S_n) \ge 0$, so, $\lim_{n \to \infty }Var(S_n)= 0$. Hence, we conclude.

## Food For Thought

Lets extend this problem a liitle bit just to increase the fun!!

Let, $X_1,….,X_n$ are independent but not identical, but still $T(X_1),T(X_2),…..,T(X_n)$, remains unbiased of $\theta$ , and $Var(T(X_i)= {\sigma_i}^2$, and

$Cov(T(X_i),T(X_j))=0$ if $i \neq j$.

Can you show that of all the estimators of form $\sum a_iT(X_i)$, where $a_i$’s are constants, and $E_{\theta}(\sum a_i T(X_i))=\theta$, the estimator,

$T*= \frac{\sum \frac{T(X_i)}{{\sigma_i}^2}}{\sum\frac{1}{{\sigma_i}^2}}$ has minimum variance.

Can you find the variance ? Think it over !!

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