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# ISI MStat PSB 2004 Problem 6 | Minimum Variance Unbiased Estimators

This is a very beautiful sample problem from ISI MStat PSB 2004 Problem 6. It's a very simple problem, and its simplicity is its beauty . Fun to think, go for it !!

## Problem- ISI MStat PSB 2004 Problem 6

Let $$Y_1,Y_2.Y_3$$, and $$Y_4$$ be four uncorrelated random variables with

$$E(Y_i) =i\theta ,$$ $$Var(Y_i)= i^2 {\sigma}^2,$$ , $$i=1,2,3,4$$ ,

where $$\theta$$ and $$\sigma$$ (>0) are unknown parameters. Find the values of $$c_1,c_2,c_3,$$ and $$c_4$$ for which $$\sum_{i=1}^4{c_i Y_i}$$ is unbiased for $$\theta$$ and has least variance.

### Prerequisites

Unbiased estimators

Minimum-Variance estimators

Cauchy-Schwarz inequality

## Solution :

This is a very simple and cute problem, just do as it is said...

for , $$\sum_{i=1}^4{c_i Y_i}$$ to be an unbiased estimator for $$\theta$$ , then it must satisfy,

$$E(\sum_{i=1}^4{c_i Y_i} )= \theta \Rightarrow \sum_{i=1}^4{c_i E(Y_i)}= \theta \Rightarrow \sum_{i=1}^4{c_i i \theta} = \theta$$

so, $$\sum_{i=1}^4 {ic_i}=1 .$$ ......................(1)

So, we have to find $$c_1,c_2,c_3,$$ and $$c_4$$, such that (1), is satisfied . But hold on there is some other conditions also.

Again, since the given estimator will also have to be minimum variance, lets calculate the variance of $$\sum_{i=1}^4{c_i Y_i}$$ ,

$$Var(\sum_{i=1}^4{c_i Y_i})= \sum_{i=1}^4{c_i}^2Var( Y_i)=\sum_{i=1}^4{i^2 {c_i}^2 {\sigma}^2 }.$$...............................................(2)

So, for minimum variance, $$\sum_{i=1}^4{i^2{c_i}^2 }$$ must be minimum in (2).

So, we must find $$c_1,c_2,c_3,$$ and $$c_4$$, such that (1), is satisfied and $$\sum_{i=1}^4{i^2{c_i}^2 }$$ in (2) is minimum.

so, minimizing $$\sum_{i=1}^4{i^2{c_i}^2 }$$ when it is given that $$\sum_{i=1}^4 {ic_i}=1$$ ,

What do you think, what should be our technique of minimizing $$\sum_{i=1}^4{i^2{c_i}^2 }$$ ???

For, me the beauty of the problem is hidden in this part of minimizing the variance. Can't we think of Cauchy-Schwarz inequality to find the minimum of, $$\sum_{i=1}^4{i^2{c_i}^2 }$$ ??

So, using CS- inequality, we have,

$$(\sum_{i=1}^4{ic_i})^2 \le n \sum_{i=1}^4{i^2{c_i}^2} \Rightarrow \sum_{i=1}^4 {i^2{c_i}^2} \ge \frac{1}{n}.$$ ...........(3). [ since $$\sum_{i=1}^4 {ic_i}=1$$ ].

now since $$\sum_{i=1}^4{i^2{c_i}^2 }$$ is minimum the equality in (3) holds, i.e. $$\sum_{i=1}^4{i^2{c_i}^2 }=\frac{1}{n}$$ .

and we know the equality condition of CS- inequality is, $$\frac{1c_1}{1}=\frac{2c_2}{1}=\frac{3c_3}{1}=\frac{4c_4}{1}=k$$ (say),

then $$c_i= \frac{k}{i}$$ for i=1,2,3,4 , where k is some constant .

Again since, $$\sum_{i=1}^4{ic_i} =1 \Rightarrow 4k=1 \Rightarrow k= \frac{1}{4}$$ . Hence the solution concludes .

## Food For Thought

Let's, deal with some more inequalities and behave Normal !

Using, Chebyshev's inequality we can find a trivial upper bound for $$P(|Z| \ge t)$$, where $$Z \sim n(0,1)$$ and t>0 ( really !! what's the bound ?). But what about some non-trivial bounds, sharper ones perhaps !! Can you show the following,

$$\sqrt{\frac{2}{\pi}}\frac{t}{1+t^2}e^{-\frac{t^2}{2}} \le P(|Z|\ge t) \le \sqrt{\frac{2}{\pi}}\frac{e^{-\frac{t^2}{2}}}{t}$$ for all t>0.

also, verify this upper bound is sharper than the trivial upper bound that one can obtain.

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### One comment on “ISI MStat PSB 2004 Problem 6 | Minimum Variance Unbiased Estimators”

1. abhishek sinha says:

In the final step, k must be equal to 1/4 (due to unbiasedness).