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.

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