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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 !!

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.


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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