# ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

## Problem- ISI MStat PSB 2006 Problem 2

Maximize $$x+y$$ subject to the condition that $$2x^2+3y^2 \le 1$$.

### Prerequisites

Cauchy-Schwarz Inequality

Tangent-Normal

Conic section

## Solution :

This is a beautiful problem, but only if one notices the trick, or else things gets ugly.

Now we need to find the maximum of $$x+y$$ when it is given that $$2x^2+3y^2 \le 1$$. Seeing the given condition we always think of using Lagrange Multipliers, but I find that thing very nasty, and always find ways to avoid it.

So let's recall the famous Cauchy-Schwarz Inequality, $$(ab+cd)^2 \le (a^2+c^2)(b^2+d^2)$$.

Now, lets take $$a=\sqrt{2}x ; b=\frac{1}{\sqrt{2}} ; c= \sqrt{3}y ; d= \frac{1}{\sqrt{3}}$$, and observe our inequality reduces to,

$$(x+y)^2 \le (2x^2+3y^2)(\frac{1}{2}+\frac{1}{3}) \le (\frac{1}{2}+\frac{1}{3})=\frac{5}{6} \Rightarrow x+y \le \sqrt{\frac{5}{6}}$$. Hence the maximum of $$x+y$$ with respect to the given condition $$2x^2+3y^2 \le 1$$ is $$\frac{5}{6}$$. Hence we got what we want without even doing any nasty calculations.

Another nice approach for doing this problem is looking through the pictures. Given the condition $$2x^2+3y^2 \le 1$$ represents a disc whose shape is elliptical, and $$x+y=k$$ is a family of straight parallel lines passing passing through that disc.

Hence the line with the maximum intercept among all the lines passing through the given disc represents the maximized value of $$x+y$$. So, basically if a line of form $$x+y=k_o$$ (say), is a tangent to the disc, then it will basically represent the line with maximum intercept from the mentioned family of line. So, we just need to find the point on the boundary of the disc, where the line of form $$x+y=k_o$$ touches as a tangent. Can you finish the rest and verify weather the maximum intercept .i.e. $$k_o= \sqrt{\frac{5}{6}}$$ or not.

## Food For Thought

Can you show another alternate solution to this problem ? No, Lagrange Multiplier Please !! How would you like to find out the point of tangency if the disc was circular ? Show us the solution we will post them in the comment.

Keep thinking !!

# 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.