Hyperbola & Tangent | ISI MStat 2016 Problem 1 | PSB Sample

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 1. This is based on finding the minimum value of a function subjected to the restriction.

ISI MStat 2016 Problem 1

Let $ x, y$ be real numbers such that $ x y=10 $ . Find the minimum value of $|x+y|$ and all also find all the points $(x, y)$ where this minimum value is achieved.
Justify your answer.

Prerequisites

Rectangular Hyperbola & it's symmetry
Tangent
Derivative test

Solution

(a) Using graph

The equation $ xy=10={(\sqrt{10})}^2 $ represents the equation of rectangular hyperbola with foci are $ (- \sqrt{10},- \sqrt{10}) $ and $ (\sqrt{10}, \sqrt{10} ) $ .

ISI MStat 2016 Problem 1 graph — **Fig-1**

Now , $ |x+y|=c \Rightarrow $ $ x+y=\pm c $ , which looks somewhat like this ,

ISI MStat 2016 Problem 1 Figure 2 — **Fig-2**

we have to find the the minimum value of $|x + y|$ subject to the restriction that $ xy=10 $ . If we move $ |x+y|=c $ along $ xy =10 $ by varying c , then local minimum can occur at the points where the level curve $ |x+y|=c $ touch $ xy =10 $ . Now as both the rectangular hyperbola and |x+y|=c are symmetric about $ x+y=0 $ for $ c \ne 0 $ and $x=y$ , the level curve will touch $ xy =10 $ when x+y=c and x+y=-c both are tangent to the curve $ xy =10 $ . And tangents occurs at the foci of $ xy =10 $ i.e at $ (- \sqrt{10},- \sqrt{10}) $ and $ (\sqrt{10}, \sqrt{10} ) $ .

ISI MStat 2016 Problem 1 Figure 3 — **Fig-3**

Hence , the minimum value of $|x + y|$ are $|\sqrt{10} +\sqrt {10} |$ and $|-\sqrt{10}- \sqrt{10}|$ both gives the same value $ 2 \sqrt{10} $ .

Therefore , the minimum value of $|x + y|$ is $ 2 \sqrt{10}$ and it attains it's minimum at $ (\sqrt{10} , \sqrt{10} ) $ and $ ( - \sqrt{10} , -\sqrt{10}) $ .

(b) Using Derivative test

$ |x+y|= |x+ \frac{10}{x} |$ as we are given that $ xy=10 $

Let, $ f(x) =|x+ \frac{10}{x}|$ then we have to find the minimum value of f(x)

Now ,as the function is not defined at x=0 and also x=0 can't give the minimum value of |x+y| due to the condition that xy=10. So, we will study f(x) for two cases when x>0 and when x<0 .

$ f(x) = \begin{cases} x+ \frac{10}{x} & ,x > 0 \\ -(x+ \frac{10}{x} ) & ,x < 0 \end{cases} $

$ f'(x)=\begin{cases} 1- \frac{10}{x^2} & ,x > 0 \\ -(1- \frac{10}{x^2} ) & ,x < 0 \end{cases} $

$ f'(x)=0 \Rightarrow $ $ x= \pm \sqrt{10} $

$ f''(x)= \begin{cases} \frac{20}{x^3} & ,x > 0 \\ -\frac{20}{x^3} & ,x < 0 \end{cases} $

So,$ f''(x) >0$ for $ x= \pm \sqrt{10} $

Hence f(x) attains it's minimum value at $ (\sqrt{10} , \sqrt{10} ) $ and $ - \sqrt{10} , -\sqrt{10}) $ and minimum value is $ 2 \sqrt{10} $.

Challenge Problem

Let $ x_1, x_2 ,..., x_n $ be be positive real numbers such that $ \prod_{i=1}^{n} x_{i} = 10 $ . Find the minimum value of $ \sum_{i=1}^{n} x_{i} $.

Cycles, Symmetry, and Counting | ISI MStat 2016 PSB | Problem 2

This problem is a beautiful and elegant application of basic counting principles, symmetry and double counting principles in combinatorics. This is Problem 2 from ISI MStat 2016 PSB.

Problem

Determine the average value of
$$
i_{1} i_{2}+i_{2} i_{3}+\cdots+i_{9} i_{10}+i_{10} i_{1}
$$ taken over all permutations $i_{1}, i_{2}, \ldots, i_{10}$ of $1,2, \ldots, 10$.

Prerequisites

Permutation and Combination
Symmetry Argument in Counting

Solution

The problem may seem mind boggling at first, when you will even try to do it for $n = 4$, instead of $n = 10$.

But, in mathematics, symmetry is really intriguing. Let's see how a symmetry argument holds here. It is just by starting to count. Let's see this problem in a geometrical manner.

$ i_{1}\to i_{2} \to i_{3} \to \cdots \to i_{9} \to i_{10} \to i_{1}$ is sort of a cycle right?

Now, the symmetry argument starts from this symmetric figure.

We will do the problem for general $n$.

Central Idea: Let's fix a pair say [ 4 - 5 ], we will see in all the permutations, in how many times, [ 4 - 5 ] can occur.

We will see that there is nothing particular about [ 4 - 5 ], and this is the symmetry argument. Therefore, the number is symmetric along with all such pairs.

Observe, along every such cycle containing [ 4 - 5 ], there are three parameters:

The position of the [ 4 - 5 ] in which edge of the cycle?
The permutation of that [ 4 - 5 ], as [ 4 - 5 ] or [ 5 - 4 ].
The number of arrangements of the remaining $ n - 2$ numbers.

The number corresponding to the above questions are the following:

$n$ options of position of edge since, there are $n$ edges.
2! = 2 ways of arranging.
$ (n-2)! $ ways of arranging the rest of the $ n -2$ numbers.

So, in total [ 4 - 5 ] edge will occur $ n \times 2! \times (n-2)! $ times.

By the symmetry argument, every edge [$ i - j$], will occur $ n \times 2! \times (n-2)! $ times.

Thus, when we sum over all such permutations, we get the following

$$ \sum_{i, j = 1; i \neq j}^{n} {ij} \times \text{number of times [i - j] pair occur} $$

$$ = \sum_{i, j = 1; i \neq j}^{n} {ij} \times n \times 2! \times (n-2)! = n \times (n-2)! \times \sum_{i, j = 1; i \neq j}^{n} {2ij} $$

Now, there are $ n!$ permutations in total. So, to take the average, we divide by $ n!$ to get $$ \frac{\sum_{i, j = 1; i \neq j}^{n} {2ij}}{n-1} $$

$$ = \frac{(\sum_{i}^{n} i)^2 - \sum_{i}^{n} i^2 }{n-1} $$

$$ = \frac{\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}}{n-1} = \frac{n(n+1)(3n+2)}{12} $$

Edit 1:

One of the readers, Vishal Routh has shared his solution using Conditional Expectation, I am sharing his solution in picture format.

Video Solution:

Restricted Regression Problem | ISI MStat 2017 PSB Problem 7

This problem is a regression problem, where we use the ordinary least square methods, to estimate the parameters in a restricted case scenario. This is ISI MStat 2017 PSB Problem 7.

Problem

Consider independent observations ${\left(y_{i}, x_{1 i}, x_{2 i}\right): 1 \leq i \leq n}$ from the regression model
$$
y_{i}=\beta_{1} x_{1 i}+\beta_{2} x_{2 i}+\epsilon_{i}, i=1, \ldots, n
$$ where $x_{1 i}$ and $x_{2 i}$ are scalar covariates, $\beta_{1}$ and $\beta_{2}$ are unknown scalar
coefficients, and $\epsilon_{i}$ are uncorrelated errors with mean 0 and variance $\sigma^{2}>0$. Instead of using the correct model, we obtain an estimate $\hat{\beta_{1}}$ of $\beta_{1}$ by minimizing
$$
\sum_{i=1}^{n}\left(y_{i}-\beta_{1} x_{1 i}\right)^{2}
$$ Find the bias and mean squared error of $\hat{\beta}_{1}$.

Prerequisites

Ordinary Least Square Method
Minimizing the Square Loss Error Function
Multiple Regression
Mean Square Error = $\text{Variance} + \text{Bias}^2$.
Bias

Solution

It is sort of a restricted regression problem because maybe we have tested the fact that $\beta_2 = 0$. Hence, we are interested in the estimate of $\beta_1$ given $\beta_2 = 0$. This is essentially the statistical significance of this problem, and we will see how it turns out in the estimate of $\beta_1$.

Let's start with some notational nomenclature.
$ \sum_{i=1}^{n} a_{i} b_{i} = s_{a,b} $

Let's minimize $ L(\beta_1) = \sum_{i=1}^{n}\left(y_{i}-\beta_{1} x_{1 i}\right)^{2}$ by differentiating w.r.t $\beta_1$ and equating to 0.

$ \frac{dL(\beta_1)}{d\beta_1}\sum_{i=1}^{n}\left(y_{i}-\beta_{1} x_{1 i}\right)^{2} = 0$

$ \Rightarrow \sum_{i=1}^{n} x_{1 i} \left(y_{i}-\beta_{1} x_{1 i}\right) = 0 $

$ \Rightarrow \hat{\beta_1} = \frac{s_{x_{1},y}}{s_{x_{1},x_{1}}} $

From, the given conditions, $ E(Y_{i})=\beta_{1} X_{1 i}+\beta_{2} X_{2 i}$.

$ \Rightarrow E(s_{X_{1},Y}) = \beta_{1}s_{X_{1},X_{1}} +\beta_{2} s_{X_{1},X_{2}} $.

Since, $x's$ are constant, $ E(\hat{\beta_1}) = \beta_{1} +\beta_{2} \frac{s_{X_{1},X_{2}}}{s_{X_{1},X_{1}}} $.

$ Bias(\hat{\beta_1}) = \beta_{2} \frac{s_{X_{1},X_{2}}}{s_{X_{1},X_{1}}} $.

Thus, observe that the more $ \beta_2 $ is close to 0, the more bias is close to 0.

From, the given conditions,

$ Y_{i} - \beta_{1} X_{1 i} - \beta_{2} X_{2 i}$ ~ Something$( 0 , \sigma^2$).

$ \hat{\beta_1} = \frac{s_{x_{1},y}}{s_{x_{1},x_{1}}}$ ~ Something$( E(\hat{\beta_{1}}) , Var(\hat{\beta_1}))$.

$Var(\hat{\beta_1}) = \frac{\sum_{i=1}^{n} x_{1i}^2 Var(Y_{i})}{s_{X_1, X_1}^2} = \frac{\sigma^2}{s_{X_1, X_1}} $

$ MSE(\hat{\beta_1}) = Variance + \text{Bias}^2 = \frac{\sigma^2}{s_{X_1, X_1}} + \beta_{2}^2(\frac{s_{X_{1},X_{2}}}{s_{X_{1},X_{1}}})^2$

Observe, that even the MSE is minimized if $\beta_2 = 0$.

Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

Function and Symmetry - AIME I 1984

A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval $-1000 \leq x\leq 1000$.

is 107
is 401
is 840
cannot be determined from the given information

Key Concepts

Functions

Symmetry

Number Theory

Check the Answer

Answer: is 401.

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

Try with Hints

by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Then least number of roots as 401.

ISI MStat 2016 Problem 1

Prerequisites

Solution

Challenge Problem

Problem

Prerequisites

Solution

Edit 1:

Video Solution:

Problem

Prerequisites

Solution

Function and Symmetry - AIME I 1984

Key Concepts

Check the Answer

Try with Hints

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