Categories

## ISI MStat PSB 2014 Problem 2 | Properties of a Function

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let’s give it a try !!

## Problem– ISI MStat PSB 2014 Problem 2

Let $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ be real numbers such
that $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R}$
Show that $m=n$ and $a_{j}=b_{j}$ for $1 \leq j \leq n$

### Prerequisites

Differentiability

Mod function

continuity

## Solution :

Let , $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R}$

Then , $f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right|$ is not differentiable at $x=a_1,a_2, \cdots , a_m$ —(1)

As we know the function $|x-a_i|$ is not differentiable at $x=a_i$ .

Again we have , $f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right|$ it also not differentiable at $x= b_1,b_2, \cdots , b_n$ —-(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ .

So , we have $a_{j}=b_{j}$ for $1 \leq j \leq n$ .

## Food For Thought

$a<b \in \mathbb{R} .$ Let $f:[a, b] \rightarrow[a, b]$ be a continuous and differentiable on (a,b) . Suppose that $\left|f^{\prime}(x)\right| \leq \alpha<1$ for all $x \in(a, b)$ for some $\alpha .$ Then prove that there exists unique $x \in[a, b]$ such that $f(x)=x$

Categories

## ISI MStat PSB 2012 Problem 3 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 3 based on finding the distribution of a random variable . Let’s give it a try !!

## Problem– ISI MStat PSB 2012 Problem 3

Let $X_{1}$ and $X_{2}$ be i.i.d. exponential random variables with mean $\lambda>0$ .Let $Y_{1}=X_{1}-X_{2}$ and $Y_{2}=R X_{1}-(1-R) X_{2},$ where $R$ is a Bernoulli random variable with parameter $1 / 2$ and is independent of $X_{1}$ and $X_{2}$
(a) Show that $Y_{1}$ and $Y_{2}$ have the same distribution.
(b) Obtain the common density function.

### Prerequisites

Cumulative Distribution Function

Bernoulli distribution

Exponential Distribution

## Solution :

Cumulative distribution of $Y_{1}$ be

$F_{Y_{1}}(y_{1})=P(Y_{1} \leq y_{1})=P(x_{1}-x_{2} \leq y_{1})$ ,$y_1 \in R$

$=P(x_{1} \leq y_{1}+x_{2})$
Now, $y_{1}+x_{2} \ge 0 \Rightarrow x_{2} \ge-y_{1}$
Now, if $y_{1} \ge 0$ then,
$P(x_{1} \le y_{1}+x_{2}) =\int_{0}^{\infty} P(x_{1} \le y_{1}+x_{2}t) \lambda e^{-\lambda x_{2}} d x_{2}$

=$\int_{0}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{1}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2}$

=$\int_{0}^{\infty} \lambda e^{\lambda x_{2}} x \lambda \times \frac{1}{\lambda} (1-e^{-(\lambda y_{1}+x_{2}) }) d x_{2}$

=$\int_{0}^{\infty} \lambda e^{-\lambda x_{2}} d x_{2}-\int_{0}^{\infty} \lambda e^{-\lambda (y_{1}+2 x_{2})} d x_{2}$

=$1-\frac{e^{-\lambda y_{1}}}{2}$

Now, $y_{1} \le 0$ then,
$P(x_{1} \leq y_{1}+x_{2}) =\int_{-y_{1}}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{4}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2}$
$=\int_{-y_{1}}^{\infty} x e^{-\lambda x_{2}}(1-e^{-\lambda(y_{1}+x_{1})}) d x_{2}$
$=\lambda \int_{-y_{1}}^{\infty} e^{-x^{2} x_{2}} d x_{2}-\int_{-y_{1}}^{\infty} \lambda e^{-\lambda(y_{1}+2 x_{2})} d x_{2}$
$=e^{+\lambda y_{1}}-\frac{e^{-\lambda y_{1}}}{2} x e^{+2 \lambda y_{1}}$
$=\frac{e^{\lambda y_{1}}}{2}$
Therefore, $F_{Y_{1}}(y_{1}) = \begin{cases} 1-\frac{e^{-\lambda y_{1}}}{2} & , i f y_{1} \ge 0 \\ \frac{e^{\lambda y_{1}}}{2} & ,if y_{1}<0 \end{cases}.$

Cumulative distribution of $Y_{2}$ be $F_{Y_{2}}(y_{2})=P(Y_{2} \le y_{2})$ , $y_2 \in R$

=$P(Y_{2} \le y_{2} \mid R=1) P(R=1)+P(Y_{2} \le y_{2} \mid R=0) P(R=0)$
$=P(x_{1} \le y_{2}) \times \frac{1}{2}+P(-x_{2} \le y_{2}) \times \frac{1}{2}$
= $\begin{cases} \frac{1}{2} [F_{x_{1}}(y_{2})+1] & , y_{2} \ge 0 \\ \frac{1}{2} [1-F_{x_{2}}(-y_{2})] & ,y_{2}<0 \end{cases}.$
=$\begin{cases} 1-\frac{e^{-\lambda y_{2}}}{2}, & \text { if } y_{2} \ge 0 \\ \frac{e^{\lambda y_{2}}}{2} \end{cases}.$
since cdf of exponential random Variable, X is $(1-e^{-\lambda x}), x \ge 0$
Thus both $Y_{1}$ and $Y_{2}$ has same distribution
(b) $f_{Y_{1}}(y_{1})=\begin{cases} \frac{d}{d y_{1}}(1-\frac{e^{-\lambda y_{1}}}{2}) & \text { if } y_{1} \ge 0 \\ \frac{d}{d y_{1}}(\frac{e^{\lambda y_{1}}}{2}) & , \text { if } y_{2}<0 \end{cases}$

= $\begin{cases} \frac{\lambda e^{-\lambda y_{1}}}{2} & \text { if } y_{1} \ge 0 \\ \frac{\lambda e^{\lambda y_{1}}}{2} & , \text { if } y_{1}<0 \end{cases}$

Similarly, for $Y_2$ .

## Food For Thought

If $\theta \sim U(0, 2 \pi )$ then find the distribution of $sin(\theta + {\theta}_{0} )$ , where ${\theta}_{0} \in (0,2 \pi)$.

Categories

## ISI MStat PSB 2014 Problem 1 | Vector Space & Linear Transformation

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 1 based on Vector space and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2014 Problem 1

Let $E={1,2, \ldots, n},$ where n is an odd positive integer. Let $V$ be
the vector space of all functions from E to $\mathbb{R}^{3}$, where the vector space
operations are given by $(f+g)(k) =f(k)+g(k)$, for $f, g \in V, k \in E \ (\lambda f)(k) =\lambda f(k),$ for $f \in V, \lambda \in \mathbb{R}, k \in E$
(a) Find the dimension of $V$
(b) Let $T: V \rightarrow V$ be the map given by $T f(k)=\frac{1}{2}(f(k)+f(n+1-k)), \quad k \in E$
Show that T is linear.
(c) Find the dimension of the null space of T.

### Prerequisites

Linear Transformation

Null Space

Dimension

## Solution :

While doing this problem we will use a standard notation for vectors of canonical basis i..e $e_j$ . In $R^{3}$ they are $e_1=(1,0,0) , e_2=(0,1,0)$ and $e_3=(0,0,1)$ .

(a) For $i \in {1 , 2 , \cdots , n}$ and $j \in {1 , 2 , 3}$ , let $f_{ij}$ be the function in $V$ which maps $i \mapsto e_j$ and $k \mapsto (0,0,0)$ where $k \in {1 , 2 , \cdots , n}$ and $k \neq i$. Then ${f_{ij} : i \in {1 , 2 , \cdots , n} , j \in {1 , 2 , 3}}$ is a basis of $V$ .

It looks somewhat like this , $f_{11}(1)={(1,0,0)} ,f_{11}(2)={(0,0,0)} , \cdots , f_{11}(n)={(0,0,0)}$

$f_{12}(1)={(0,1,0)} ,f_{12}(2)={(0,0,0)} , \cdots , f_{12}(n)={(0,0,0)}$ , $\cdots , f_{n3}(1)={(0,0,0)} ,f_{n3}(2)={(0,0,0)} , \cdots , f_{n3}(n)={(0,0,1)}$

Hence , dimension of $V$ is 3n.

(b) To show T is linear we have to show that $T(af(k)+bg(k)) =aT(f(k))+bT(g(k))$ for some scalar a,b .

$T(af(k)+bg(k))=\frac{ af(k)+bg(k)+af(n+1-k)+bg(n+1-k)}{2} = a \frac{f(k)+f(n+1-k)}{2} + b \frac{g(k)+g(n+1-k)}{2} = aT(f(k))+bT(g(k))$.

Hence proved .

(c) $f\in ker T$ gives $f(k)=-f(n+1-k)$ so, the values of $f$ for the last $\frac{n-1}{2}$ points are opposite to first $\frac{n-1}{2}(i.e. f(n)=-f(1) ~\text{etc.})$ so we can freely assign the values of f for first $\frac{n-1}{2}$ to any of $e_j$ .Hence, the null space has dimension $\frac{3(n-1)}{2}.$

## Food For Thought

let $T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ be a non singular linear transformation.Prove that there exists a line passing through the origin that is being mapped to itself.

Prerequisites : eigen values & vectors and Polynomials

Categories

## ISI MStat PSB 2010 Problem 1 | Tricky Linear Algebra Question

This is a very beautiful sample problem from ISI MStat PSB 2010 Problem 1 based on Matrix multiplication and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2010 Problem 1

Let $A$ be a $4 \times 4$ matrix with non-negative entries such that the sum of the entries in each row of $A$ equals 1 . Find the sum of all entries in matrix $A^{5}$ .

### Prerequisites

Matrix Multiplication

Eigen Values

Eigen Vectors

## Solution :

Doing this problem you have to use the hint given in the question . Here the hint is that the sum of the entries in each row of $A$ equals 1 . How can you use that ? Think about it!

Here comes the trick .

Let V be a vector such that $V={[1,1,1,1]}^{T}$ . Now if we multiply A by V then we will get V i.e $AV=V$ .

This is because it is given that the sum of the entries in each row of $A$ equals 1 .

So, from $AV=V$ we can say that 1 is an eigen value of A .

Hence $A^5V=A^4(AV)=A^4V=A^3(AV)= \cdots = V$ . From here we can say the sum of all the entries of each rows of $A^5$ is 1.

Therefore the sum of all the entries of $A^5$ is also 4 .

## Food For Thought

Let $A$ and B be $n \times n$ matrices with real entries satisfying $tr(A A^{T}+B B^{T})=tr(A B+A^{T} B^{T})$ .
Prove that $A=B^{T}$ .

Hint : Use properties of trace that’s the trick here .

Categories

## ISI MStat PSB 2012 Problem 2 | Dealing with Polynomials using Calculus

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 2 based on calculus . Let’s give it a try !!

## Problem– ISI MStat PSB 2012 Problem 2

Let $f$ be a polynomial. Assume that $f(0)=1, \lim _{x \rightarrow \infty} f”(x)=4$ and $f(x) \geq f(1)$ for all $x \in \mathbb{R} .$ Find $f(2)$ .

Limit

Derivative

Polynomials

## Solution :

Here given $f(x)$ is a polynomial and $\lim _{x \rightarrow \infty} f”(x)=4$

So, Case 1: If f(x) is a polynomial of degree 1 then f”(x)=0 hence limit can’t be 4.

Case 2: If f(x) is a polynomial of degree 2 ,say $f(x) = ax^2+bx+c$ then $f”(x)= 2a$ .Hence taking limit we get $2a=4 \Rightarrow a=2$

Case 3: If f(x) is a polynomial of degree >2 then $f”(x) = O(x)$ . So, it tends to infinity or – infinity as x tends to infinity .

Therefore the only case that satisfies the condition is Case 2 .

So , f(x) = $2x^2+bx+c$ ,say . Now given that $f(0)=1 \Rightarrow c=1$ .

Again , it is given that $f(x) \geq f(1)$ for all $x \in \mathbb{R}$ which implies that f(x) has minimum at x=1 .

That is f'(x)=0 at x=1 . Here we have $f'(x)=4x+b=0 \Rightarrow x=\frac{-b}{4}=1 \Rightarrow b=-4$

Thus we get $f(x)=2x^2-4x+1$ . Putting x=2 , we get $f(2)=1$ .

## Food For Thought

Assume f is differentiable on $(a, b)$ and is continuous on $[a, b]$ with $f(a)=f(b)=0$. Prove that for every real $\lambda$ there is some c in $(a, b)$ such that $f'(c)=\lambda f(c)$.

Categories

## ISI MStat PSB 2012 Problem 5 | Application of Central Limit Theorem

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 5 based on central limit theorem . Let’s give it a try !!

## Problem– ISI MStat PSB 2012 Problem 5

Let $X_{1}, X_{2}, \ldots, X_{j} \ldots$ be i.i.d. $N(0,1)$ random variables. Show that for any $a>0$

$\lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) = 0$

### Prerequisites

Limit

Central Limit Theorem

Normal Distribution

Chi- Square Distribution

## Solution :

$X_{1}, X_{2}, \ldots, X_{j} \ldots$ are i.i.d. $N(0,1)$ random variables .

Let $S_n = \sum_{i=1}^{n} {X_i}^2$ , then $S_n \sim \chi^{2}(n)$ , where $\chi^{2}(n)$ is Chi-Square distribution with n degrees of freedom .

Therefore , $E(S_n)= n$ and $Var(S_n)=2n$ .

In this type of problems obvious thing that would come to our mind is to apply Central Limit Theorem right ! Let’s try to apply it .

Now by Lindeberg Levy Central Limit Theorem we can say $\frac{S_n-E(S_n)}{\sqrt{Var(S_n)}}$ = $\frac{S_n-n}{\sqrt{2n}} {\to }^{d} N(0,1)$ , as n approaches infinity.

So, $\lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a)$

= $\lim {n \rightarrow \infty} P( \frac{S_n-n}{\sqrt{2n}} \le \frac{a-n}{\sqrt{2n}} )$

= $\lim {n \rightarrow \infty} \Phi(\frac{a-n}{\sqrt{2n}})$

= $\lim {n \rightarrow \infty} \Phi(\frac{a}{\sqrt{2n}}- \sqrt{\frac{n}{2}}) = \Phi(0- \infty)$ (Since $\Phi(x)$ is right continuous ) $= 0$ .

Hence Proved .

## Food For Thought

Let $\{X_{1}: i \geq 1 \}$ be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5.Then $(\frac{1}{n} \sum_{i=1}^{n} x_{i})^{2}$ converges in probability to ?

Categories

## ISI MStat 2019 PSA Problem 15 | Trigonometry Problem

This is the problem from ISI MStat 2019 PSA Problem 15. First, try it yourself and then go through the sequential hints we provide.

## Trigonometry – ISI MStat Year 2019 PSA Question 15

How many solutions does the equation $cos ^{2} x+3 \sin x \cos x+1=0$ have for $x \in[0,2 \pi)$ ?

• 1
• 3
• 4
• 2

### Key Concepts

Trigonometry

Factorization

ISI MStat 2019 PSA Problem 15

Precollege Mathematics

## Try with Hints

Factorize and Solve.

$\cos ^{2} x+3 \sin x \cos x + 1 = (2\cos x + \sin x)(\cos x +\sin x) = 0$.
$tanx = -2, tanx = -1$.
Draw the graph.

So, if you see the figure you will find there are 4 such x for $x \in[0,2 \pi)$.

Categories

## ISI MStat 2019 PSA Problem 14 | Reflection of a point

This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.

## Reflection of a point – ISI MStat Year 2019 PSA Question 14

The reflection of the point (1,2) with respect to the line $x+2 y=15$ is

• (3,6)
• (6,3)
• (10,5)
• (5,10)

### Key Concepts

Straight line

ISI MStat 2019 PSA Problem 14

Precollege Mathematics

## Try with Hints

Find an algorithm to find the reflection,

Find the line perpendicular to $x+2 y=15$ through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.

The line perpendicular to $x+2 y=15$ is of the form $-2x+y=k$ .Now it passes through (1,2) . So, $-2+2=k \Rightarrow k=0$

Hence the line perpendicular to $x+2 y=15$ through (1,2) is y=2x.

Now we will find point of intersection (Foot of Perpendicular )

(3,6) is the point of intersection i.e the foot of perpendicular.

Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)

$(3,6)=( \frac{x+1}{2} , \frac{y+2}{2} )$ $\Rightarrow x=5 , y=10$

Therefore the reflection point is (5,10) .

Categories

## ISI MStat 2015 PSA Problem 17 | Basic Inequality

This is a problem from ISI MStat 2015 PSA Problem 17. First, try the problem yourself, then go through the sequential hints we provide.

## Basic Inequality – ISI MStat Year 2015 PSA Question 17

Let $X=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}$. Then,

• x<1
• $x>\frac{3}{2}$
• $1<x< \frac{3}{2}$
• None of these

### Key Concepts

Basic Inequality

Answer: is $1<x< \frac{3}{2}$

ISI MStat 2015 PSA Problem 17

Precollege Mathematics

## Try with Hints

Take it easy. Group things up. Use $\frac{1}{n+k} < \frac{1}{n}$ for all natural (k)

$\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000}$
$\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}$

$\Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2}$

Again see that we can write , x=$(\frac{1}{1001}+\frac{1}{3001}) + (\frac{1}{1002}+\frac{1}{3000})+ \dots + (\frac{1}{2000}+\frac{1}{2002}) + \frac{1}{2001}$ > $\frac{2}{2001} +\frac{2}{2001} + \dots + \frac{2}{2001} +\frac{1}{2001} > \frac{2001}{2001}=1$

Hence , $1<x< \frac{3}{2}$.

Categories

## ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

## Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given $\theta$ in the range $0 \leq \theta<\pi,$ the equation $2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0$ represents a circle for all $\theta$ in the interval

• $0 < \theta <\frac{\pi}{3}$
• $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$
• $0 < \theta <\frac{\pi}{2}$
• $0 \le \theta <\frac{\pi}{2}$

### Key Concepts

Equation of a circle

Trigonometry

Basic Inequality

Answer: is $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

## Try with Hints

Complete the Square.

We get ,

$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$
$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$

We are given that $0 \leq \theta<\pi,$ . So, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$.