Categories

## ISI MStat PSB 2009 Problem 1 | Nilpotent Matrices

This is a very simple sample problem from ISI MStat PSB 2009 Problem 1. It is based on basic properties of Nilpotent Matrices and Skew-symmetric Matrices. Try it !

## Problem– ISI MStat PSB 2009 Problem 1

(a) Let $A$ be an $n \times n$ matrix such that $(I+A)^4=O$ where $I$ denotes the identity matrix. Show that $A$ is non-singular.

(b) Give an example of a non-zero $2 \times 2$ real matrix $A$ such that $\vec{x’}A \vec{x}=0$ for all real vectors $\vec{x}$.

### Prerequisites

Nilpotent Matrix

Eigenvalues

Skew-symmetric Matrix

## Solution :

The first part of the problem is quite easy,

It is given that for a $n \times n$ matrix $A$, we have $(I+A)^4=O$, so, $I+A$ is a nilpotet matrix, right !

And we know that all the eigenvalues of a nilpotent matrix are $0$. Hence all the eigenvalues of $I+A$ are 0.

Now let $\lambda_1, \lambda_2,……,\lambda_k$ be the eigenvalues of the matrix $A$. So, the eigenvalues of the nilpotent matrix $I+A$ are of form $1+\lambda_k$ where, $k=1,2…..,n$. Now since, $1+\lambda_k=0$ which implies $\lambda_k=-1$, for $k=1,2,…,n$.

Since all the eigenvalues of $A$ are non-zero, infact $|A|=(-1)^n$. Hence our required propositon.

(b) Now this one is quite interesting,

If for any $2\times 2$ matrix, the Quadratic form of that matrix with respect to a vector $\vec{x}=(x_1,x_2)^T$ is of form,

$a{x_1}^2+ bx_1x_2+cx_2x_1+d{x_2}^2$ where $a,b,c$ and $d$ are the elements of the matrix. Now if we equate that with $0$, what condition should it impose on $a, b, c$ and $d$ !! I leave it as an exercise for you to complete it. Also Try to generalize it you will end up with a nice result.

## Food For Thought

Now, extending the first part of the question, $A$ is invertible right !! So, can you prove that we can always get two vectors from $\mathbb{R}^n$, say $\vec{x}$ and $\vec{y}$, such that the necessary and sufficient condition for the invertiblity of the matrix $A+\vec{x}\vec{y’}$ is “ $\vec{y’} A^{-1} \vec{x}$ must be different from $1$” !!

This is a very important result for Statistics Students !! Keep thinking !!

Categories

## 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 !!

Categories

## ISI MStat PSB 2007 Problem 6 | Counting Principle & Expectations

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 6 based on counting principle . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 6

18 boys and 2 girls are made to stand in a line in a random order. Let $X$ be the number of boys standing in between the girls. Find
(a) P(X=5)
(b) $E(X)$

### Prerequisites

Basic Counting Principle

Probability

Discrete random variable

## Solution :

If there are j boys in between 2 girls then first we have to choose j boys out of 18 boys in ${18 \choose j}$ ways now this j boys can arrange among themselves in j! ways and 2 girls can arrange among themselves in 2! ways and now consider these j boys and 2 girls as a single person then this single person along with remaining (18-j) boys can arrange among themselves in (18-j+1)! ways .

Giving all total $2! {18 \choose j} j! (18-j+1)!$ possible arrangements .

Again without any restrictions there are (18+2)!=20! arrangements .

Hence $P(X=j)=\frac{ 2! {18 \choose j} j! (18-j+1)! }{20!} = \frac{19-j}{190}$

(a) $P(X=5)=\frac{19-5}{190}=\frac{7}{95}$

(b) $E(X)= \sum_{j=0}^{18} \frac{j (19-j )}{190} = 9$ just computing the sum of first 18 natural number and sum of squares of first 18 natural numbes.

## Food For Thought

Find the same under the condition that 18 boys and 2 girls sit in a circular table in a random order .

Categories

## ISI MStat PSB 2005 Problem 3 | The Orthogonal Matrix

This is a very subtle sample problem from ISI MStat PSB 2005 Problem 3. Given that one knows the property of orthogonal matrices its just a counting problem. Give it a thought!

## Problem– ISI MStat PSB 2005 Problem 3

Let $A$ be a $n \times n$ orthogonal matrix, where $n$ is even and suppose $|A|=-1$, where $|A|$ denotes the determinant of $A$. Show that $|I-A|=0$, where $I$ denotes the $n \times n$ identity matrix.

### Prerequisites

Orthogonal Matrix

Eigenvalues

Characteristic Polynomial

## Solution :

This is a very simple problem, when you are aware of the basic facts.

We, know that, the eigenvalues of a orthogonal matrix is $-1$ and $1$ .($i$ and $-i$ if its skew-symmetric). But this given matrix $A$ is not skew-symmetric.(Why??).So let for the matrix $A$, the algebraic multiplicity of $-1$ and $1$ be $m$ and $n$, respectively.

So, since $|A|=-1$, hence the algebraic multiplicity of $-1$ is definitely odd, since we know by the property of eigenvalues determinant of a matrix is just the product of its eigenvalues.

Now since, $n$ is even and the algebraic multiplicity of $-1$ i.e. $m$ is odd, hence $n$ is also odd and $n \ge 1$.

Hence, the Characteristic Polynomial of $A$, is $|I\lambda – A |=0$, where $\lambda$ is the eigenvalue of $A$, and in this problem $\lambda=-1$ or $1$.

Hence, putting $\lambda=1$, we conclude that, $|I-A|=0$. Hence we are done !!

## Food For Thought

Now, suppose $M$ is any non-singular matrix, such that $M^2=-I$. What can you say about the column space of $M$ ?

Keep thinking !!

Categories

## ISI MStat PSB 2006 Problem 6 | Counting Principle & Expectations

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 6 based on counting principle . Let’s give it a try !!

## Problem– ISI MStat PSB 2006 Problem 6

Let $Y_{1}, Y_{2}, Y_{3}$ be i.i.d. continuous random variables. For i=1,2, define $U_{i}$ as

$U_{i}= \begin{cases} 1 , & Y_{i+1}>Y_{i} \\ 0 , & \text{otherwise} \end{cases}$
Find the mean and variance of $U_{1}+U_{2}$ .

### Prerequisites

Basic Counting Principle

Probability

Continuous random variable

## Solution :

$E(U_1+U_2)=E(U_1)+E(U_2)$

Now $E(U_1)=1 \times P(Y_2>Y_1) = \frac{1}{2}$, as there are only two cases either $Y_2>Y_1$ or $Y_2<Y_1$.

Similarly , $E(U_2)= \frac{1}{2}$

So, $E(U_1+U_2)= 1$

$Var(U_1+U_2)=Var(U_1)+Var(U_2)+2Cov(U_1,U_2)$ .

$Var(U_1)=E({U_1}^2)-{E(U_1)}^2=1^2 \times P(Y_2>Y_1) – {\frac{1}{2}}^2 = \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Similarly ,$Var(U_2)=\frac{1}{4}$

$Cov(U_1,U_2)=E(U_1 U_2)-E(U_1)E(U_2)=1 \times P(Y_3>Y_2>Y_1) – {\frac{1}{2}}^2 = \frac{1}{3!}-{\frac{1}{2}}^2$ ( as there are 3! possible arrangements of $Y_i’s$ keeping inequalities fixed .

Therefore , $Var(U_1+U_2) = 2 \times \frac{1}{4} + 2 \times (\frac{1}{6}- \frac{1}{4}) = \frac{1}{3}$

## Food For Thought

Find the same under the condition that $Y_i’s$ are iid poission random variables .

Categories

## ISI MStat PSB 2006 Problem 5 | Binomial Distribution

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 5 based on use of binomial distribution . Let’s give it a try !!

## Problem– ISI MStat PSB 2006 Problem 5

Suppose $X$ is the number of heads in 10 tossses of a fair coin. Given $X=5,$ what is the probability that the first head occured in the third toss?

### Prerequisites

Basic Counting Principle

Conditional Probability

Binomial Distribution

## Solution :

As $X$ is the number of heads in 10 tossses of a fair coin so $X \sim binom(10, \frac{1}{2} )$

A be the event that first head occured in third toss

B be the event that X=5

We have to find that $P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{ {7 \choose 4} {\frac{1}{2}}^{10} }{ {10 \choose 5} {\frac{1}{2}}^{10}}$

As , $P(A \cap B)$ = Probability that out of 5 heads occur at 10 tosses 1st head occur at 3rd throw

=Probability that first two tails $\times$ probability that 3rd one is head $\times$ probability that out of 7 toss 4 toss will give head

= ${\frac{1}{2}}^2 \times \frac{1}{2} \times {7 \choose 4} {\frac{1}{2}}^{7}$

Hence our required probability is $\frac{5}{36}$

## Food For Thought

Under the same condition find the probability that X= 3 given 1st head obtained from 2nd throw .

Categories

## ISI MStat PSB 2006 Problem 1 | Inverse of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 1 based on Inverse of a matrix. Let’s give it a try !!

## Problem– ISI MStat PSB 2006 Problem 1

Let A and B be two invertible $n \times n$ real matrices. Assume that $A+B$ is invertible. Show that $A^{-1}+B^{-1}$ is also invertible.

### Prerequisites

Matrix Multiplication

Inverse of a matrix

## Solution :

We are given that A,B,A+B are all invertible real matrices . And in this type of problems every information given is a hint to solve the problem let’s give a try to use them to show that $A^{-1}+B^{-1}$ is also invertible.

Observe that ,$A(A^{-1}+B^{-1})B= (B+A) \Rightarrow |A^{-1}+B^{-1}|=\frac{|A+B|}{|A| |B| }$ taking determinant is both sides as A+B , A and B are invertible so |A+B| , |A| and |B| are non-zero . Hence $A^{-1}+B^{-1}$ is also non-singular .

Again we have , $A(A^{-1}+B^{-1})B= (B+A) \Rightarrow B^{-1} {(A^{-1}+B^{-1})}^{-1} A^{-1} = {(A+B)}^{-1}$ , taking inverse on both sides .

Now as A+B , A and B are invertible so , we have ${(A^{-1}+B^{-1})}^{-1}=B {(A+B)}^{-1} A$ . Hence we are done .

## Food For Thought

If $A \& B$ are non-singular matrices of the same order such that $(A+B)$ and $\left(A+A B^{-1} A\right)$ are also non-singular, then find the value of $(A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1}$.

Categories

## ISI MStat PSB 2007 Problem 2 | Rank of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 2 based on Rank of a matrix. Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 2

Let $A$ and $B$ be $n \times n$ real matrices such that $A^{2}=A$ and $B^{2}=B$
Suppose that $I-(A+B)$ is invertible. Show that rank(A)=rank(B).

### Prerequisites

Matrix Multiplication

Inverse of a matrix

Rank of a matrix

## Solution :

Here it is given that $I-(A+B)$ is invertible which implies it’s a non-singular matrix .

Now observe that ,$A(I-(A+B))=A-A^2-AB= -AB$ as $A^2=A$

Again , $B(I-(A+B))=B-BA-B^2=-BA$ as $B^2=B$ .

Now we know that for non-singular matrix M and another matrix N , $rank(MN)=rank(N)$ . We will use it to get that

$rank(A)=rank(A(I-(A+B)))=rank(-AB)=rank(AB)$ and $rank(B)=rank(B(I-(A+B)))=rank(-BA)=rank(BA)$ .

And it’s also known that $rank(AB)=rank(BA)$ . Hence $rank(A)=rank(B)$ (Proved) .

## Food For Thought

Try to prove the same using inequalities involving rank of a matrix.

Categories

## ISI MStat PSB 2007 Problem 1 | Determinant and Eigenvalues of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 1 based on Determinant and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 1

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{2}=0 .$ Find the determinant of $I+A$ where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

## Solution :

Let ${\lambda}_{1} , {\lambda}_{2}$ be two eigen values of A then , ${{\lambda}_{1}}^2 , {{\lambda}_{2}}^2$ .

Now it’s given that $A^2=0$ , so we have ${{\lambda}_{1}}^2=0 , {{\lambda}_{2}}^2 =0$ . You may verify it ! (Hint : use the theorem that $\lambda$ is a eigen value of matrix B and $\vec{x}$ is it’s corresponding eigen value then we can write $Bx=\lambda \vec{x}$ or , use $det(B- \lambda I )=0$ ).

Hence we have ${\lambda}_{1} =0 , {\lambda}_{2}=0$ .

Now , eigen values of Identity matrix I are 1 . So, we can write for eigen value $\vec{x}$ of (A+I) , $(A+I) \vec{x}= Ax+I\vec{x}=0+\vec{x}=\vec{x}$.

Thus we get that both the eigen values of (A+1) are 1 . Again we know that determinant of a matrix is product of it’s eigen values .

So, we have $|A+I|=1$.

Do you think this solution is correct ?

If yes , then you are absolutely wrong . The mistake is in assuming A and I has same eigen vectors $Ax+I\vec{x} \ne \vec{x}$

Correct Solution

We have shown in first part of wrong solution that A has eigen values 0 . Hence the characteristic polynomial of A can be written as , $|A- \lambda I|= {\lambda}^2$ .

Now taking $\lambda =-1$ we get $|A+ I|={(-1)}^2 \implies |A+I|= 1$ .

## Food For Thought

If we are given that $A^{n} = 0$ for positive integer n , instead of $A^2=0$ then find the same .

Categories

## ISI MStat PSB 2007 Problem 4 | Application of Newton Leibniz theorem

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 4 based on use of Newton Leibniz theorem . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 4

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function. Define $g:[0, \infty) \rightarrow \mathbb{R}$ by,
$g(x)=\int_{-x}^{x}(2 x t+1) f(t) dt$
Show that g is differentiable on $(0, \infty)$ and find the derivative of g.

### Prerequisites

Riemann integrability

Continuity

Newton Leibniz theorem

## Solution :

As $f: \mathbb{R} \rightarrow \mathbb{R}$ be a bounded continuous function hence the function

$|\Phi(t)|=|(2xt+1)f(t)|=|2xt+1||f(t)|<(|2xt|+1)M<(2|x|^2+1)M$ , which is finite for a particular x so it’s a riemann integrable function on t.

Now, by fundamental theorem we have g(x)=F(x)-F(-x) , where F is antiderivative of $\Phi(t)$ .

Hence from above we can say that g(x) is differentiable function over x .
Now by Leibniz integral rule we have $g'(x)=(2x^2+1)f(x)+f(-x)(1-2x^2) + \int_{-x}^{x} (2t)f(t) dt$.

## Food For Thought

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Now, we define $g(x)$ such that $g(x)=f(x) \int_{0}^{x} f(t) d t$
Prove that if g is a non increasing function, then f is identically equal to 0.