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## ISI MStat PSB 2008 Problem 7 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 7 based on finding the distribution of a random variable . Let’s give it a try !!

## Problem– ISI MStat PSB 2008 Problem 7

Let $X$ and $Y$ be exponential random variables with parameters 1 and 2 respectively. Another random variable $Z$ is defined as follows.

A coin, with probability p of Heads (and probability 1-p of Tails) is
tossed. Define $Z$ by $Z=\begin{cases} X & , \text { if the coin turns Heads } \\ Y & , \text { if the coin turns Tails } \end{cases}$
Find $P(1 \leq Z \leq 2)$

### Prerequisites

Cumulative Distribution Function

Exponential Distribution

## Solution :

Let , $F_{i}$ be the CDF for i=X,Y, Z then we have ,

$F_{Z}(z) = P(Z \le z) = P( Z \le z | coin turns Head )P(coin turns Head) + P( Z \le z | coin turns Tail ) P( coin turns Tail)$

=$P( X \le z)p + P(Y \le z ) (1-p)$ = $F_{X}(z)p+F_{Y}(y) (1-p)$

Therefore pdf of Z is given by $f_{Z}(z)= pf_{X}(z)+(1-p)f_{Y}(z)$ , where $f_{X} and f_{Y}$ are pdf of X,Y respectively .

So , $P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4}$

## Food For Thought

Find the the distribution function of $K=\frac{X}{Y}$ and then find $\lim_{K \to \infty} P(K >1 )$

Categories

## ISI MStat PSB 2008 Problem 2 | Definite integral as the limit of the Riemann sum

This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 2 based on definite integral as the limit of the Riemann sum . Let’s give it a try !!

## Problem– ISI MStat PSB 2008 Problem 2

For $k \geq 1,$ let $a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right)$

Find $\lim_{k \rightarrow \infty} a_{k}$ .

### Prerequisites

Integration

Gamma function

Definite integral as the limit of the Riemann sum

## Solution :

$a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} \exp \left(-\frac{1}{2} \frac{m^{2}}{n^{2}}\right) = \int_{0}^{k} e^{\frac{-y^2}{2}} dy$ , this can be written you may see in details Definite integral as the limit of the Riemann sum .

Therefore , $lim_{k \to \infty} a_{k}= \int_{0}^{ \infty} e^{\frac{-y^2}{2}} dy$ —-(1) , let $\frac{y^2}{2}=z \Rightarrow dy= \frac{dz}{\sqrt{2z}}$

Substituting we get , $\int_{0}^{ \infty} z^{\frac{1}{2} -1} e^{z} \frac{1}{\sqrt{2}} dz =\frac{ \gamma(\frac{1}{2}) }{\sqrt{2}} = \sqrt{\frac{\pi}{2}}$

Statistical Insight

Let $X \sim N(0,1)$ i.e X is a standard normal random variable then,

$Y=|X|$ called folded Normal has pdf $f_{Y}(y)= \begin{cases} \frac{2}{\sqrt{2 \pi }} e^{\frac{-x^2}{2}} & , y>0 \\ 0 &, otherwise \end{cases}$ . (Verify!)

So, from (1) we can say that $\int_{0}^{ \infty} e^{\frac{-y^2}{2}} dy = \frac{\sqrt{2 \pi }}{2} \int_{0}^{ \infty}\frac{2}{\sqrt{2 \pi }} f_{Y}(y) dy$

$=\frac{\sqrt{2 \pi }}{2} \times 1$ ( As that a PDF of folded Normal distribution ) .

## Food For Thought

Find the same when $a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} {(\frac{m}{n})}^{5} \exp \left(-\frac{1}{2} \frac{m}{n}\right)$.

Categories

## ISI MStat PSB 2008 Problem 3 | Functional equation

Content
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This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 3 based on Functional equation . Let’s give it a try !!

## Problem– ISI MStat PSB 2008 Problem 3

Let $g$ be a continuous function with $g(1)=1$ such that $g(x+y)=5 g(x) g(y)$ for all $x, y .$ Find $g(x)$.

### Prerequisites

Continuity & Differentiability

Differential equation

Cauchy’s functional equation

## Solution :

We are g is continuous function such that$g(x+y)=5 g(x) g(y)$ for all $x, y$ and g(1)=1.

Now putting x=y=0 , we get $g(0)=5{g(0)}^2 \Rightarrow g(0)=0$ or , $g(0)= \frac{1}{5}$ .

If g(0)=0 , then g(x)=0 for all x but we are given that g(1)=1 . Hence contradiction .

So, $g(0)=\frac{1}{5}$ .

Now , we can write $g'(x)= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \frac{5g(x)g(h)-g(x)}{h}$

$= 5g(x) \lim_{h \to 0} \frac{g(h)- \frac{1}{5} }{ h} = 5g(x) \lim_{h \to 0} \frac{g(h)- g(0) }{ h} = 5g(x)g'(0)$ (by definition)

Therefore , $g(x)=5g'(0)g(x)= Kg(x)$ , for some constant k ,say.

Now we will solve the differential equation , let y=g(x) then we have from above

$\frac{dy}{dx} = ky \Rightarrow \frac{dy}{y}=k{dx}$ . Integrating both sides we get ,

$ln(y)=kx+c$ c is integrating constant . So , we get $y=e^{kx+c} \Rightarrow g(x)=e^{kx+c}$

Solve the equation g(0)=1/5 and g(1)=1 to get the values of K and c . Finally we will get , $g(x)=\frac{1}{5} e^{(ln(5)) x} =5^{x-1}$.

But there is a little mistake in this solution .

What’s the mistake ?

Ans- Here we assume that g is differentiable at x=0 , which may not be true .

Correct Solution comes here!

We are given that $g(x+y)=5 g(x) g(y)$ for all $x, y .$ Now taking log both sides we get ,

$log(g(x+y))=log5+log(g(x))+log(g(y)) \Rightarrow log_5 (g(x+y))=1+log_5 (g(x))+log_5 (g(y))$

$\Rightarrow log_5 (g(x+y)) +1= log_5 (g(x))+1+log_5 (g(y)) +1 \Rightarrow \phi(x+y)=\phi(x)+\phi(y)$ , where $\phi(x)=1+log_5 (g(x))$

It’s a cauchy function as $\phi(x)$ is also continuous . Hence , $\phi(x)=cx$ , c is a constant $\Rightarrow 1+log_5 (g(x))=cx \Rightarrow g(x)=5^{cx-1}$.

Now $g(1)=1 \Rightarrow 5^{c-1}=1 \Rightarrow c=1$.

Therefore , $g(x)=5^{x-1}$

## Food For Thought

Let $f:R to R$ be a non-constant , 3 times differentiable function . If $f(1+ \frac{1}{n})=1$ for all integer n then find $f”(1)$ .

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 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.

### Prerequisites

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.