Categories

## Likelihood & the Moment | ISI MStat 2016 PSB Problem 7

This problem is a beautiful example when the maximum likelihood estimator is same as the method of moment estimator. Infact, we have proposed a general problem, is when exactly, they are equal? This is from ISI MStat 2016 PSB Problem 7, Stay Tuned.

## Problem

Let $X_{1}, X_{2}, \ldots, X_{n}$ be independent and identically distributed random variables ~ $X$ with probability mass function
$$f(x ; \theta)=\frac{x \theta^{x}}{h(\theta)} \quad \text { for } x=1,2,3, \dots$$
where $0<\theta<1$ is an unknown parameter and $h(\theta)$ is a function of $\theta$ Show that the maximum likelihood estimator of $\theta$ is also a method of moments estimator.

## Solution

This $h(\theta)$ looks really irritating.

#### Find the $h(\theta)$.

$\sum_{x = 1}^{\infty} f(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x \theta^{x}}{h(\theta)} = 1$

$\Rightarrow h(\theta) = \sum_{x = 1}^{\infty} {x \theta^{x}}$

$\Rightarrow (1 – \theta) \times h(\theta) = \sum_{x = 1}^{\infty} {\theta^{x}} = \frac{\theta}{1 – \theta} \Rightarrow h(\theta) = \frac{\theta}{(1 – \theta)^2}$.

#### Maximum Likelihood Estimator of $\theta$

$L(\theta)=\prod_{i=1}^{n} f\left(x_{i} | \theta\right)$

$l(\theta) = log(L(\theta)) = \sum_{i=1}^{n} \log \left(f\left(x_{i} | \theta\right)\right)$

Note: All irrelevant stuff except the thing associated with $\theta$ is kept as constant ($c$).

$\Rightarrow l(\theta) = c + n\bar{X}log(\theta) – nlog(h(\theta))$

$l^{\prime}(\theta) = 0 \overset{Check!}{\Rightarrow} \hat{\theta}_{mle} = \frac{\bar{X} -1}{\bar{X} +1}$

#### Method of Moments Estimator

We need to know the $E(X)$.

$E(X) = \sum_{x = 1}^{\infty} xf(x ; \theta) = \sum_{x = 1}^{\infty} \frac{x^2 \theta^{x}}{h(\theta)}$.

$E(X)(1 – \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x}}{h(\theta)}$.

$E(X)\theta(1 – \theta) = \sum_{x = 1}^{\infty} \frac{(2x-1)\theta^{x+1}}{h(\theta)}$

$E(X)((1 – \theta) – \theta(1 – \theta)) =\frac{\sum_{x = 1}^{\infty} 2\theta^{x} – \theta }{h(\theta)} = \frac{\theta(1 + \theta)}{(1 – \theta)h(\theta)}$.

$\Rightarrow E(X) = \frac{\theta(1 + \theta)}{(1 – \theta)^3h(\theta)} = \frac{1+\theta}{1-\theta}.$

$E(X) = \bar{X} \Rightarrow \frac{1+\theta_{mom}}{1-\theta_{mom}}= \bar{X} \Rightarrow \hat{\theta}_{mom} = \frac{\bar{X} -1}{\bar{X} +1}$

## Food For Thought and Research Problem

Normal (unknown mean and variance), exponential, and Poisson all have sufficient statistics equal to their moments and have MLEs and MoM estimators the same (not strictly true for things like Poisson where there are multiple MoM estimators).

So, when do you think, the Method of Moments Estimator = Maximum Likelihood Estimator?

Pitman Kooper Lemma tells us that it is an exponential family.

Also, you can prove that that there exists a specific form of the exponential family.

Stay tuned for more exciting such stuff!

Categories

## Start Date: 13th October, 2019

Class timing: Sunday, 8:30 PM to 10 PM.

##### Faculty: Srijit Mukherjee, B.Stat (2019), M.Stat (2021), Indian Statistical Institute
Research Interests: Probability Theory, Dynamical Systems, Discrete Mathematics

##### Course pre-requisites
Algebraic Manipulation
Trigonometry, Euclidean Geometry
##### Target Exams
Pre-RMO, RMO, INMO, IMO ( India Olympiad )

ISI – CMI Entrance ( India Undergraduate Entrance )

AMC 8, AMC 10,  AMC 12, AIME, SMO, etc ( USA Olympiad, Non – Indian Olympiad )

##### Watch the video

https://youtu.be/CdvZFbm2G0g

Categories

## AM GM inequality in ISI Entrance

Arithmetic Mean and Geometric Mean inequality form a foundational principle. This problem from I.S.I. Entrance is an application of that.

## I.S.I., C.M.I. Entrance Program at Cheenta

Taught by Olympians, I.S.I. students, alumni, and researchers from leading universities in the world.

Group + One-on-One Classes + 24/7 Doubt Clearing Forum

Categories

## How are Bezout’s Theorem and Inverse related? – Number Theory

The inverse of a number (modulo some specific integer) is inherently related to GCD (Greatest Common Divisor). Euclidean Algorithm and Bezout’s Theorem forms the bridge between these ideas. We explore them in a very lucid manner.

Categories

## How to use Invariance in Combinatorics – ISI Entrance Problem

Invariance is a fundamental phenomenon in mathematics. In this combinatorics problem from ISI Entrance, we discuss how to use invariance.

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Categories

# Understand the problem

If $a,b$ are positive reals such that $a+b<2$ ,then prove that  $$\displaystyle \frac {1}{1+a^2} + \frac {1}{1+b^2} \leq \frac {2}{1+ab}$$
##### Source of the problem
Mathematical Circles
##### Topic
Inequality involving AM-GM
Medium
##### Suggested Book
Mathematical Circles

Do you really need a hint? Try it first!

Assume that the given inequality is true for $a>0 , b>0$ and $a+b<2$ . Then proceed .
In order to simplyfy the given inequality multiply both the sides by $(1+a^2)(1+b^2)(1+ab)$ (as its a positive quantity and it is directly coming from   $a>0 , b>0$  ) .
Come up with the simplest form of inequality  i.e. $(a-b)^2 (1-ab) \geq 0$ .
$(a-b)^2 \geq 0$ as $a,b \in{R}$ . And to get $(1-ab)>0$ use the well known inequality for positive reals i.e. $AM \geq GM$ and the still unused inequality i.e $a+b <2$ also . $$\displaystyle a>0 , b>0 \Rightarrow \sqrt{ab}>0 \Rightarrow( 1+ \sqrt{ab})>0 \\ a>0 , b>0 , a+b <2 \Rightarrow 1 > \frac{a+b}{2} \geq \sqrt {ab} \\ \Rightarrow 1 > \sqrt{ab} \\ \Rightarrow ( 1 – \sqrt{ab}) >0 \\ \Rightarrow (1 – \sqrt{ab}) (1+ \sqrt{ab}) >0 \\ \Rightarrow (1 – ab)>0$$

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

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## Pigeonhole Principle

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## Value of Sum | PRMO – 2018 | Question 16

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## Chessboard Problem | PRMO-2018 | Problem No-26

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## Measure of Angle | PRMO-2018 | Problem No-29

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## Good numbers Problem | PRMO-2018 | Question 22

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## Digits Problem | PRMO – 2018 | Question 19

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Categories

# Understand the problem

Let A be a 3 x 3 real matrix with all the diagoanl entries equal to 0 . If 1 + i is an eigen value of A , the determinant of A equal ?
##### Source of the problem
Sample Questions (MMA ) :2019
Linear Algebra
Medium
##### Suggested Book
Schaum’s Outline of Linear Algebra

Do you really need a hint? Try it first!

Let A be a 3 x 3 real matrix with trace 0

Now( 1+i) be an eigen value . (Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots & Eigen vectors are In linear algebra,)

(An eigenvector  or characteristic vector of a linear transformation is a non-zero vector that changes by only a scalar factor when that linear transformation is applied to it. More formally, if T is a linear transformation from a vector space V over a field F into itself and v is a vector in V that is not the zero vector, then v is an eigenvector of T if T(v) is a scalar multiple of v. This condition can be written as the equation )

So ,( 1 + i) be the roots of the characteristic poly of A

Now A is a real matrix, char poly of A  $\epsilon \mathbb{R}[x]$ [Right!]

Therefore ( 1 – i) is also a root of char poly of A

deg( char poly of A ) =3

So , it has two imaginary roots & one real root

Let real root be r

Note tr(A) = 0 => r + 1 +i +1 -i = 0 , => r = -2

Can you play with the determinants now ?

We know the determinant of A is the product of all eigen values

(-2)(1+i)(1-i) = detA

=> det(A) = -2[1 +1] = -4

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

# Similar Problems

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

## ISI MStat PSB 2009 Problem 8 | How big is the Mean?

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

## ISI MStat PSB 2009 Problem 4 | Polarized to Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 4. It is based on the idea of Polar Transformations, but need a good deal of observation o realize that. Give it a Try it !

## ISI MStat PSB 2009 Problem 6 | abNormal MLE of Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 6. It is based on the idea of Restricted Maximum Likelihood Estimators, and Mean Squared Errors. Give it a Try it !

## ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

This is a very simple but beautiful sample problem from ISI MStat PSB 2009 Problem 3. It is based on recognizing density function and then using CLT. Try it !

## 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 !

## 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 on Inequality | ISI – MSQMS – B, 2018 | Problem 2a

Try this problem from ISI MSQMS 2018 which involves the concept of Inequality. You can use the sequential hints provided to solve the problem.

## Data, Determinant and Simplex

This problem is a beautiful problem connecting linear algebra, geometry and data. Go ahead and dwelve into the glorious connection.

## Problem on Integral Inequality | ISI – MSQMS – B, 2015

Try this problem from ISI MSQMS 2015 which involves the concept of Integral Inequality and real analysis. You can use the sequential hints provided to solve the problem.

Categories

# Understand the problem

A new flag of ISI club is to be designed with 5 vertical strips using some or all the four colours : green , naroon , red and yellow . In how many ways this can be done  so that no two adjacent strips have the same colour ?
##### Source of the problem
Sample Questions ( MMA ) :2019
Combinatorics
Medium
##### Suggested Book
Schaum’s outline of combinatorics by  Balakrishnan

Do you really need a hint? Try it first!

This is an application of multiplication property in combinatorics. The rule of product or multiplication principle is a basic counting principle (a.k.a. the fundamental principle of counting). Stated simply, it is the idea that if there are a ways of doing something and b ways of doing another thing, then there are a · b ways of performing both actions.

ISI club has 5 vertical strips . We have to colour them using 4 colours . So , the first strip can be coloured in 4 ways . WLOG we take it to be green . Can the second strip be coloured green ?

No  ! Right ?

So , we have to choose the second strip from rest of the colours . [ Because two adjacent strip has same colour ]
Similarly , third strip can be coloured into 3 ways , fourth strips can be coloured into 3 ways and fifth strips can be coloured into 3 ways .[ We have to exclude the colour of the second one]

Therefore , the total number  of probabilities are – $3^4$ x 4 = 324

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

# Similar Problems

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

## ISI MStat PSB 2009 Problem 8 | How big is the Mean?

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

## ISI MStat PSB 2009 Problem 4 | Polarized to Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 4. It is based on the idea of Polar Transformations, but need a good deal of observation o realize that. Give it a Try it !

## ISI MStat PSB 2009 Problem 6 | abNormal MLE of Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 6. It is based on the idea of Restricted Maximum Likelihood Estimators, and Mean Squared Errors. Give it a Try it !

## ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

This is a very simple but beautiful sample problem from ISI MStat PSB 2009 Problem 3. It is based on recognizing density function and then using CLT. Try it !

## 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 !

## 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 on Inequality | ISI – MSQMS – B, 2018 | Problem 2a

Try this problem from ISI MSQMS 2018 which involves the concept of Inequality. You can use the sequential hints provided to solve the problem.

## Data, Determinant and Simplex

This problem is a beautiful problem connecting linear algebra, geometry and data. Go ahead and dwelve into the glorious connection.

## Problem on Integral Inequality | ISI – MSQMS – B, 2015

Try this problem from ISI MSQMS 2015 which involves the concept of Integral Inequality and real analysis. You can use the sequential hints provided to solve the problem.

Categories

# Understand the problem

For x ≥ 0 define
f(x) =$\frac{1}{x+2cos x}$
.
Determine the set {y ∈ R : y = f(x), x ≥ 0}.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 2.

##### Topic
Calculus , Function

Medium

# Problems In CALCULUS OF ONE VARIABLE

Do you really need a hint? Try it first!

This problem simply ask for the range of the function defined by f(x)=$\frac {1}{x+2cosx}$

compute the derivative of the function = $\frac {2sinx-1}{(x+2cosx)^2}$

First extrema occurs at $x$= $\frac{\pi}{6}$

The first derivative is negetive in the interval [ 0, $\frac{\pi}{6}$]

hence the function is decreasing in this interval

f(0)=$\frac{1}{2}$ ; f($\frac{\pi}{6}$)=$\frac{1}{\frac{\pi}{6}+ {\sqrt{3}}}$

For x>$\frac{\pi}{6}$

the derivative becomes positive , and remain so upto x=$\frac{5\pi}{6}$ after which it becomes negative

thus we have minima at x= $\frac{\pi}{6}$ and maxima at x= $\frac{5\pi}{6}$

f($\frac{5\pi}{6}$)= $\frac{1}{\frac{5\pi}{6}+\sqrt{3}}$

note that as $x\rightarrow \infty$the denominator of the function increases

Hence we can conclude that    $f(x)\rightarrow0$

clearly x=$\frac{5\pi}{6}$ gives the global maxima

so , the range is (0,$\frac{1}{\frac{5\pi}{6}+\sqrt{3}}$

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Similar Problem

## IIT JAM Stat Mock Test Toppers

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## Pigeonhole Principle

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## Mahalanobis National Statistics Competition

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## Bijection Principle Problem | ISI Entrance TOMATO Obj 22

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Categories

# Understand the problem

In a badminton singles tournament, each player played against all the others
exactly once and each game had a winner. After all the games, each player
listed the names of all the players she defeated as well as the names of all the
players defeated by the players defeated by her. For instance, if A defeats B
and B defeats C, then in the list of A both B and C are included. Prove that
at least one player listed the names of all other players.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 4.

combinatorics

7.5 out of 10

### Problem Solving Strategies by Engel

Do you really need a hint? Try it first!

Do you know what is Well-ordering principle ?

it says that Every nonempty set A of nonnegative integers has a minimal element and a maximal element , which need not to be unique .

now some time this simple property help us to get very nice solution to a problem , can you some how apply this property .

Use the method of contradiction , first of all assume that there is no player which have the given property .

Now try to use the property of hint 1 .

If there is no such list ,

A’s list has the maximum no. of players

Now , if

A does not have the certain property then there exist another another player B , who has won against A . Now B’s list contain the name of A [ by the 1st condition ]

and all the names of the players defeated by A [ by the 2nd condition]

Now , can you find out some contradiction ,

yes exactly ….. B’s list contain more number of element than A

So, A’s list must have the certain property .

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Similar Problem

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## Pigeonhole Principle

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Here is a video solution for a Problem based on using Vectors and Carpet Theorem in Geometry 1? This problem is helpful for Math Olympiad, ISI & CMI Entrance, and other math contests. Watch and Learn! Here goes the question… Given ABCD is a quadrilateral and P and...

## Mahalanobis National Statistics Competition

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## Carpet Strategy in Geometry | Watch and Learn

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## What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

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## AM-GM Inequality Problem | ISI Entrance

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