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Calculus I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Statistics

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 \(f(x)=x\)


ISI MStat PSB 2008 Problem 10
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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 \( {\theta}_{0} \in (0,2 \pi) \).


ISI MStat PSB 2008 Problem 10
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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


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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 .


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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 \( f(2)\) .

Prerequisites


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 \( f'(c)=\lambda f(c) \).


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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 ?


ISI MStat PSB 2008 Problem 10
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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

Check the Answer


Answer: is 4

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.

Trigonometry problem graph - ISI MStat 2019 PSA Problem 15
Fig:1

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

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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

Check the Answer


Answer: is (5,10)

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) .

ISI MStat 2019 PSA Problem 14
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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

Check the Answer


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 , \( 1<x< \frac{3}{2} \).

ISI MStat 2015 PSA Problem 17
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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

Check the Answer


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, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

ISI MStat 2016 PSA Problem 9
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