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

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 \( \lim_{K \to \infty} P(K >1 ) \)


ISI MStat PSB 2008 Problem 10
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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 \( a_{k}=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{m=1}^{kn} {(\frac{m}{n})}^{5} \exp \left(-\frac{1}{2} \frac{m}{n}\right) \).


ISI MStat PSB 2008 Problem 10
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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 \( f”(1) \) .


    ISI MStat PSB 2008 Problem 10
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    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 .


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    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 .


    ISI MStat PSB 2008 Problem 10
    Outstanding Statistics Program with Applications

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    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 .


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    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 \( (A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1} \).


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    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.


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    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 .


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    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.


    ISI MStat PSB 2008 Problem 10
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