Categories
Cheenta Probability Series Descriptive Statistics Experiment Design and Sample Survey IIT JAM MS ISI M.Stat PSB ISI MSAT ISI MSTAT Probability Regression Techniques Statistics Testing of Hypothesis Theory of Estimation

ISI MStat Entrance 2020 Problems and Solutions

This post contains Indian Statistical Institute, ISI MStat Entrance 2020 Problems and Solutions. Try to solve them out.

This is a work in progress.

Subjective Paper – ISI MStat Entrance 2020 Problems and Solutions

  • Let \(f(x)=x^{2}-2 x+2\). Let \(L_{1}\) and \(L_{2}\) be the tangents to its graph at \(x=0\) and \(x=2\) respectively. Find the area of the region enclosed by the graph of \(f\) and the two lines \(L_{1}\) and \(L_{2}\).

    Solution
  • Find the number of \(3 \times 3\) matrices \(A\) such that the entries of \(A\) belong to the set \(\mathbb{Z}\) of all integers, and such that the trace of \(A^{t} A\) is 6 . \(\left(A^{t}\right.\) denotes the transpose of the matrix \(\left.A\right)\).

    Solution
  • Consider \(n\) independent and identically distributed positive random variables \(X_{1}, X_{2}, \ldots, X_{n},\) Suppose \(S\) is a fixed subset of \({1,2, \ldots, n}\) consisting of \(k\) distinct elements where \(1 \leq k<n\)
    (a) Compute \(\mathbb{E}\left[\frac{\sum_{i \in S} X_{i}}{\sum_{i=1}^{n} X_{i}}\right]\)

    (b) Assume that \(X_{i}\) ‘s have mean \(\mu\) and variance \(\sigma^{2}, 0<\sigma^{2}<\infty\). If \(j \notin S,\) show that the correlation between \(\left(\sum_{i \in S} X_{i}\right) X_{j}\) and \(\sum_{i \in S} X_{i}\) lies between -\(\frac{1}{\sqrt{k+1}} \text { and } \frac{1}{\sqrt{k+1}}\).

    Solution
  • Let \(X_{1,} X_{2}, \ldots, X_{n}\) be independent and identically distributed random variables. Let \(S_{n}=X_{1}+\cdots+X_{n}\). For each of the following statements, determine whether they are true or false. Give reasons in each case.

    (a) If \(S_{n} \sim E_{x p}\) with mean \(n,\) then each \(X_{i} \sim E x p\) with mean 1 .

    (b) If \(S_{n} \sim B i n(n k, p),\) then each \(X_{i} \sim B i n(k, p)\)

    Solution
  • Let \(U_{1}, U_{2}, \ldots, U_{n}\) be independent and identically distributed random variables each having a uniform distribution on (0,1) . Let \( X=\min \{U_{1}, U_{2}, \ldots, U_{n}\} \), \( Y=\max \{U_{1}, U_{2}, \ldots, U_{n}\} \)

    Evaluate \(\mathbb{E}[X \mid Y=y]\) and \( \mathbb{E}[Y \mid X=x] \).

    Solution
  • Suppose individuals are classified into three categories \(C_{1}, C_{2}\) and \(C_{3}\) Let \(p^{2},(1-p)^{2}\) and \(2 p(1-p)\) be the respective population proportions, where \(p \in(0,1)\). A random sample of \(N\) individuals is selected from the population and the category of each selected individual recorded.

    For \(i=1,2,3,\) let \(X_{i}\) denote the number of individuals in the sample belonging to category \(C_{i} .\) Define \(U=X_{1}+\frac{X_{3}}{2}\)

    (a) Is \(U\) sufficient for \(p ?\) Justify your answer.

    (b) Show that the mean squared error of \(\frac{U}{N}\) is \(\frac{p(1-p)}{2 N}\)

    Solution
  • Consider the following model: \( y_{i}=\beta x_{i}+\varepsilon_{i} x_{i}, \quad i=1,2, \ldots, n \), where \(y_{i}, i=1,2, \ldots, n\) are observed; \(x_{i}, i=1,2, \ldots, n\) are known positive constants and \(\beta\) is an unknown parameter. The errors \(\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{n}\) are independent and identically distributed random variables having the probability density function \[ f(u)=\frac{1}{2 \lambda} \exp \left(-\frac{|u|}{\lambda}\right), \quad-\infty<u<\infty \] and \(\lambda\) is an unknown parameter.

    (a) Find the least squares estimator of \(\beta\).

    (b) Find the maximum likelihood estimator of \(\beta\).

    Solution
  • Assume that \(X_{1}, \ldots, X_{n}\) is a random sample from \(N(\mu, 1),\) with \(\mu \in \mathbb{R}\). We want to test \(H_{0}: \mu=0\) against \(H_{1}: \mu=1\). For a fixed integer \(m \in{1, \ldots, n},\) the following statistics are defined:

    \begin{aligned}
    T_{1} &= \frac{\left(X_{1}+\ldots+X_{m}\right)}{m} \\
    T_{2} &= \frac{\left(X_{2}+\ldots+X_{m+1}\right)}{m} \\
    \vdots &=\vdots \\
    T_{n-m+1} &= \frac{\left(X_{n-m+1}+\ldots+X_{n}\right)}{m}
    \end{aligned}

    \(\operatorname{Fix} \alpha \in(0,1) .\) Consider the test

    Reject \(H_{0}\) if \( \max \{T_{i}: 1 \leq i \leq n-m+1\}>c_{m, \alpha}\)

    Find a choice of \(c_{m, \alpha} \in \mathbb{R}\) in terms of the standard normal distribution function \(\Phi\) that ensures that the size of the test is at most \(\alpha\).

    Solution
  • A finite population has \(N\) units, with \(x_{i}\) being the value associated with the \(i\) th unit, \(i=1,2, \ldots, N\). Let \(\bar{x}{N}\) be the population mean. A statistician carries out the following experiment.

    Step 1: Draw an SRSWOR of size \(n({1}\) and denote the sample mean by \( \bar{X}{n}\)

    Step 2: Draw a SRSWR of size \(m\) from \(S_{1}\). The \(x\) -values of the sampled units are denoted by \(\{Y_{1}, \ldots, Y_{m}\}\)

    An estimator of the population mean is defined as,

    \[ \widehat{T}{m}=\frac{1}{m} \sum{i=1}^{m} Y_{i} \]

    (a) Show that \(\widehat{T}{m}\) is an unbiased estimator of the population mean.

    (b) Which of the following has lower variance: \(\widehat{T}{m}\) or \(\bar{X}_{n} ?\)

    Solution
Cheena Statistics Logo
Cheenta Statistics Department
ISI MStat and IIT JAM Training Program

Objective Paper


Objective Answer Key

Please suggest changes in the comment section.

1. C2. D3. A4. B5. A
6. B7. C8. A9. C10. A
11. C12. D13. C14. B15. B
16. C17. D18. B19. B20. C
21. C22. D23. A24. B25. D
26. B27. D28. D29. B30. C

Watch videos related to the ISI MStat Problems here.

Categories
IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Probability Testing of Hypothesis Theory of Estimation

Testing of Hypothesis | ISI MStat 2016 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination, 2016 involving the basic idea of Type 1 error of Testing of Hypothesis but focussing on the fundamental relationship of Exponential Distribution and the Geometric Distribution.

The Problem:

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from an exponential distribution with mean \(\lambda\).

Assume that the observed data is available on \(\left[X_{1}\right], \ldots,\left[X_{n}\right]\), instead of \(X_{1}, \ldots, X_{n},\) where \([x]\) denotes the largest integer less than or equal to \(x\).

Consider a test for \(H_{0}: \lambda=1\) vs \(H_{1}: \lambda>1\) which rejects \(H_{0}\) when \(\sum_{i=1}^{n}\left[X_{i}\right]>c_{n} .\)

Given \(\alpha \in(0,1),\) obtain values of \(c_{n}\) such that the size of the test converges to \(\alpha\) as \(n \rightarrow \infty\).

Prerequisites:

(a) Testing of Hypothesis

(b) Type 1 Error

(c) Exponential Distribution

(d) Relationship of Exponential Distribution and Geometric Distribution

(e) Central Limit Theorem

Solution:

  • X ~ Exponential(\(\lambda\)), then \(Y = [\frac{X}{a}]\) ~ Geom(\(p\)), where \( p = 1-e^{-\lambda a} \in(0,1) \)

Proof:

\(Y\) is clearly discrete taking values in the set of non-negative integers, due to the flooring. Then, for any integer \(n \geq 0\) we have
\(
\begin{array}{c}
P(Y=n)=P(X \in[\text {an, } a(n+1))) \
=\int_{a n}^{a(n+1)} \lambda \mathrm{e}^{-\lambda x} d x=(1-p)^{n} p
\end{array}
\)
where \(p=1-e^{-\lambda a} \in(0,1),\) as \(\lambda>0\) and \(a>0\).

  • \(X_i\) ~ Geom(\(p\)), then \(\sum_{i = 1}^{n} \) ~ NBinom(n,p)
  • \(X_i\) ~ Exponential(\(\lambda\)), then \(S_n = \sum_{i=1}^{n}\left[X_{i}\right]\) ~ NBinom(\((n,p)\)), where \( p = 1-e^{-\lambda} \in(0,1) \)

Testing of Hypothesis

\(H_{0}: \lambda=1\) vs \(H_{1}: \lambda>1\)

We reject \(H_{0}\) when \(\sum_{i=1}^{n}\left[X_{i}\right]>c_{n} .\)

Here, the size of the test i.e the Type 1 error (for simple hypothesis), \( \alpha_n\) = \( P(S_n > c_{n} | \lambda=1)\).

We want to select \(c_n\) such that \(\alpha_n \to \alpha\).

\(S_n\) ~ NBinom(\(n,p\)), where \( p = 1-e^{-1} \) under \(H_0\).

Now, \(\frac{\sqrt{n}(\frac{S_n}{n} – \frac{1}{p})}{\sqrt{\frac{1-p}{p^2}}} \rightarrow Z = N(0,1)\) by Central Limit Theorem.

Observe that thus, \( \alpha_n = P(S_n > c_{n} | \lambda=1) \rightarrow P(Z > \frac{\sqrt{n}(\frac{c_n}{n} – \frac{1}{p})}{\sqrt{\frac{1-p}{p^2}}}) = \alpha\).

Thus, \( \frac{\sqrt{n}(\frac{c_n}{n} – \frac{1}{p})}{\sqrt{\frac{1-p}{p^2}}} = Z_{\alpha} \).

We can solve this to find \(c_n\), where \( p = 1-e^{-1} \)

Food for Thought

If X ~ Exponential(\(\lambda\)), then what is the distribution of {X} [ The fractional part of X]. This question is crucial is getting back Exponential Distrbution from Geometric Distribution.

Rather, the food for thought, asks you how do we get Exponential Distribution from Geometric Distribution.

Stay Tuned. Stay Blessed! See you in the next post.

Categories
College Mathematics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Statistics Theory of Estimation

ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 8. It is based on basic idea of Maximum Likelihood Estimators, but with a bit of thinking. Give it a thought !

Problem– ISI MStat PSB 2006 Problem 8


Let \((X_1,Y_1),……,(X_n,Y_n)\) be a random sample from the discrete distributions with joint probability

\(f_{X,Y}(x,y) = \begin{cases} \frac{\theta}{4} & (x,y)=(0,0) \ and \ (1,1) \\ \frac{2-\theta}{4} & (x,y)=(0,1) \ and \ (1,0) \end{cases}\)

with \(0 \le \theta \le 2\). Find the maximum likelihood estimator of \(\theta\).

Prerequisites


Maximum Likelihood Estimators

Indicator Random Variables

Bernoulli Trials

Solution :

This is a very beautiful Problem, not very difficult, but her beauty is hidden in her simplicity, lets explore !!

Observe, that the given pmf is as good as useless while taking us anywhere, so we should think out of the box, but before going out of the box, lets collect whats in the box !

So, from the given pmf we get, \(P( \ of\ getting\ pairs \ of\ form \ (1,1) \ or \ (0,0))=2\times \frac{\theta}{4}=\frac{\theta}{2}\),

Similarly, \(P( \ of\ getting\ pairs \ of\ form \ (0,1) \ or \ (1,0))=2\times \frac{2-\theta}{4}=\frac{2-\theta}{2}=1-P( \ of\ getting\ pairs \ of\ form \ (1,1) \ or \ (0,0))\)

So, clearly it is giving us a push towards involving Bernoulli trials, isn’t it !!

So, lets treat the pairs with match, .i.e. \(x=y\), be our success, and the other possibilities be failure, then our success probability is \(\frac{\theta}{2}\), where \(0\le \theta \le 2\). So, if \(S\) be the number of successful pairs in our given sample of size \(n\), then it is evident \(S \sim Binomial(n, \frac{\theta}{2})\).

So, now its simplified by all means, and we know the MLE of population proportion in binomial is the proportion of success in the sample,

Hence, \(\frac{\hat{\theta_{MLE}}}{2}= \frac{s}{n}\), where \(s\) is the number of those pairs in our sample where \(X_i=Y_i\).

So, \(\hat{\theta_{MLE}}=\frac{2(number\ of \ pairs \ in\ the\ sample\ of \ form\ (0,0)\ or \ (1,1))}{n}\).

Hence, we are done !!


Food For Thought

Say, \(X\) and \(Y\) are two independent exponential random variable with means \(\mu\) and \(\lambda\) respectively. But you observe two other variables, \(Z\) and \(W\), such that \(Z=min(X,Y)\) and \(W\) takes the value \(1\) when \(Z=X\) and \(0\) otherwise. Can you find the MLEs of the parameters ?

Give it a try !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Categories
Beautiful Mathematics videos Cheenta Probability Series Descriptive Statistics Experiment Design and Sample Survey I.S.I. and C.M.I. Entrance IIT JAM MS IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Probability Statistics

How to roll a Dice by tossing a Coin ? Cheenta Statistics Department

How can you roll a dice by tossing a coin? Can you use your probability knowledge? Use your conditioning skills.

Suppose, you have gone to a picnic with your friends. You have planned to play the physical version of the Snake and Ladder game. You found out that you have lost your dice.

The shit just became real!

Now, you have an unbiased coin in your wallet / purse. You know Probability.

Aapna Time Aayega

starts playing in the background. :p

Can you simulate the dice from the coin?

Ofcourse, you know chances better than others. :3

Take a coin.

Toss it 3 times. Record the outcomes.

HHH = Number 1

HHT = Number 2

HTH = Number 3

HTT = Number 4

THH = Number 5

THT = Number 6

TTH = Reject it, don’t ccount the toss and toss again

TTT = Reject it, don’t ccount the toss and toss again

Voila done!

What is the probability of HHH in this experiment?

Let X be the outcome in the restricted experiment as shown.

How is this experiment is different from the actual experiment?

This experiment is conditioning on the event A = {HHH, HHT, HTH, HTT, THH, THT}.

\(P( X = HHH) = P (X = HHH | X \in A ) = \frac{P (X = HHH)}{P (X \in A)} = \frac{1}{6}\)


Beautiful right?

Can you generalize this idea?

Food for thought

  • Give an algorithm to simulate any conditional probability.
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( m \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( n \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{n}\), where \( m \leq n \leq 2^k \) using conditional probability.

Watch the Video here:

Some Useful Links:

Books for ISI MStat Entrance Exam

How to Prepare for ISI MStat Entrance Exam

ISI MStat and IIT JAM Stat Problems and Solutions

Cheenta Statistics Program for ISI MStat and IIT JAM Stat

Simple Linear Regression – Playlist on YouTube

Categories
College Mathematics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Statistics Testing of Hypothesis Theory of Estimation

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 !

Problem– ISI MStat PSB 2009 Problem 8


Let \(X_1,…..,X_n\) be i.i.d. observation from the density,

\(f(x)=\frac{1}{\mu}exp(-\frac{x}{\mu}) , x>0\)

where \(\mu >0\) is an unknown parameter.

Consider the problem of testing the hypothesis \(H_o : \mu \le \mu_o\) against \(H_1 : \mu > \mu_o\).

(a) Show that the test with critical region \([\bar{X} \ge \mu_o {\chi_{2n,1-\alpha}}^2/2n]\), where \( {\chi^2}_{2n,1-\alpha} \) is the \((1-\alpha)\)th quantile of the \({\chi^2}_{2n}\) distribution, has size \(\alpha\).

(b) Give an expression of the power in terms of the c.d.f. of the \({\chi^2}_{2n}\) distribution.

Prerequisites


Likelihood Ratio Test

Exponential Distribution

Chi-squared Distribution

Solution :

This problem is quite regular and simple, from the given form of the hypotheses , it is almost clear that using Neyman-Pearson can land you in trouble. So, lets go for something more general , that is Likelihood Ratio Testing.

Hence, the Likelihood function of the \(\mu\) for the given sample is ,

\(L(\mu | \vec{X})=(\frac{1}{\mu})^n exp(-\frac{\sum_{i=1}^n X_i}{\mu}) , \mu>0\), also observe that sample mean \(\vec{X}\) is the MLE of \(\mu\).

So, the Likelihood Ratio statistic is,

\(\lambda(\vec{x})=\frac{\sup_{\mu \le \mu_o}L(\mu |\vec{x})}{\sup_\mu L(\mu |\vec{x})} \\ =\begin{cases} 1 & \mu_o \ge \bar{X} \\ \frac{L(\mu_o|\vec{x})}{L(\bar{X}|\vec{x})} & \mu_o < \bar{X} \end{cases} \)

So, our test function is ,

\(\phi(\vec{x})=\begin{cases} 1 & \lambda(\vec{x})<k \\ 0 & otherwise \end{cases}\).

We, reject \(H_o\) at size \(\alpha\), when \(\phi(\vec{x})=1\), for some \(k\), \(E_{H_o}(\phi) \le \alpha\),

Hence, \(\lambda(\vec{x}) < k \\ \Rightarrow L(\mu_o|\vec{x})<kL(\bar{X}|\vec{x}) \\ \ln k_1 -\frac{1}{\mu_o}\sum_{i=1}^n X_i < \ln k -n \ln \bar{X} -\frac{1}{n} \\ n \ln \bar{X}-\frac{n\bar{X}}{\mu_o} < K* \).

for some constant, \(K*\).

Let \(g(\bar{x})=n\ln \bar{x} -\frac{n\bar{x}}{\mu_o}\), and observe that \(g\) is,

Here, \(K*, \mu_o\) are fixed quantities.

decreasing function of \(\bar{x}\) for \(\bar{x} \ge \mu_o\),

Hence, there exists a \(c\) such that \(\bar{x} \ge c \),we have \(g(\bar) < K*\). See the figure.

So, the critical region of the test is of form \(\bar{X} \ge c\), for some \(c\) such that,

\(P_{H_o}(\bar{X} \ge c)=\alpha \), for some \(0 \le \alpha \le 1\), where \(\alpha\) is the size of the test.

Now, our task is to find \(c\), and for that observe, if \(X \sim Exponential(\theta)\), then \(\frac{2X}{\theta} \sim {\chi^2}_2\),

Hence, in this problem, since the \(X_i\)’s follows \(Exponential(\mu)\), hence, \(\frac{2n\bar{X}}{\mu} \sim {\chi^2}_{2n}\), we have,

\(P_{H_o}(\bar{X} \ge c)=\alpha \\ P_{H_o}(\frac{2n\bar{X}}{\mu_o} \ge \frac{2nc}{\mu_o})=\alpha \\ P_{H_o}({\chi^2}{2n} \ge \frac{2nc}{\mu_o})=\alpha \),

which gives \(c=\frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}\),

Hence, the rejection region is indeed, \([\bar{X} \ge \frac{\mu_o {\chi^2}_{2n;1-\alpha}}{2n}\).

Hence Proved !

(b) Now, we know that the power of the test is,

\(\beta= E_{\mu}(\phi) \\ = P_{\mu}(\lambda(\bar{x})>k)=P(\bar{X} \ge \frac{\mu_o {\chi_{2n;1-\alpha}}^2}{2n}) \\ \beta = P_{\mu}({\chi^2}_{2n} \ge \frac{mu_o}{\mu}{\chi^2}_{2n;1-\alpha}) \).

Hence, the power of the test is of form of a cdf of chi-squared distribution.


Food For Thought

Can you use any other testing procedure to conduct this test ?

Think about it !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Categories
Calculus College Mathematics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Miscellaneous Probability Statistics

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 !

Problem– ISI MStat PSB 2009 Problem 4


Let \(R\) and \(\theta\) be independent and non-negative random variables such that \(R^2 \sim {\chi_2}^2 \) and \(\theta \sim U(0,2\pi)\). Fix \(\theta_o \in (0,2\pi)\). Find the distribution of \(R\sin(\theta+\theta_o)\).

Prerequisites


Convolution

Polar Transformation

Normal Distribution

Solution :

This problem may get nasty, if one try to find the required distribution, by the so-called CDF method. Its better to observe a bit, before moving forward!! Recall how we derive the probability distribution of the sample variance of a sample from a normal population ??

Yes, you are thinking right, we need to use Polar Transformation !!

But, before transforming lets make some modifications, to reduce future complications,

Given, \(\theta \sim U(0,2\pi)\) and \(\theta_o \) is some fixed number in \((0,2\pi)\), so, let \(Z=\theta+\theta_o \sim U(\theta_o,2\pi +\theta_o)\).

Hence, we need to find the distribution of \(R\sin Z\). Now, from the given and modified information the joint pdf of \(R^2\) and \(Z\) are,

\(f_{R^2,Z}(r,z)=\frac{r}{2\pi}exp(-\frac{r^2}{2}) \ \ R>0, \theta_o \le z \le 2\pi +\theta_o \)

Now, let the transformation be \((R,Z) \to (X,Y)\),

\(X=R\cos Z \\ Y=R\sin Z\), Also, here \(X,Y \in \mathbb{R}\)

Hence, \(R^2=X^2+Y^2 \\ Z= \tan^{-1} (\frac{Y}{X}) \)

Hence, verify the Jacobian of the transformation \(J(\frac{r,z}{x,y})=\frac{1}{r}\).

Hence, the joint pdf of \(X\) and \(Y\) is,

\(f_{X,Y}(xy)=f_{R,Z}(x^2+y^2, \tan^{-1}(\frac{y}{x})) J(\frac{r,z}{x,y}) \\ =\frac{1}{2\pi}exp(-\frac{x^2+y^2}{2})\) , \(x,y \in \mathbb{R}\).

Yeah, Now it is looking familiar right !!

Since, we need the distribution of \(Y=R\sin Z=R\sin(\theta+\theta_o)\), we integrate \(f_{X,Y}\) w.r.t to \(X\) over the real line, and we will end up with, the conclusion that,

\(R\sin(\theta+\theta_o) \sim N(0,1)\). Hence, We are done !!


Food For Thought

From the above solution, the distribution of \(R\cos(\theta+\theta_o)\) is also determinable right !! Can you go further investigating the occurrence pattern of \(\tan(\theta+\theta_o)\) ?? \(R\) and \(\theta\) are the same variables as defined in the question.

Give it a try !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Categories
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
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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

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
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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

ISI MStat PSB 2008 Problem 3 | Functional equation

Content
 [hide]

    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
    Outstanding Statistics Program with Applications

    Outstanding Statistics Program with Applications

    Subscribe to Cheenta at Youtube


    Categories
    Calculus College Mathematics I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT ISI MSTAT Probability Statistics Theory of Estimation

    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 !

    Problem-ISI MStat PSB 2009 Problem 6


    Suppose \(X_1,…..,X_n\) are i.i.d. \(N(\theta,1)\), \(\theta_o \le \theta \le \theta_1\), where \(\theta_o < \theta_1\) are two specified numbers. Find the MLE of \(\theta\) and show that it is better than the sample mean \(\bar{X}\) in the sense of having smaller mean squared error.

    Prerequisites


    Maximum Likelihood Estimators

    Normal Distribution

    Mean Squared Error

    Solution :

    This is a very interesting Problem ! We all know, that if the condition “\(\theta_o \le \theta \le \theta_1\), for some specified numbers \(\theta_o < \theta_1\)” had been not given, then the MLE would have been simply \(\bar{X}=\frac{1}{n}\sum_{k=1}^n X_k\), the sample mean of the given sample. But due to the restriction over \(\theta\) things get interestingly complicated.

    So, simplify a bit, lets write the Likelihood Function of \(theta\) given this sample, \(\vec{X}=(X_1,….,X_n)’\),

    \(L(\theta |\vec{X})={\frac{1}{\sqrt{2\pi}}}^nexp(-\frac{1}{2}\sum_{k=1}^n(X_k-\theta)^2)\), when \(\theta_o \le \theta \le \theta_1\)ow taking natural log both sides and differentiating, we find that ,

    \(\frac{d\ln L(\theta|\vec{X})}{d\theta}= \sum_{k=1}^n (X_k-\theta) \).

    Now, verify that if \(\bar{X} < \theta_o\), then \(L(\theta |\vec{X})\) is always a decreasing function of \(\theta\), [ where, \(\theta_o \le \theta \le \theta_1\)], Hence the maximum likelihood attains at \(\theta_o\) itself. Similarly, when, \(\theta_o \le \bar{X} \le \theta_1\), the maximum likelihood attains at \(\bar{X}\), lastly the likelihood function will be increasing, hence the maximum likelihood will be found at \(\theta_1\).

    Hence, the Restricted Maximum Likelihood Estimator of \(\theta\), say

    \(\hat{\theta_{RML}} = \begin{cases} \theta_o & \bar{X} < \theta_o \\ \bar{X} & \theta_o\le \bar{X} \le \theta_1 \\ \theta_1 & \bar{X} > \theta_1 \end{cases}\)

    Now, to check that, \(\hat{\theta_{RML}}\) is a better estimator than \(\bar{X}\), in terms of Mean Squared Error (MSE).

    Now, \(MSE_{\theta}(\bar{X})=E_{\theta}(\bar{X}-\theta)^2=\int^{-\infty}_\infty (\bar{X}-\theta)^2f_X(x)\,dx\)

    \(=\int^{-\infty}_{\theta_o} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_o}_{\theta_1} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_1}_\infty (\bar{X}-\theta)^2f_X(x)\,dx\).

    \(\ge \int^{-\infty}_{\theta_o} (\theta_o-\theta)^2f_X(x)\,dx+\int^{\theta_o}_{\theta_1} (\bar{X}-\theta)^2f_X(x)\,dx+\int^{\theta_1}_\infty (\theta_1-\theta)^2f_X(x)\,dx\)

    \(=E_{\theta}(\hat{\theta_{RML}}-\theta)^2=MSE_{\theta}(\hat{\theta_{RML}})\).

    Hence proved !!


    Food For Thought

    Now, can you find an unbiased estimator, for \(\theta^2\) ?? Okay!! now its quite easy right !! But is the estimator you are thinking about is the best unbiased estimator !! Calculate the variance and also compare weather the Variance is attaining Cramer-Rao Lowe Bound.

    Give it a try !! You may need the help of Stein’s Identity.


    ISI MStat PSB 2008 Problem 10
    Outstanding Statistics Program with Applications

    Outstanding Statistics Program with Applications

    Subscribe to Cheenta at Youtube