ISI MStat Entrance 2021 Problems and Solutions PSA & PSB

This post contains ISI MStat Entrance PSA and PSB 2021 Problems and Solutions that can be very helpful and resourceful for your ISI MStat Preparation.

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PSA Paper
PSB Paper
ISI MStat 2021 PSA Answer Key and Solutions

Click on the links to learn about the detailed solution. (Coming Soon)

  1. 49 (Rolle's Theorem)

2. 2 (4 - number of linear constraints)

3. k = 2 (a = -d, and form a biquadratic which has two real solutions)

4. 0 (divide by $x^4$, use $\frac{sinx}{x}$ limit result)

5. $\frac{p}{q}$ must be a rational number. (The product must be a rational number.)

6. $\alpha = 1, \beta =1$ (Use sandwich theorem on an easy inequality on ceiling of x)

7. $\frac{2n}{n+1}$ (Use geometry and definite integration)

8. $2+ \sqrt{5}$ (Just write down the pythagoras theorem in terms of the variables and solve)

9. 10 (Use the roots of unity)

10. $\frac{3}{8}$ (Find out the cases when it is non zero, and use classical probability)

11. $\frac{(n+1)^n}{n!}$ (Use ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$)

12. $P(\pi)$ is even for all $\pi$. (Observe that there is one more odd than number of evens, so there will be one odd-odd match)

13. is equal to 12. (The $i,j$th element is $a_{ii}b{ij}c{jj}$. Use gp series then.)

14. 160 (Use the fact any permutation can be written as compositions of transpositions. Observe that the given condition is equivalent to that 2 transpositions are not possible)

15. $m_t < \infty$ for all $t \geq 0$ (All monotone functions are bounded on [a,b])

16.$H(x) = \frac{1-F(-x)+ F(x)}{2}$ (If $F(x)$ is right continuous, $F(-x)$ is left continuous.).

17. $\frac{1}{25}$ (Use the distribution function of $\frac{X}{Y}$)

18. 3 (Find the distribution of order statistic, and find the expectation)

19. (II) but not (I) (If $F(x)$ is right continuous, $F(-x)$ is left continuous.).

20. $20\lambda^4$ (Use gamma integral to find the $E(X_{1}^4)$.)

21. The two new observations are 15 and 5. (Use the condition to find two linear equations to find the observations).

22. It is less than 2. (Use the beta coefficients in terms of sample covariance and sample variance, and compare)

23. 4:3 (Use Bayes' Theorem)

24. The two-sample t-test statistic and the ANOVA statistics yield the same power for any non-zero value of $\mu_1 - \mu_2$ and for any $n,m$. (Both the test statistic are one to one function of one another)

25. t³-1 - 2(t-1)

26. $\frac{2 \sum_{i=1}^{n} X_i}{n(n+1)}$ (Use the invariance property of MLE)

27. $Y_1^2 + Y_2^2 + Y_1Y_2$ (Write the bivariate normal distribution in terms of $Y_1, Y_2$ and use Neyman Factorization Theorem.)

28. can be negative (Simson's Paradox)

29. $2z$ (There are three random variables, $N$ = stopping time to get $Y=1$, $Y$ and $X$. Use the conditioning properly. Take your time)

30. $\frac{40}{3}$ (Use the property that Poisson | Poisson in the given problem follows Binomial)


ISI MStat 2021 PSB Solutions
Coming soon.

ISI MStat PSB 2021 Problem 1

Solution

ISI MStat PSB 2021 Problem 2

Solution

ISI MStat PSB 2021 Problem 3

Solution

ISI MStat PSB 2021 Problem 4

Solution

ISI MStat PSB 2021 Problem 5

Solution

ISI MStat PSB 2021 Problem 6

Solution

ISI MStat PSB 2021 Problem 7

Solution

ISI MStat PSB 2021 Problem 8

Solution

ISI MStat PSB 2021 Problem 9

Solution

Please suggest changes in the comment section.

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ISI MStat Entrance 2020 Problems and Solutions PSA & PSB

This post contains ISI MStat Entrance PSA and PSB 2020 Problems and Solutions that can be very helpful and resourceful for your ISI MStat Preparation.

ISI MStat Entrance 2020 Problems and Solutions - Subjective Paper


ISI MStat 2020 Problem 1

Let f(x)=x2−2x+2. Let L1 and L2 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 L1 and L2.

Solution

ISI MStat 2020 Problem 2

Find the number of 3×3 matrices A such that the entries of A belong to the set Z of all integers, and such that the trace of AtA is 6 . (At denotes the transpose of the matrix A).

Solution

ISI MStat 2020 Problem 3

Consider $n$ independent and identically distributed positive random variables $X_{1}, X_{2}, \ldots, X_{n}$. Suppose $S$ is a fixed subect of ${1,2, \ldots, n}$ consisting of $k$ distinct ekements where $1 \leq k<n$.
(a) Compute
$$
\mathrm{E}\left[\frac{\sum_{i \in s} X_{i}}{\sum_{i=1}^{\infty} X_{i}}\right]
$$
(b) Assume that $X_{i}$ is 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}X_{i} $ lies between $-\frac{1}{\sqrt{k+1}}$ and $\frac{1}{\sqrt{k+1}}$.

Solution

ISI MStat 2020 Problem 4

Let X1,X2,…,Xn be independent and identically distributed random variables. Let Sn=X1+⋯+Xn. For each of the following statements, determine whether they are true or false. Give reasons in each case.

(a) If Sn∼Exp with mean n, then each Xi∼Exp with mean 1 .

(b) If Sn∼Bin(nk,p), then each Xi∼Bin(k,p)

Solution

ISI MStat 2020 Problem 5

Let U1,U2,…,Un be independent and identically distributed random variables each having a uniform distribution on (0,1) . Let X=min{U1,U2,…,Un}, Y=max{U1,U2,…,Un}

Evaluate E[X∣Y=y] and E[Y∣X=x].

Solution

ISI MStat 2020 Problem 6

Suppose individuals are classified into three categories C1,C2 and C3 Let p2,(1−p)2 and 2p(1−p) be the respective population proportions, where p∈(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 Xi denote the number of individuals in the sample belonging to category Ci. Define U=X1+X32

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

(b) Show that the mean squared error of UN is p(1−p)2N

Solution

ISI MStat 2020 Problem 7

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
$$
f(u)=\frac{1}{2 \lambda} \exp \left(-\frac{|u|}{\lambda}\right),-\infty<u<\infty
$$
and $\lambda$ is an unknown parameter.
(a) Find the least squares estimator of $\beta$.
(b) Find the maximum likelihood estimator of $\beta$.

Solution

ISI MStat 2020 Problem 8

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

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

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

Solution

ISI MStat 2020 Problem 9

ISI MStat 2020 - Objective Paper


ISI MStat 2020 PSA Answer Key

Click on the links to learn about the detailed solution.

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

Please suggest changes in the comment section.

ISI MStat 2020 Probability Problems Discussion [Recorded Class]

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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

Watch the Video here:

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ISI MStat and IIT JAM Stat Problems and Solutions

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Simple Linear Regression - Playlist on YouTube

ISI MStat PSB 2013 Problem 5 | Simple Random Sampling

This is a sample problem from ISI MStat PSB 2013 Problem 5. It is based on the simple random sampling model, finding the unbiased estimates of the population size. But think over the "Food for Thought" any kind of discussion will be appreciated. Give it a try!

Problem- ISI MStat PSB 2013 Problem 5


A box has a unknown number of tickets serially numbered 1,2,.....,N. Two tickets are drawn using simple random sampling without replacement (SRSWOR) from the box. If X and Y are the numbers on the tickets and Z=max(X,Y), show that

(a) Z is not ubiased for N.

(b) \( aX+ bY+ c \) is unbiased for N if and only if \(a+b=2 \) and \( c=-1 \).

Prerequisites


Naive Probability

Counting priciples

Unbiased estimators.

Simple random sampling .

Solution :

For this problem, first let us find the pmf of Z, where we will need some counting techniques.

Since we are drawing balls at a random and not replacing the drawn ball after each draw (SRSWOR), so, clearly its about choosing two numbers from the se of N elements {1,.....,N}. So, all possible sample of size 2 , that than be drawn from the population of N units is \( { N \choose 2}\) .

now Z defined here as the maximum of the two chosen numbers, so, all possible values of Z are 2,3,....,N.

Now lets assume that Z=k, so now we just need to find out what are the possible pairs, such that k comes the max among both, or in other words if k is the maximum of the drawn numbers, what are the possible values that the other number can take ? Well, its simple the other ticket can carry any number less than k, an since there are k-1 such numbers. So there are (k-1) such pairs where the maximum numbered ticket is k. (not concerned on the ordering, of the two observation)

So, the pmf of Z=max(X,Y) , i.e. \( P(Z=k) = \begin{cases} \frac{k-1}{{N \choose 2}} & k=2,3,....,N \\ 0 & otherwise \end{cases} \)

(a) So, now to check whether Z is unbiased for N, we need to check E(Z),

\(E(Z)= \sum_{k=2}^N{k}{\frac{k-1}{{N \choose 2}}} =\frac{1}{{N \choose 2}}\sum_{k=2}^N{k(k-1)}=(\frac{2}{3}) (N+1) \).

so, \( E(Z)=\frac{2}{3} (N+1) \neq N \). Hence Z is not Unbiased for the population size, N.

(b) Similarly, we find the expectation of T=aX+bY+c,

\( E(T)=aE(X)+bE(Y)+c= a \sum_{i=1}^N i P(X=i) + b \sum_{j=1}^N j P(Y=j) + c,\)

now here \( P(X=i)=P(Y=i)= \frac{1}{N} \), so, \( E(T) = a \frac{N+1}{2}+ b\frac{N+1}{2}+c = (a+b) \frac{N+1}{2} +c,\)

clearly, E(T) = N, i.e T will be unbiased for N, iff a+b=2 and c=-1.

Hence we are done !


Food For Thought

Now, suppose that the numbers on the tickets are random, that is it can be any positive integer, ( like say 220 or 284), but thankfully you now the total number of tickets .i.e. N is known . Now you are collecting tickets for yourself and k-1 of your friends, and the number c is lucky for you and you wish to keep it in your collection, and select the remaining k-1 tickets out of N-1 tickets, and you calculate a sample mean(of the collected numbers) \( \bar{y'}\), Can I claim that \(c+ (N-1)\bar{y'}\) is an unbiased estimator of the population total ? Do you know this estimator shows less variance than the conventional unbiased estimator of the population total? Can you show that too?? why do you think the variance minimizes??

By the Way, Do you know, in mathematics 220 and 284 are quite special ?? They are the first "amicable numbers". One can obtain the other by summing over its own divisors !! So, to become amicable one needs to increase the size of their mind and heart !! Keep increasing both!! Till then.... bye.


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Unbiased, Pascal and MLE | ISI MStat 2019 PSB Problem 7

This is a problem from the ISI MStat Entrance Examination,2019 involving the MLE of the population size and investigating its unbiasedness.

The Problem:

Suppose an SRSWOR of size n has been drawn from a population labelled \(1,2,3,...,N \) , where the population size \(N\) is unknown.

(a)Find the maximum likelihood estimator \( \hat{N} \) of \(N\).

(b)Find the probability mass function of \( \hat{N} \).

(c)Show that \( \frac{n+1}{n}\hat{N} -1\) is an unbiased estimator of \(N\).

Prerequisites:

(a) Simple random sampling (SRSWR/SRSWOR)

(b)Maximum Likelihood estimator and how to find it.

(c)Unbiasedness of an estimator.

(d)Identities involving Binomial coefficients. (For this, you may refer to any standard text on Combinatorics like R.A.Brualdi,Miklos Bona etc.)

Solution:

(a) Let \(X_1,X_2,..X_n \) be the sample to be selected. In the SRSWOR scheme,

the selection probability of a sample of size \(n\) is given by \(P(s)=\frac{1}{{N \choose n}} \).

As, \(X_1,..,X_n \in \{1,2,...,N \} \) , we have the maximum among them , that is the \( n \) th order statistic, \(X_{(n)} \) is always less than \(N\).

Now, \( {N \choose n} \) is an increasing function of \(N\). So, of course, \( {X_{(n)} \choose n} \le {N \choose n } \) , thus on reciprocating, we have \(P(s) \le \frac{1}{ {X_{(n)} \choose n}} \). Hence the maximum likelihood estimator of \(N\) i.e. \( \hat{N} \) is \( X_{(n)} \).

(b) We need to find the pmf of \( \hat{N} \).

See that \(P(\hat{N}=m) = \frac{ {m \choose n} - {m-1 \choose n } }{ {N \choose n }} \) , where \(m=n,n+1,...,N \).

Can you convince yourself why?

(c) We use a well known identity , the Pascal's Identity to rewrite the distribution of $\hat{N}=X_{(n)}$ a bit more precisely:

We write \( P(\hat{N}=m) = \frac{ {m-1 \choose n-1}}{ {N \choose n} } ; \text{whenever m=n,n+1,...,N } \)

Thus, we have :

\( \begin{align}
E(\hat{N})&=\sum_{m=n}^N m P(\hat{N}=m)
=\frac{n}{\binom{N}{n}}\sum_{m=n}^N \frac{m}{n}\binom{m-1}{n-1}
=\frac{n}{\binom{N}{n}}\sum_{m=n}^N \binom{m}{n}
\end{align} \)

Also, use the Hockey Stick Identity to see that \( \sum_{m=n}^{N} {m \choose n} = {N+1 \choose n+1} \)

So, we have \( E(\hat{N})=\frac{n}{ {N \choose n}} {N+1 \choose n+1}=\frac{n(N+1)}{n+1} \).

Thus, we get \( E( \frac{n+1}{n}\hat{N} -1) = N \)

Video Solution:

Useful Exercise:

Look up the many proofs of the Hockey Stick Identity. But make sure you at least learn the proof by a combinatorial argument and an alternative proof involving visualizing the identity via the Pascal's Triangle.

Collect All The Toys | ISI MStat 2013 PSB Problem 9

Remember, we used to collect all the toy species from our chips' packets. We were all confused about how many more chips to buy? Here is how, probability guides us through in this ISI MStat 2013 Problem 9.

Toys - ISI MStat 2013 Problem 9
Collect all the toys offer

Problem

Chips are on sale for Rs. 30 each. Each chip contains exactly one toy, which can be one of four types with equal probability. Suppose you keep on buying chips and stop when you collect all the four types of toys. What will be your expected expenditure?

Prerequisites

Solution

I am really excited to find the solution. Remember, in your childhood, you were excited to buy packets of chips to collect all the toys and show them to your friends. There was a problem, that you don't have enough money to buy the chips. A natural question was how much money you have to spend?

See, the problem is right here! Let's dive into the solution. Remember we have to be a little bit mathematical here.

Let's see how we get the four new toys.

1st new toy \( \rightarrow \) 2nd new toy \( \rightarrow \) 3rd new toy \( \rightarrow \) 4th new toy.

\( N_1\) = The number of chips to be bought to get the first new toy.

\( N_2\) = The number of chips to be bought to get the second new toy after you got the first new toy.

\( N_3\) = The number of chips to be bought to get the third new toy after you got the second new toy.

\( N_4\) = The number of chips to be bought to get the fourth new toy after you got the third new toy.

Observe that the expected number of chips to be bought = \( E ( N_1 + N_2 + N_3 + N_4 \) = \(E (N_1) + E(N_2) + E(N_3) + E(N_4) \).

Now, can you guess what are the random variables \( N_1, N_2, N_3, N_4 \)?

There are all geometric random variables. ( Why? )

[ Hint : Success is when you don't get the already observed toys ]

Observe that \( N_i\) ~ Geom(\(\frac{4-i}{4}\)) ( Why? )

If \(N\) ~ Geom(\(p\)), then \(E(N) = \frac{1}{p}\).

Therefore, the required number of chips to be brought is \( 4 \times \sum_{i=1}{4} \frac{1}{4} = 8 + \frac{1}{3}\).

Therefore, you can guess, what will be the answer if there are \(n\) toys? ( \(n H_n\) ) , where \(H_n\) is the nth harmonic number.

Simulation and Verification

v = NULL
N = 0
n = 4  #number of toys
init = rep(0,n)
toy = rep(0,n)
True = rep(TRUE,n)
for (j in 1:10000)
{
  for (i in 1:100) 
  {
    s = sample(4,1)
    toy[s] = toy[s] + 1
    N = N + 1
    if (identical (toy > init, True))
    {
      break
    }
  }
v = c(v,N)
N = 0
toy = rep(0,n)
}
mean(v) = # 8.3214

Therefore, it is verified by simulation.

This image has an empty alt attribute; its file name is image-4.png
This is the histogram of the mean, which is expected to be around 8.33

Stay Tuned! Stay Shambho!