🏛️ ISI Advanced Examination Practice

📌 Q1 Uniform Convergence of Sequences of Functions (12 Marks)

🧠 Approach & Key Concepts

This problem tests your understanding of Real Analysis, specifically continuity and uniform convergence. Because $x^n$ approaches $0$ everywhere except at $x=1$, the behavior of the sequence is entirely governed by what happens near $x=1$. We are given that $g(1) = 0$, which suppresses the "spike" at the boundary. The proof strategy utilizes the $\epsilon-\delta$ definition of continuity to split the domain $[0,1]$ into two parts: a region close to $1$ where $g(x)$ is arbitrarily small, and a region away from $1$ where $x^n$ decays exponentially to zero.

✍️ Step-by-Step Proof / Derivation

Step 1: Utilizing the Continuity at $x=1$

Let $\epsilon > 0$ be given. Since $g$ is continuous at $x=1$ and $g(1) = 0$, by the definition of continuity, there exists a $\delta > 0$ (where $\delta < 1$) such that for all $x \in (1 - \delta, 1]$:

Step 2: Utilizing the Boundedness of Continuous Functions

Because $g$ is a continuous function on a closed and bounded interval $[0,1]$, by the Extreme Value Theorem, $g$ is bounded. Let $M$ be the supremum of $|g(x)|$ on $[0,1]$:

Step 3: Partitioning the Domain to Bound the Supremum

We now evaluate the term $|x^n g(x)|$ by splitting the interval $[0,1]$ into two sub-intervals: $[0, 1-\delta]$ and $(1-\delta, 1]$.

Case A: On the interval $(1-\delta, 1]$

For any $x$ in this range, we know $|g(x)| < \epsilon$. Furthermore, since $x \in [0,1]$, we have $x^n \leq 1$. Therefore:

Case B: On the interval $[0, 1-\delta]$

For any $x$ in this range, the maximum value $x$ can take is $1-\delta$. Thus, $|x^n| \leq (1-\delta)^n$. Using the global bound $M$ for $g(x)$, we have:

Since $0 < 1-\delta < 1$, the sequence $(1-\delta)^n$ converges to $0$ as $n \to \infty$. Therefore, there exists an integer $N$ such that for all $n > N$:

Consequently, for all $n > N$ and $x \in [0, 1-\delta]$, $|x^n g(x)| < \epsilon$.

Step 4: Conclusion

Combining Case A and Case B, we see that for any given $\epsilon > 0$, there exists an $N$ such that for all $n > N$, $|x^n g(x)| < \epsilon$ for every $x \in [0,1]$. This implies:

📌 Q2 Monotonicity of Averaged Integral Functions (12 Marks)

🧠 Approach & Key Concepts

This problem examines the behavior of the running average of an increasing function. Intuitively, if a function is always increasing, its current value will always be greater than or equal to its historical average. Therefore, adding a new, larger value to the average will pull the average up. We can rigorously prove this without assuming $f$ is differentiable by directly comparing $g(t_2) - g(t_1)$ for $t_2 > t_1 > 0$ using bounding properties of the integral.

✍️ Step-by-Step Proof / Derivation

Step 1: Setting up the Inequality

Let $0 < t_1 < t_2$. We wish to determine the sign of $g(t_2) - g(t_1)$.

We split the integral from $0$ to $t_2$ at the point $t_1$:

Group the integrals from $0$ to $t_1$ together:

Step 2: Bounding the Integrals using Monotonicity

We are given that $f(x)$ is an increasing function. This allows us to strictly bound both integrals using the boundary value $f(t_1)$.

Bound 1 (The lower integral): For all $x \in [0, t_1]$, $f(x) \leq f(t_1)$. Thus:

Bound 2 (The upper integral): For all $x \in [t_1, t_2]$, $f(x) \geq f(t_1)$. Thus:

Step 3: Evaluating the Difference

Substitute these bounds back into our equation for the difference. Because we are subtracting the first integral, its upper bound becomes a lower bound for the overall expression:

Cancel out $t_1$ in the second term:

📌 Q3 Linear Algebra & Subspace Intersections (12 Marks)

🧠 Approach & Key Concepts

This is a pure Linear Algebra problem centering on the Rank-Nullity Theorem and the dimensionality of subspace intersections. We are asked to prove that the intersection of the null space of $\mathbf{A}$ and the null space of $\mathbf{B}$ contains at least one non-zero vector. We will use the dimension formula for the sum of vector subspaces: $\dim(U \cap V) = \dim(U) + \dim(V) - \dim(U + V)$.

✍️ Step-by-Step Proof / Derivation

Step 1: Applying the Rank-Nullity Theorem

Let $N(\mathbf{A}) = \{\mathbf{x} \in \mathbb{R}^n : \mathbf{A}\mathbf{x} = \mathbf{0}\}$ be the null space of matrix $\mathbf{A}$. Similarly, let $N(\mathbf{B}) = \{\mathbf{x} \in \mathbb{R}^n : \mathbf{B}\mathbf{x} = \mathbf{0}\}$ be the null space of matrix $\mathbf{B}$. By the Rank-Nullity Theorem for $n \times n$ matrices:

Step 2: Using the Dimension Theorem for Subspaces

We are interested in the intersection $N(\mathbf{A}) \cap N(\mathbf{B})$, which represents the set of vectors annihilated by both matrices. According to the standard dimension formula for subspaces:

Substitute the rank-nullity expressions into this formula:

Step 3: Bounding the Sum Subspace

The subspace $N(\mathbf{A}) + N(\mathbf{B})$ is a subspace of $\mathbb{R}^n$. Therefore, its maximum possible dimension is $n$.

Substitute this bound back into our equation:

Step 4: Utilizing the Problem Hypothesis

We are given that $\text{rank}(\mathbf{A}) + \text{rank}(\mathbf{B}) < n$. Since ranks are integers, this implies:

Therefore, combining this with Step 3, we get:

📌 Q4 Conditional Probability in a Best-of-Five Match (12 Marks)

🧠 Approach & Key Concepts

This is a Discrete Probability & Combinatorics problem involving the Negative Binomial framework, though restricted to a finite maximum of 5 trials.

• For part (a): We evaluate the condition "match lasts 5 sets." This strictly happens if and only if the score is tied 2-2 after 4 sets. Because the sets are independent, the winner of the match is simply the winner of the 5th set.
• For part (b): We use Bayes' Theorem / Conditional Probability formula: $P(A|B) = P(A \cap B) / P(B)$. We must explicitly map out the sample space for Bimal winning in exactly 3, 4, or 5 sets.

✍️ Step-by-Step Proof / Derivation

Step 1: Defining the Probabilities

Let $p = 0.6$ be the probability Bikas wins a set. Let $q = 0.4$ be the probability Bimal wins a set.

Step 2: Solving Part (a)

We want to find $P(\text{Bikas wins} \mid \text{Match lasts 5 sets})$.

A match lasts exactly 5 sets if and only if neither player reaches 3 wins in the first 4 sets. This means the score must be exactly $2-2$ after the 4th set. Let $E_5$ be the event that the match reaches 5 sets. Let $W_A$ be the event Bikas wins.

The event $W_A \cap E_5$ implies the score was $2-2$ after 4 sets, and Bikas won the 5th set. Because the 5th set is an independent event:

Dividing by $P(E_5)$ which is exactly $P(\text{Score is } 2-2)$:

Step 3: Calculating Sample Space for Part (b)

We want to find $P(\text{Match lasts 5 sets} \mid \text{Bimal wins}) = \frac{P(\text{Bimal wins in 5 sets})}{P(\text{Bimal wins total})}$.

Let's calculate the disjoint probabilities of Bimal winning in exactly 3, 4, or 5 sets. Bimal wins the match if he wins his 3rd set on the $k$-th set played.

Case 1: Bimal wins in 3 sets

He must win all 3 initial sets (WWW).

Case 2: Bimal wins in 4 sets

He must win exactly 2 out of the first 3 sets, AND win the 4th set.

Case 3: Bimal wins in 5 sets

He must win exactly 2 out of the first 4 sets (making it 2-2), AND win the 5th set.

Step 4: Calculating Conditional Probability

The total probability that Bimal wins the match is the sum of these three disjoint cases:

Now, apply the conditional probability formula:

To simplify, multiply numerator and denominator by 100,000 to get integers, then divide by common factors (e.g., dividing both by 512):

📌 Q5 UMVUE for Discrete Uniform Distributions (12 Marks)

🧠 Approach & Key Concepts

This problem deals with exact point estimation in discrete parameter spaces.

• For part (a): Evaluating the expectation of $\Psi(X_1)$ reduces to computing a finite sum. Because of the clever construction of $\Psi(x)$ as a forward difference, the sum telescopes, collapsing massively into a single term.
• For part (b): To find the UMVUE, we leverage the Lehmann-Scheffé Theorem. We first identify the complete and sufficient statistic for $\theta$, which is the maximum order statistic $X_{(n)}$. Then, we analytically solve for a function $h(X_{(n)})$ whose expectation is exactly $e^\theta$ by matching coefficients of the discrete probability mass function (PMF).

✍️ Step-by-Step Proof / Derivation

Step 1: Evaluating the Telescoping Sum for Part (a)

The random variable $X_1$ follows a Discrete Uniform distribution on $\{1, 2, \dots, \theta\}$. Thus, its PMF is $P(X_1 = x) = \frac{1}{\theta}$ for $x \in \{1, 2, \dots, \theta\}$.

By the definition of expected value:

We factor out the constant $\frac{1}{\theta}$ and expand the summation to observe the telescoping nature:

Every intermediate term perfectly cancels out, leaving only the final term and the zero term:

Step 2: Finding the Complete Sufficient Statistic for Part (b)

The joint PMF of the sample is:

By the Neyman Factorization Theorem, $T = X_{(n)} = \max(X_1, \dots, X_n)$ is a sufficient statistic for $\theta$. Because it's a family of uniform distributions bounding the parameter space, $T$ is also a complete statistic.

Step 3: Finding the Distribution of $X_{(n)}$

To construct an unbiased estimator $h(T)$, we need the PMF of $T$. We start with the Cumulative Distribution Function (CDF):

The PMF is the difference between consecutive CDF values:

Step 4: Constructing the Unbiased Estimator $h(T)$

We want to find a function $h(T)$ such that $\mathbb{E}[h(T)] = e^\theta$ for all $\theta \in \mathbb{Z}^+$. Setting up the expectation equation:

Multiply both sides by $\theta^n$ to isolate the summation:

This equation must hold for any integer $\theta$. Therefore, it must also hold for $\theta - 1$:

Subtracting the $\theta-1$ equation from the $\theta$ equation eliminates all terms in the summation except the $y = \theta$ term:

Solving for the function $h$ at any generic value $y$:

By the Lehmann-Scheffé Theorem, since $h(X_{(n)})$ is unbiased and a function of a complete sufficient statistic, it is the unique UMVUE.

📌 Q6 Consistency of the Geometric Mean (12 Marks)

🧠 Approach & Key Concepts

This problem analyzes Asymptotic Consistency through log-transformations. The geometric mean of a product transforms into the arithmetic mean of a sum when we apply the natural logarithm. This allows us to invoke the Weak Law of Large Numbers (WLLN). Finally, applying the Continuous Mapping Theorem (CMT) translates the convergence in probability back to the original scale.

✍️ Step-by-Step Proof / Derivation

Step 1: Log-Transformation of the Geometric Mean

Let $G_n$ be the geometric mean of the sample:

Taking the natural logarithm of both sides converts the product into a sum:

Step 2: Applying the Weak Law of Large Numbers (WLLN)

Let $Y_i = \ln(X_i)$. Because the $X_i$ are i.i.d., the transformed variables $Y_i$ are also i.i.d. random variables. By the Weak Law of Large Numbers, the sample mean of $Y_i$ converges in probability to the expected value of $Y_1$:

Step 3: Evaluating the Expected Value

We must compute $\mathbb{E}[\ln(X_1)]$ where $X_1 \sim U(0, \theta)$. The probability density function is $f(x) = \frac{1}{\theta}$ for $x \in (0, \theta)$.

We solve the integral using integration by parts ($\int \ln(x) dx = x\ln(x) - x$):

Evaluating at the limits (noting that $\lim_{x \to 0^+} x\ln(x) = 0$ by L'Hôpital's rule):

Using logarithm properties, since $1 = \ln(e)$, we can rewrite this as:

Thus, we have established that $\ln(G_n) \xrightarrow{p} \ln(\theta/e)$.

Step 4: Reversing the Transformation via CMT

To find the probability limit of $G_n$ itself, we exponentiate. The exponential function $h(x) = e^x$ is a strictly continuous function everywhere on $\mathbb{R}$. By the Continuous Mapping Theorem, if $Z_n \xrightarrow{p} c$, then $h(Z_n) \xrightarrow{p} h(c)$.

📌 Q7 Sufficiency and MLE for a Bivariate Uniform Region (12 Marks)

🧠 Approach & Key Concepts

This problem evaluates 2D support boundaries in Maximum Likelihood Estimation.

• Geometric Support: The boundary of the given square region is mathematically defined by the absolute value inequality $|X| + |Y| \leq \theta$. We must calculate the area of this region to find the density constant.
• For part (a): The Neyman Factorization Theorem directly reveals the sufficient statistic by analyzing the indicator functions mapping the boundary condition across all $n$ sample points.
• For part (b): The MLE for a boundary parameter associated with a decreasing likelihood function is always the tightest bound supported by the observed data.

✍️ Step-by-Step Proof / Derivation

Step 1: Formulating the Joint Density Function

The specified vertices form a rhombus (a rotated square) centered at the origin. The geometric boundary of this region is $|x| + |y| \leq \theta$. The lengths of its diagonals are both $2\theta$. The area of a rhombus is $\frac{1}{2} d_1 d_2$:

Because the distribution is uniform over this region, the joint probability density function for a single observation $(x,y)$ is the reciprocal of the area, multiplied by an indicator function:

Step 2: Finding the Sufficient Statistic for Part (a)

For $n$ independent and identically distributed copies, the joint likelihood function is the product of the individual densities:

The product of the indicators is $1$ if and only if every observation satisfies the boundary condition. This is strictly equivalent to stating that the maximum observed sum of absolute coordinates must be less than or equal to $\theta$:

By the Neyman Factorization Theorem, we can decompose this likelihood into $g(T(\mathbf{X}, \mathbf{Y}) \mid \theta) \cdot h(\mathbf{X}, \mathbf{Y})$. Setting $h(\mathbf{X}, \mathbf{Y}) = 1$, the function strictly depends on the data only through the statistic $T$. Thus:

is a real-valued sufficient statistic for $\theta$.

Step 3: Finding the MLE for Part (b)

We analyze the behavior of the likelihood function $L(\theta)$ derived in Step 2.

For $\theta \geq T$, the function $1/\theta^{2n}$ is a strictly decreasing function of $\theta$. To maximize $L(\theta)$, we must choose the smallest possible value for $\theta$ that does not violate the support boundary (which would cause the likelihood to drop to zero).

The smallest permissible value is exactly the boundary condition itself.

📌 Q8 Finite Population Sampling and Variance Bounds (12 Marks)

🧠 Approach & Key Concepts

This problem tests Survey Sampling theory alongside discrete optimization.

• For part (a): We simply apply the standard formula for the variance of a sample mean under Simple Random Sampling Without Replacement (SRSWOR), which incorporates the finite population correction (FPC) factor.
• For part (b): We must find the absolute minimum and maximum possible values for the population variance $\sigma^2$, given the strict constraints that the values must be integers between 0 and 9, and their sum is fixed. Variance is minimized when the data points are as clustered around the mean as possible, and maximized when they are pushed to the extreme boundaries.

✍️ Step-by-Step Proof / Derivation

Step 1: Applying the SRSWOR Variance Formula for Part (a)

Under SRSWOR, the variance of the sample mean is given by the formula:

Here, the population size is $N = 25$ and the sample size is $n = 5$. We plug these values in:

Step 2: Establishing Constraints for Part (b)

We are given $\mu = 5.4$ and $N = 25$. Therefore, the strict sum of all elements in the population is:

We want to bound $\sigma^2 = \frac{1}{N} \sum_{i=1}^{25} X_i^2 - \mu^2$. Since $\mu^2 = 5.4^2 = 29.16$ is fixed, bounding the variance is strictly equivalent to bounding the sum of squares $\sum X_i^2$ subject to $\sum X_i = 135$ and $X_i \in \{0, 1, \dots, 9\}$.

Step 3: Finding the Minimum Variance Bound

To minimize the sum of squares, the integer values must be as concentrated around the mean ($5.4$) as mathematically possible. The closest available integers are $5$ and $6$. Assume the population consists of $a$ fives and $b$ sixes.

Substituting $a = 25 - b$ into the sum constraint: $5(25 - b) + 6b = 135 \implies 125 - 5b + 6b = 135 \implies b = 10$. Thus, $a = 15$. The population has fifteen $5$s and ten $6$s.

Using the relationship from part (a), the minimum variance of the sample mean is:

Step 4: Finding the Maximum Variance Bound

To maximize the sum of squares, the values must be pushed to the absolute extremes of the permissible interval, which are $0$ and $9$. Assume the population consists of $c$ nines and $d$ zeros.

Solving the second equation gives $9c = 135 \implies c = 15$. Thus, $d = 10$. The population has fifteen $9$s and ten $0$s. This perfectly satisfies the constraints.

Using the relationship from part (a), the maximum variance of the sample mean is:

📌 Q9 Minimizing Expected Risk / Likelihood Ratios (12 Marks)

🧠 Approach & Key Concepts

This problem moves away from fixed-size tests (Neyman-Pearson) and instead asks to minimize an arbitrary linear combination of errors. This translates perfectly into a Bayesian Decision Theory framework, where we minimize the expected risk $R = L_0 \alpha + L_1 \beta$. To minimize this integral over the rejection region, we place any point $x$ into the rejection region if and only if its contribution to the integral is negative. This directly yields a Likelihood Ratio threshold.

✍️ Step-by-Step Proof / Derivation

Step 1: Setting up the Objective Function

Let $W$ be the critical region where we reject $H_0$.

• $\mathbb{P}(\text{Type I error}) = \mathbb{P}(X \in W \mid H_0) = \int_W f_0(x)dx$
• $\mathbb{P}(\text{Type II error}) = \mathbb{P}(X \notin W \mid H_1) = 1 - \mathbb{P}(X \in W \mid H_1) = 1 - \int_W f_1(x)dx$

We want to minimize the objective function $L(W)$:

Step 2: Identifying the Optimal Rejection Region

To make $L(W)$ as small as possible, the integral component must be as negative as possible. Therefore, we should include a point $x$ in the rejection region $W$ if and only if the integrand evaluated at $x$ is strictly negative:

This reveals that the optimal test is a likelihood ratio test with threshold $1/2$.

Step 3: Calculating the Likelihood Ratio for Normal Densities

The densities for $N(0, 1)$ and $N(1, 1)$ are $f_0(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ and $f_1(x) = \frac{1}{\sqrt{2\pi}} e^{-(x-1)^2/2}$. We form the ratio:

Step 4: Solving for the Critical Value

Substitute the ratio back into the optimal test inequality:

Take the natural logarithm of both sides:

📌 Q10 Multiple Regression, Correlation, and Rayleigh Quotients (12 Marks)

🧠 Approach & Key Concepts

This problem leverages linear algebra within standardized OLS regression frameworks.

• For part (a): Because the variables are centered and standardized, the $\mathbf{X}^\top\mathbf{Y}$ vector directly represents the sample covariances (which are strictly proportional to the sample correlations). We invert the $\mathbf{X}^\top\mathbf{X}$ matrix to map the relationship between the $\hat{\beta}$ estimates and these correlations.
• For part (b): Finding the extreme bounds of a quadratic form $\mathbf{l}^\top \mathbf{A} \mathbf{l}$ subject to a unit length constraint $\mathbf{l}^\top\mathbf{l} = 1$ is an elegant application of the Rayleigh Quotient. The minimum variance is mathematically dictated by the smallest eigenvalue of the covariance matrix of the estimators.

✍️ Step-by-Step Proof / Derivation

Step 1: OLS Setup and Relationships for Part (a)

The standard OLS estimator is $\hat{\boldsymbol{\beta}} = (\mathbf{X}^\top\mathbf{X})^{-1}\mathbf{X}^\top\mathbf{Y}$. Because the sums of squares are 1 and the cross-product is $\alpha$, the design matrix structure is:

Let $c_1 = \sum X_{1i}Y_i$ and $c_2 = \sum X_{2i}Y_i$. Since the variables are zero-mean, $c_1$ and $c_2$ are the sample covariances. The sample correlation $r_{y,x_j}$ is $c_j / \sqrt{S_{yy} S_{xx}} = c_j / \sqrt{S_{yy}}$. Therefore, comparing the correlations is strictly equivalent to comparing $c_1$ and $c_2$.

The OLS estimates are given by the multiplication:

Step 2: Proving $r_{y,x_1} > r_{y,x_2}$

We are given that $\hat{\beta}_1 > \hat{\beta}_2$. Substituting our equations from Step 1:

Because $\alpha \in (-1, 1)$, the denominator $(1-\alpha^2)$ is strictly positive. We multiply both sides to eliminate it:

Since $\alpha > -1$, the term $(1+\alpha)$ is strictly positive. Dividing both sides yields $c_1 > c_2$. Because the correlations are proportional to $c_1$ and $c_2$ by the exact same positive constant $1/\sqrt{S_{yy}}$, it follows that $r_{y,x_1} > r_{y,x_2}$.

Step 3: Variance Formulation and Rayleigh Quotient for Part (b)

Let $\mathbf{l} = (\ell_1, \ell_2)^\top$. We seek the variance of the linear combination $\mathbf{l}^\top \hat{\boldsymbol{\beta}}$. The covariance matrix of OLS estimators is $\sigma^2(\mathbf{X}^\top\mathbf{X})^{-1}$. Thus:

We want to find the minimum of this quadratic form subject to the constraint $\mathbf{l}^\top\mathbf{l} = 1$. By the properties of the Rayleigh Quotient, the minimum value is identically the smallest eigenvalue of the matrix $\sigma^2(\mathbf{X}^\top\mathbf{X})^{-1}$.

Step 4: Eigenvalue Decomposition

Let's find the eigenvalues of $(\mathbf{X}^\top\mathbf{X})^{-1}$. The eigenvalues of an inverse matrix are the reciprocals of the original matrix's eigenvalues. We find the eigenvalues $\lambda$ of $\mathbf{X}^\top\mathbf{X} = \begin{pmatrix} 1 & \alpha \\ \alpha & 1 \end{pmatrix}$:

The eigenvalues of $\mathbf{X}^\top\mathbf{X}$ are $(1+\alpha)$ and $(1-\alpha)$. Therefore, the eigenvalues of $(\mathbf{X}^\top\mathbf{X})^{-1}$ are $\frac{1}{1+\alpha}$ and $\frac{1}{1-\alpha}$.

To find the absolute minimum possible variance, we select the smallest of these two inverse eigenvalues. The smaller fraction occurs when the denominator is largest. Between $(1+\alpha)$ and $(1-\alpha)$, the maximum possible denominator is exactly $(1+|\alpha|)$.

Therefore, the minimum eigenvalue of $(\mathbf{X}^\top\mathbf{X})^{-1}$ is exactly $\frac{1}{1+|\alpha|}$.

📚 Paper Summary & Key Focus Areas

🎯 Core Concepts Tested in This Paper

• Real Analysis (Q1, Q2): Demonstrating uniform convergence over bounded domains utilizing $\epsilon-\delta$ continuity arguments. Proving monotonicity of integral averages without relying on strict differentiability constraints.
• Linear Algebra (Q3, Q10): Applying the Rank-Nullity Theorem and the Dimension formula for subspace intersections to prove the existence of shared null vectors. Formulating constrained variance optimization problems as Rayleigh Quotients to seamlessly extract bounds using eigenvalues.
• Discrete Probability & Optimization (Q4, Q8): Parsing complex conditional sample spaces in boundary-constrained Negative Binomial setups (Best-of-Five series). Leveraging extreme-value constraints within finite integer domains to generate strict absolute bounds for population variances.
• Estimation & Decision Theory (Q5, Q6, Q9): Utilizing the Continuous Mapping Theorem alongside WLLN to prove consistency under log transformations. Deriving UMVUEs for discrete parameter bounds via the Lehmann-Scheffé theorem and telescoping series. Translating non-standard risk functions (e.g. minimizing Type I + 2*Type II errors) directly into exact Likelihood Ratio threshold tests via Bayesian decision boundaries.
• Support Boundaries in MLE (Q7): Analyzing multi-dimensional boundary supports (e.g., rhombus shapes defined by $|x| + |y| \leq \theta$) to correctly construct joint indicators and apply the Neyman Factorization theorem.