🏛️ ISI Advanced Examination Practice

📌 Q1 Markov Chains on a Cycle Graph (15 Marks)

🧠 Approach & Key Concepts

This problem models a Random Walk on a Cycle Graph ($C_4$).

• For part (a): While a binomial combinatorial sum is possible, the most rigorous and elegant approach involves utilizing the spectral properties of the transition matrix. Since the state space forms a ring, the transition matrix is a Circulant Matrix, whose eigenvalues are easily found using roots of unity.
• For part (b): We verify double-stochasticity for the stationary distribution. For convergence, we must analyze the Periodicity of the chain. Bipartite graphs strictly oscillate, precluding point-wise convergence.

✍️ Step-by-Step Proof / Derivation

Step 1: Setting up the Transition Matrix

Let us index the state space $S = \{A, B, C, D\}$ as $\{0, 1, 2, 3\}$. The particle moves clockwise ($+1 \pmod 4$) with probability $p=0.6$, and anticlockwise ($-1 \pmod 4$) with probability $q=0.4$. The 1-step transition matrix $\mathbf{P}$ is:

Step 2: Spectral Decomposition of the Circulant Matrix

Matrix $\mathbf{P}$ is a circulant matrix defined by its first row vector $\mathbf{c} = [0, 0.6, 0, 0.4]$. The eigenvalues $\lambda_k$ of a $4 \times 4$ circulant matrix are given by the Discrete Fourier Transform of $\mathbf{c}$:

Substituting $c_1 = p = 0.6$ and $c_3 = q = 0.4$, we compute the four eigenvalues:

• $\lambda_0 = p + q = 1$
• $\lambda_1 = p(i) + q(-i) = (0.6 - 0.4)i = 0.2i$
• $\lambda_2 = p(-1) + q(-1) = -1$
• $\lambda_3 = p(-i) + q(i) = -(0.6 - 0.4)i = -0.2i$

Step 3: Calculating $\mathbb{P}(X_{12} = A)$

Because the eigenvectors of a circulant matrix are independent of its entries (they are the columns of the Fourier matrix), the $n$-step transition probability returning to the starting state $P_{00}^{(n)}$ is simply the average of the $n$-th powers of its eigenvalues:

For $n = 12$, we evaluate the 12th power of each eigenvalue:

• $\lambda_0^{12} = 1^{12} = 1$
• $\lambda_1^{12} = (0.2i)^{12} = (0.2)^{12} (i^4)^3 = 0.2^{12} \cdot 1 = 0.2^{12}$
• $\lambda_2^{12} = (-1)^{12} = 1$
• $\lambda_3^{12} = (-0.2i)^{12} = 0.2^{12} (i^4)^3 = 0.2^{12}$

Step 4: Finding the Stationary Distribution for Part (b)

A stationary distribution $\boldsymbol{\pi} = (\pi_A, \pi_B, \pi_C, \pi_D)$ must satisfy $\boldsymbol{\pi} \mathbf{P} = \boldsymbol{\pi}$ and $\sum \pi_i = 1$. By inspecting the transition matrix $\mathbf{P}$, we observe that both the row sums and the column sums are exactly $1$.

Because $\mathbf{P}$ is a doubly stochastic matrix over a finite state space, the unique stationary distribution is the uniform distribution:

Step 5: Analyzing Convergence and Periodicity

For a discrete-time Markov chain to converge pointwise to its stationary distribution ($\lim_{n \to \infty} P_{ij}^{(n)} = \pi_j$), the chain must be irreducible, positive recurrent, and aperiodic.

Let us define the period $d$ of state $A$. The period is the greatest common divisor of all $n$ such that returning to $A$ is possible: $d = \gcd\{n : P_{00}^{(n)} > 0\}$. Because the graph is a square (a bipartite graph), any path that starts at $A$ and returns to $A$ must take an even number of steps.

Because the period $d = 2 > 1$, the Markov chain is periodic. Specifically, $P_{00}^{(2k+1)} = 0$ for all integers $k$, meaning the probability sequence strictly oscillates and does not converge to $1/4$.

📌 Q2 Discrete MLE and Asymptotic Degeneracy (15 Marks)

🧠 Approach & Key Concepts

This problem introduces a boundary-constrained optimization where the parameter space is restricted to integers ($\theta \in \mathbb{N}$).

• For part (a): The likelihood function is strictly decreasing with respect to $\theta$. Therefore, the MLE is the smallest permissible integer that covers all observed data.
• For part (b): We evaluate the probability of hitting the exact true parameter using the Cumulative Distribution Function (CDF) of the maximum order statistic $X_{(n)}$.
• For part (c): We apply the definitions of convergence in probability and convergence in distribution to show that an estimator with an exact hit probability approaching $1$ forces the scaled error to converge to a degenerate point mass.

✍️ Step-by-Step Proof / Derivation

Step 1: Finding the MLE for Part (a)

The probability density function for a single observation is $f(x_i | \theta) = \frac{1}{\theta} \mathbb{I}(0 < x_i \leq \theta)$. The joint likelihood function for the sample is the product of the marginals:

where $X_{(n)} = \max(X_1, \dots, X_n)$ is the maximum order statistic. To maximize $L(\theta)$, we must minimize $\theta$ subject to two constraints:

1. $\theta \geq X_{(n)}$ (to keep the indicator function non-zero).
2. $\theta \in \{1, 2, 3, \dots\}$ (since $\theta$ is a positive integer).

The smallest integer that is greater than or equal to $X_{(n)}$ is mathematically defined as the ceiling function. Thus, the MLE is:

Step 2: Proving Asymptotic Consistency for Part (b)

We want to find the probability that our estimator exactly matches the true integer parameter $\theta$.

For the ceiling of a number to equal integer $\theta$, the number must fall strictly between $\theta - 1$ and $\theta$. Therefore:

Since the true distribution of $X_i$ is bounded strictly at $\theta$, the upper condition $X_{(n)} \leq \theta$ happens with probability 1. We can rewrite the probability using the complement rule:

The CDF of the maximum order statistic $X_{(n)}$ for uniform data is $F_{(n)}(x) = \left(\frac{x}{\theta}\right)^n$ for $x \in (0, \theta)$. Substituting $x = \theta - 1$:

Since $\theta \geq 1$, the term $(1 - 1/\theta)$ is strictly less than $1$ (for $\theta > 1$) or exactly $0$ (for $\theta = 1$). In either case, as $n \to \infty$:

Step 3: Finding the Limiting Distribution for Part (c)

Let $Z_n = n(\theta - \hat{\theta}_n)$ be the sequence of scaled error random variables. We analyze the probability mass at exactly zero.

From Part (b), we know that $\lim_{n \to \infty} \mathbb{P}(\hat{\theta}_n = \theta) = 1$. Therefore:

Because the probability mass concentrates entirely at the single point $0$ as $n$ approaches infinity, the sequence of random variables $Z_n$ converges in probability to the constant $0$ ($Z_n \xrightarrow{p} 0$).

It is a standard theorem in asymptotic statistics that convergence in probability to a constant implies convergence in distribution to a degenerate random variable at that constant.

📌 Q3 Mixture Distributions and Popoviciu's Inequality (15 Marks)

🧠 Approach & Key Concepts

The variable $Y$ is a Gaussian Mixture Model conditioned on the latent variable $U$.

• For part (a): The variance of a mixture is elegantly decomposed using the Law of Total Variance (EVE's Law). The difference between the total variance and the expected conditional variance is the variance of the conditional means.
• For parts (b & c): We must maximize the variance of the bounded discrete random variable $U \in \{1, \dots, n\}$. By Popoviciu's Inequality on Variances (or a simple extreme-point proof), the variance of a bounded distribution is maximized when the entire probability mass is split evenly between the extreme absolute bounds.

✍️ Step-by-Step Proof / Derivation

Step 1: Applying the Law of Total Variance for Part (a)

By definition, $Y$ conditioned on $U = i$ has the distribution of $X_i$, which means $Y | U \sim N(U, \sigma_U^2)$. From this conditional distribution, we extract the conditional mean and conditional variance:

We use the Law of Total Variance to compute $V_1 = \text{Var}(Y)$:

Let us analyze the two components on the right-hand side:

1. The expected conditional variance is $\mathbb{E}[\sigma_U^2] = \sum_{i=1}^n \mathbb{P}(U=i) \sigma_i^2 = \sum_{i=1}^n \lambda_i \sigma_i^2$. This is exactly defined as $V_2$.
2. The variance of the conditional mean is $\text{Var}(U)$.

Substituting these into the equation yields:

Since the variance of any real-valued random variable is strictly non-negative ($\text{Var}(U) \geq 0$), it must follow that $V_1 \geq V_2$. Therefore, $V_1$ is bigger (or equal if $U$ is a degenerate constant).

Step 2: Formulating the Optimization Problem for Part (b)

We need to maximize the absolute difference $|V_1 - V_2|$. From Step 1, we know:

Our goal is to find the probability vector $\boldsymbol{\lambda}$ that maximizes $\text{Var}(U)$ subject to the constraint that $U$ is supported on the integers $\{1, 2, \dots, n\}$. Let us expand $\text{Var}(U)$:

Step 3: Proving the Extreme Point Mass Maximum

Notice that for any $1 \leq i \leq n$, the quadratic product $(i - 1)(n - i) \geq 0$. Expanding this geometric bound gives:

Taking the expectation on both sides with respect to $\boldsymbol{\lambda}$:

Substitute this upper bound back into the variance equation to create a function $g(x)$ where $x = \mathbb{E}[U]$:

To maximize this bounding quadratic function $g(x) = -x^2 + (n+1)x - n$, we take the derivative with respect to $x$ and set it to zero:

Thus, the variance is maximized when the expected value is exactly at the midpoint of the interval. For the inequality $i^2 \leq (n+1)i - n$ to achieve strict equality (which is required to reach the maximum theoretical variance), the probability mass must be $0$ everywhere the inequality is strict.

The inequality is strictly $0$ if and only if $i = 1$ or $i = n$. This forces $\lambda_i = 0$ for all $1 < i < n$.

To satisfy $\mathbb{E}[U] = \frac{n+1}{2}$ with only $\lambda_1$ and $\lambda_n$ active:

Solving this system yields $\lambda_1 = 1/2$ and $\lambda_n = 1/2$.

Step 4: Evaluating the Maximum Value for Part (c)

We evaluate $\text{Var}(U)$ using our optimal discrete uniform distribution on the extremes:

📌 Q4 Loss Functions and Weighted Medians (15 Marks)

🧠 Approach & Key Concepts

This problem tests the properties of $L_1$ loss functions (Absolute Deviations).

• For part (a): It is a standard statistical theorem that the sum of absolute deviations $\sum f_i |x_i - \theta|$ is minimized when $\theta$ is the median of the frequency distribution.
• For part (b): By factoring out $x_i$, the loss function transforms into a Weighted Median problem. Because the weights grow exponentially ($x_i = 2^i$), the final weight $x_n$ dominates the sum of all preceding weights. This structural dominance strictly anchors the minimum to the final data point.

✍️ Step-by-Step Proof / Derivation

Step 1: Minimizing Absolute Loss for Part (a)

We need to minimize $\psi(\theta) = \sum_{i=1}^{10} f_i |x_i - \theta|$. The minimizer of the sum of absolute deviations is the median of the data. First, we compute the total number of observations, $N$:

Since $N = 200$ (an even number), the median is the average of the 100th and 101st ordered observations. Let us calculate the cumulative frequencies to locate these positions:

• $x = 1$: $CF = 17$
• $x = 2$: $CF = 17 + 35 = 52$
• $x = 3$: $CF = 52 + 46 = 98$
• $x = 4$: $CF = 98 + 38 = 136$

Because the cumulative frequency reaches $98$ at $x=3$ and jumps to $136$ at $x=4$, both the 100th and 101st observations fall perfectly into the category where $x_i = 4$.

Step 2: Formulating the Weighted Median for Part (b)

We want to minimize $\Delta(\theta) = \sum_{i=1}^n |y_i - \theta x_i|$. Because $x_i = 2^i > 0$, we can factor $x_i$ out of the absolute value:

Let $w_i = x_i = 2^i$ act as weights, and let $z_i = y_i / x_i$ act as the data points. The function $\Delta(\theta) = \sum_{i=1}^n w_i |z_i - \theta|$ is precisely minimized when $\theta$ is the weighted median of the sequence $z_i$ with corresponding weights $w_i$.

Step 3: Analyzing the Exponential Weights

The weighted median is the value $z_k$ at which the cumulative weight crosses $50\%$ of the total weight $W$. Let us compute the total weight:

Now, let us examine the weight of the $n$-th observation, $w_n = 2^n$. We compare this single weight to the sum of all other weights combined:

Notice the strict inequality:

Step 4: Proving the Minimizer is at $z_n$

Because the single weight $w_n$ is strictly greater than the sum of all other weights, $z_n$ carries more than $50\%$ of the total distribution's mass. Regardless of where $z_n$ falls in the sorted order of $z_i$, any movement of $\theta$ away from $z_n$ will incur a heavier penalty from the $w_n |z_n - \theta|$ term than it could possibly save from the remaining $\sum_{i=1}^{n-1} w_i |z_i - \theta|$ terms.

Therefore, the minimum of the loss function is strictly anchored at $z_n$.

📌 Q5 SRSWR and Order Statistics (15 Marks)

🧠 Approach & Key Concepts

This question bridges Survey Sampling and Non-parametric order statistics.

• For part (a): In SRSWR, drawing samples is akin to distributing $n$ items into $N$ bins, which strictly follows a Multinomial Distribution. We will use the covariance structure of the multinomial to derive the variance of the linear combination.
• For part (b): We must calculate the probability that the 3rd order statistic of a sample of 5 aligns perfectly with the 8th order statistic of a population of 15. This transforms into a Binomial probability problem evaluating the counts of items drawn below and above the median.

✍️ Step-by-Step Proof / Derivation

Step 1: The Multinomial Representation for Part (a)

Let $C_i$ be the number of times the $i$-th population unit $X_i$ is drawn in the sample of size $n=5$. Since the sampling is with replacement from $N=15$ items, each draw is independent with probability $p_i = 1/15$. Therefore, the vector of counts follows a Multinomial distribution:

The sample mean $\bar{x}$ is the sum of the drawn values divided by $n=5$. This can be rewritten by grouping identical draws:

By defining $W_i = C_i / 5$ (the sample proportion), we get $\bar{x} = \sum_{i=1}^{15} W_i X_i$. Note that $W_i$ is a scaled multinomial variable.

Step 2: Deriving Expectation and Variance of $\bar{x}$

From the properties of the multinomial distribution:

• $\mathbb{E}[C_i] = n p_i = 5 \times \frac{1}{15} = \frac{1}{3}$
• $\text{Var}(C_i) = n p_i (1 - p_i) = 5 \left(\frac{1}{15}\right) \left(\frac{14}{15}\right) = \frac{14}{45}$
• $\text{Cov}(C_i, C_j) = -n p_i p_j = -5 \left(\frac{1}{15}\right) \left(\frac{1}{15}\right) = -\frac{1}{45}$ (for $i \neq j$)

Using the linearity of expectation:

Given $\sum X_i = 30$, $\mathbb{E}[\bar{x}] = 30 / 15 = 2$.

For the variance, we expand the quadratic form $\text{Var}\left(\frac{1}{5} \sum C_i X_i\right)$:

We know $(\sum X_i)^2 = \sum X_i^2 + \sum_{i \neq j} X_i X_j$. Plugging in the known sums: $30^2 = 75 + \sum_{i \neq j} X_i X_j$, meaning $\sum_{i \neq j} X_i X_j = 900 - 75 = 825$.

Step 3: Calculating $P(\tilde{x} = \tilde{X})$ for Part (b)

Since the population has 15 distinct items, the population median $\tilde{X}$ is the 8th order statistic, $X_{(8)}$. Exactly $7$ items are smaller than $\tilde{X}$, and $7$ are larger. On any single draw:

• $p_< = \mathbb{P}(X_i < \tilde{X}) = 7/15$
• $p_= = \mathbb{P}(X_i = \tilde{X}) = 1/15$
• $p_> = \mathbb{P}(X_i > \tilde{X}) = 7/15$

The sample has size $n=5$, so the sample median $\tilde{x}$ is the 3rd order statistic $x_{(3)}$. For the sample median to exactly equal the population median ($x_{(3)} = \tilde{X}$), the number of draws strictly less than $\tilde{X}$ (let's call this $N_1$) must be $\leq 2$, AND the number of draws strictly greater than $\tilde{X}$ (let's call this $N_3$) must be $\leq 2$.

Since $N_1 + N_2 + N_3 = 5$ (where $N_2$ is the number of times $\tilde{X}$ is drawn), the condition $N_3 \leq 2$ is mathematically equivalent to $N_1 + N_2 \geq 3$.

Step 4: Binomial Computations

Notice that $N_1 \sim \text{Bin}(5, 7/15)$ and $(N_1 + N_2) \sim \text{Bin}(5, 8/15)$. Let $p = 7/15$ and $q = 8/15$. The probability becomes:

Since $1-p = q$ and $1-q = p$, we group the binomial terms:

Computing each term exactly:

• $k=0$: $(8/15)^5 - (7/15)^5 = (32768 - 16807) / 15^5 = 15961 / 15^5$
• $k=1$: $5pq(q^3 - p^3) = 5\left(\frac{56}{225}\right) \left(\frac{512 - 343}{3375}\right) = \frac{280 \times 169}{15^5} = \frac{47320}{15^5}$
• $k=2$: $10p^2q^2(q - p) = 10\left(\frac{3136}{50625}\right) \left(\frac{1}{15}\right) = \frac{31360}{15^5}$

Adding them together:

📌 Q6 Linear Models and Estimability in Block Designs (15 Marks)

🧠 Approach & Key Concepts

This problem analyzes an Unbalanced Two-Way ANOVA without interaction (an incomplete block design).

• For part (a): We evaluate the absolute estimability of individual parameters in an overparameterized linear model.
• For part (b): Estimability of treatment contrasts depends entirely on the connectedness of the design's incidence graph. If there is a path between any two diets through shared age groups, the contrast is estimable.
• For part (c): We construct the standard F-test using the Adjusted Treatment Sum of Squares (obtained via the $C$-matrix from intra-block analysis) and the Error Sum of Squares.

✍️ Step-by-Step Proof / Derivation

Step 1: Assessing Absolute Estimability for Part (a)

The specified model is $y_{ijk} = \mu + \tau_i + \delta_j + \epsilon_{ijk}$. This is a classic overparameterized linear model. Let the design matrix be $\mathbf{X}$. The expected value is $\mathbb{E}[\mathbf{Y}] = \mathbf{X}\boldsymbol{\beta}$, where $\boldsymbol{\beta} = (\mu, \tau_1, \tau_2, \tau_3, \delta_1, \delta_2, \delta_3)^\top$.

A parameter vector $\mathbf{c}^\top \boldsymbol{\beta}$ is estimable if and only if $\mathbf{c}^\top$ lies in the row space of $\mathbf{X}$. However, the sum of the columns corresponding to the diet effects ($\tau_i$) equals the column for $\mu$, and the sum of the columns for the age effects ($\delta_j$) also equals the column for $\mu$.

Because of these linear dependencies, the design matrix $\mathbf{X}$ is not of full column rank. Therefore, the individual absolute parameters ($\mu, \tau_1, \dots, \delta_3$) are not estimable unless artificial constraints (like $\sum \tau_i = 0$ and $\sum \delta_j = 0$) are imposed.

Step 2: Checking Connectedness for Part (b)

A treatment contrast $\tau_i - \tau_j$ is estimable if the design is connected. We define a graph where vertices are diets and age groups, and edges exist if there is an observation in that diet-age cell. Let's trace the connections from the incidence table:

• Diet 1 connects to Age 1.
• Age 1 connects to Diet 2 and Diet 3.
• Diet 2 connects to Age 2.
• Age 2 connects to Diet 3.
• Diet 3 connects to Age 3.

Since we can trace a path from any diet to any other diet (e.g., Diet 1 $\to$ Age 1 $\to$ Diet 2), the design is connected. Therefore, all elementary contrasts $\tau_i - \tau_j$ are estimable.

Step 3: Deriving the BLUEs of the Contrasts

By the Gauss-Markov Theorem, the BLUEs of estimable functions are obtained by substituting the solutions of the normal equations $\mathbf{X}^\top \mathbf{X} \hat{\boldsymbol{\beta}} = \mathbf{X}^\top \mathbf{Y}$. For an incomplete block design, we use the intra-block equations:

where $\mathbf{Q}_i = T_i - \sum_j \frac{n_{ij}}{n_{.j}} B_j$ (adjusted diet totals), and $\mathbf{C} = \text{diag}(n_{i.}) - \mathbf{N} \text{diag}(1/n_{.j}) \mathbf{N}^\top$. Here, $\mathbf{N}$ is the $3 \times 3$ incidence matrix $n_{ij}$.

Solving this reduced system yields the estimates $\hat{\tau}_i$. Since the $\mathbf{C}$ matrix has rank $v-1 = 2$, we impose $\sum \hat{\tau}_i = 0$ to get a unique generalized inverse $\mathbf{C}^-$. The BLUE for the contrast is:

Step 4: Constructing the F-Statistic for Part (c)

We are testing $H_0: \tau_1 = \tau_2 = \tau_3$. The test statistic compares the Adjusted Treatment Sum of Squares ($SST_{adj}$) to the Error Sum of Squares ($SSE$).

The total degrees of freedom is $N - 1 = 49$. The regression model has $1 (\mu) + 2 (\text{diets}) + 2 (\text{ages}) = 5$ independent parameters. Thus, the error degrees of freedom is $\nu_E = 50 - 5 = 45$. The numerator degrees of freedom is $v - 1 = 2$.

Under the null hypothesis $H_0$, this statistic follows an $F$-distribution with $2$ and $45$ degrees of freedom. The critical region for a test of level $\alpha$ is to reject $H_0$ if:

📌 Q7 UMVUE and the Rao-Blackwell Theorem (15 Marks)

🧠 Approach & Key Concepts

This is a classic application of the Lehmann-Scheffé Theorem.

• For part (a): We use the simple raw moments of the uniform distribution.
• For part (b): We find the expectation of a function of the maximum order statistic $X_{(n)}$, which is known to be the complete and sufficient statistic for $\theta$. We then scale it to achieve unbiasedness, yielding the Uniformly Minimum Variance Unbiased Estimator (UMVUE).
• For part (c): Rather than calculating the complex conditional density $f(x_1 | x_{(n)})$, we invoke the Rao-Blackwell Theorem combined with uniqueness to prove the identity elegantly.

✍️ Step-by-Step Proof / Derivation

Step 1: Unbiased Estimator from a Single Observation for Part (a)

For a single observation $X_1 \sim U(0, \theta)$, the probability density function is $f(x) = 1/\theta$ for $0 < x < \theta$. The $r$-th raw moment is:

To make this an unbiased estimator of $\theta^r$, we multiply both sides by $(r+1)$:

Thus, $T_1 = (r+1)X_1^r$ is an unbiased estimator of $\theta^r$.

Step 2: Finding the UMVUE using $X_{(n)}$ for Part (b)

The maximum order statistic $X_{(n)}$ is the complete and sufficient statistic for $\theta$. Its probability density function is $g(y) = \frac{n y^{n-1}}{\theta^n}$ for $0 < y < \theta$. We find the expected value of $X_{(n)}^r$:

To construct an unbiased estimator, we multiply by the reciprocal of the constant term:

Since this estimator is a function of the complete sufficient statistic, by the Lehmann-Scheffé theorem, $T_2 = \frac{n+r}{n} X_{(n)}^r$ is the unique UMVUE for $\theta^r$.

Step 3: Elegant Proof via Rao-Blackwell for Part (c)

We are asked to evaluate the conditional expectation $\mathbb{E}[X_1^r | X_{(n)}]$. According to the Rao-Blackwell Theorem, if we condition an unbiased estimator on a sufficient statistic, the result is a new unbiased estimator that is a function of the sufficient statistic.

Let's condition our unbiased estimator from Part (a), $T_1 = (r+1)X_1^r$, on the sufficient statistic $X_{(n)}$:

By the properties of conditional expectation, $\mathbb{E}[\phi(X_{(n)})] = \mathbb{E}[(r+1)X_1^r] = \theta^r$. Thus, $\phi(X_{(n)})$ is an unbiased estimator of $\theta^r$ based strictly on the complete sufficient statistic $X_{(n)}$.

Because $X_{(n)}$ is complete, the unbiased estimator based on it is strictly unique (Lehmann-Scheffé uniqueness). Therefore, $\phi(X_{(n)})$ must perfectly equal the UMVUE we derived in Part (b):

By linearity of conditional expectation, we factor out the constant $(r+1)$ and divide:

📌 Q8 Asymptotic Rank Statistics and Correlations (15 Marks)

🧠 Approach & Key Concepts

This problem evaluates non-parametric rank sums via Indicator Variables. Since the labeling is random and independent of the continuous values, the rank $k$ of an observation is independent of its label.

• For part (a): We define indicator variables for the labels, sum them up, and apply the Weak Law of Large Numbers (WLLN) to find the asymptotic ratio of their expected values.
• For part (b): We compute the variance and covariance of the rank sums. Because the total number of labels is a multinomial process across ranks, the covariance will be negative. We extract the dominant terms $O(M^3)$ as $M \to \infty$.

✍️ Step-by-Step Proof / Derivation

Step 1: Defining the Indicator Variables for Part (a)

Let $I_k^{(i)}$ be an indicator variable that equals $1$ if the observation with rank $k$ (where $k = 1, \dots, M$) receives label $A_i$, and $0$ otherwise. Since the labeling is completely random and independent of the actual values, the probability is simply $p_i$:

The sum of the ranks for group $A_i$ is $R_i = \sum_{k=1}^M k I_k^{(i)}$. By linearity of expectation:

By the Law of Large Numbers, as $M \to \infty$, $R_i$ converges in probability to its expected value. Therefore, the ratio converges to the ratio of their expectations:

Step 2: Variances and Covariances for Part (b)

Because the labels are assigned independently to each observation, $I_k^{(i)}$ and $I_j^{(i)}$ are independent for $k \neq j$. Therefore, the variance of the sum is the sum of the variances:

The variance of the Bernoulli indicator is $p_i(1 - p_i)$. Thus:

To find the covariance between $R_1$ and $R_2$, we note that $\text{Cov}(I_k^{(1)}, I_j^{(2)}) = 0$ for $k \neq j$. However, for the same rank $k$, the indicators are negatively correlated because an observation cannot simultaneously be labeled $A_1$ and $A_2$:

Thus, the covariance of the rank sums is:

Step 3: Calculating the Correlation Coefficient

The correlation coefficient $\rho(R_1, R_2)$ is defined as the covariance divided by the product of the standard deviations:

Notice that the massive sum term $\frac{M(M+1)(2M+1)}{6}$ appears linearly in the covariance, and as a square root product in the denominator $\sqrt{\left(\frac{M(M+1)(2M+1)}{6}\right)^2}$. Therefore, this structural term perfectly cancels out!

Simplifying the denominator:

📚 Paper Summary & Key Focus Areas

🎯 Core Concepts Tested in This Paper

• Stochastic Processes (Q1): Utilizing the Spectral Theorem for circulant transition matrices over combinatorial brute-force methods. Understanding how the Periodicity of bipartite graph topologies prevents point-wise convergence to a stationary distribution.
• Asymptotic Theory & Inference (Q2, Q8): Handling boundary conditions in MLE where the parameter space is restricted to integers ($\mathbb{N}$), and proving convergence to a degenerate distribution. Utilizing indicator variables and the Weak Law of Large Numbers (WLLN) to find asymptotic limits and correlations of non-parametric rank sums.
• Mixture Models & Inequalities (Q3, Q4): Breaking down variance using the Law of Total Variance. Applying Popoviciu's Inequality to rigorously prove maximum variance bounds. Proving that $L_1$ loss functions are minimized by medians and extending this to exponentially dominating weighted medians.
• Sampling & Order Statistics (Q5): Translating Simple Random Sampling with Replacement (SRSWR) into a Multinomial Distribution framework to compute variances. Calculating exact alignment probabilities between sample and population medians using Binomial structures.
• Linear Models & Design of Experiments (Q6): Evaluating absolute versus contrast Estimability in overparameterized ANOVA models. Checking graph connectedness in incomplete block designs before deriving BLUEs and F-statistics via intra-block analysis.
• Advanced Point Estimation (Q7): Applying the Lehmann-Scheffé Theorem to find UMVUEs using complete sufficient statistics ($X_{(n)}$). Elegantly bypassing complex conditional integrations by invoking the Rao-Blackwell Theorem.