🏛️ ISI Advanced Examination Practice

📌 Q3 Matrix Theory & Positive Semi-Definiteness (12 Marks)

🧠 Approach & Key Concepts

This problem tests the understanding of quadratic forms and projection matrices. We will use the definition of a Positive Semi-Definite (PSD) matrix ($v^\top A v \geq 0$) combined with the Cauchy-Schwarz Inequality. For the rank, we will utilize the property that for an idempotent matrix, the rank is equal to its trace.

✍️ Step-by-Step Proof / Derivation

Step 1: Verification of Positive Semi-Definiteness

A symmetric matrix $\mathbf{A}$ is PSD if for any non-zero vector $\mathbf{v} \in \mathbb{R}^d$, the quadratic form is non-negative. We expand $\mathbf{v}^\top \mathbf{A} \mathbf{v}$:

Using inner product notation, where $\mathbf{v}^\top \mathbf{v} = \|\mathbf{v}\|^2$ and $\mathbf{v}^\top \mathbf{u} = \mathbf{u}^\top \mathbf{v}$:

Step 2: Applying Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality states that $(\mathbf{u}^\top \mathbf{v})^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2$. Since $\mathbf{u}$ is a unit vector, $\|\mathbf{u}\|^2 = 1$. Therefore:

Thus, $\mathbf{v}^\top \mathbf{A} \mathbf{v} \geq 0$ for all $\mathbf{v}$, meaning $\mathbf{A}$ is positive semi-definite.

Step 3: Finding the Rank of A via Idempotency

We first check if $\mathbf{A}$ is idempotent ($\mathbf{A}^2 = \mathbf{A}$).

Since $\mathbf{u}$ is a unit vector, $\mathbf{u}^\top \mathbf{u} = 1$, reducing the equation to $\mathbf{A}^2 = \mathbf{I} - \mathbf{u}\mathbf{u}^\top = \mathbf{A}$. For idempotent matrices, $\text{rank}(\mathbf{A}) = \text{tr}(\mathbf{A})$.

📌 Q4 Combinatorial Probability on Graphs (12 Marks)

🧠 Approach & Key Concepts

This is a conditional probability problem modeled as a graph coloring problem. The four pillars form a Cycle Graph ($C_4$). We need to find the probability of using exactly 4 colors given that it is a proper graph coloring (adjacent nodes have different colors). We will use the Chromatic Polynomial for the sample space and permutations for the favorable events.

✍️ Step-by-Step Proof / Derivation

Step 1: Define the Sample Space ($S$)

The total number of ways to color a cycle graph $C_n$ with $k$ colors such that adjacent vertices are distinct is given by the chromatic polynomial $P(C_n, k) = (k-1)^n + (-1)^n(k-1)$. For $n=4$ pillars and $k=4$ colors:

Step 2: Define the Favorable Event ($E$)

The event is that "all 4 colors are used." This implies every pillar gets a unique color, which is a simple permutation of 4 colors over 4 spots:

We must ensure these 24 permutations satisfy the adjacency condition. Since all colors are mutually distinct, no two adjacent pillars can share a color. Thus, all 24 ways are valid.

Step 3: Calculate the Conditional Probability

Since the colors are chosen uniformly at random, the probability is the ratio of favorable outcomes to the valid sample space:

📌 Q5 Bounds on Variance and Expectations (12 Marks)

🧠 Approach & Key Concepts

This problem tests your ability to establish bounds on moments of bounded random variables.

• For part (a): We will use the algebraic property of bounded variables. Since $X \in [a, b]$, the product $(X-a)(X-b)$ is always non-positive. Taking the expectation of this quadratic provides a strict upper bound on the second moment, $\mathbb{E}(X^2)$, and consequently the variance. (This is related to the Bhatia-Davis inequality).
• For part (b): We will exploit the independence of the variables to separate the expectations, and then apply Jensen's Inequality for the strictly convex function $g(x) = 1/x$.

✍️ Step-by-Step Proof / Derivation

Step 1: Establishing the Quadratic Bound for Part (a)

We are given that $X$ takes values in the interval $[1, 10]$. Therefore, with probability 1:

This inequality implies that $(X - 1) \geq 0$ and $(X - 10) \leq 0$. Multiplying these two terms yields a non-positive result:

Expanding this expression gives:

Step 2: Bounding the Variance and Standard Deviation

Taking the expectation on both sides of the inequality, and using the linearity of expectation, we get:

We are given that $\mathbb{E}(X) = 2.8$. Substituting this value into our inequality:

Now, we calculate the maximum possible variance of $X$ using the standard variance formula $\text{Var}(X) = \mathbb{E}(X^2) - [\mathbb{E}(X)]^2$:

The standard deviation $\sigma_X$ is the square root of the variance. Since variance is bounded by $12.96$:

This completes the proof for part (a).

Step 3: Separating the Expectation for Part (b)

We are given two independent and identically distributed (i.i.d.) random variables, $X_1$ and $X_2$, both distributed as $X$. We need to evaluate the expectation of their ratio.

Because $X_1$ and $X_2$ are independent, any function of $X_1$ is independent of any function of $X_2$. Therefore, the expectation of their product can be factored:

Since $X_1$ and $X_2$ are identically distributed, $\mathbb{E}(X_1) = \mathbb{E}(X_2) = \mathbb{E}(X)$. The equation simplifies to:

Step 4: Applying Jensen's Inequality

Consider the function $g(x) = \frac{1}{x}$. We evaluate its convexity on the support of $X$, which is $[1, 10]$ (hence $x > 0$).

The second derivative of $g(x)$ is $g''(x) = \frac{2}{x^3}$. For all $x \in [1, 10]$, $g''(x) > 0$. Since the second derivative is strictly positive, $g(x) = \frac{1}{x}$ is a strictly convex function on the interval.

By Jensen's Inequality, for any convex function $g$ and random variable $X$:

Applying this to our specific function $g(x) = 1/x$:

Substitute this lower bound back into our factored expectation from Step 3:

📌 Q6 Conditional Distributions and Bivariate Normality (12 Marks)

🧠 Approach & Key Concepts

This problem explores the relationship between conditional independence and marginal independence, as well as the strict properties required for a Bivariate Normal Distribution.

• For part (a): Two variables can be conditionally independent given $Z$, but marginally dependent if they both share the latent factor $Z$. We will use the Law of Total Covariance to show that their marginal covariance is non-zero, disproving independence.
• For part (b): A fundamental property of the Bivariate Normal distribution is that its marginal distributions must also be strictly univariate Normal. We will compute the Moment Generating Function (MGF) of the marginal distribution of $X$ to show it is a mixture, not a pure Normal distribution.

✍️ Step-by-Step Proof / Derivation

Step 1: Setting up the Conditional Distributions for Part (a)

We are given the latent variable $Z \sim U(0,1)$. Thus, its expectation and variance are:

We are also given that conditional on $Z=z$, $X$ and $Y$ are independent and identically distributed (i.i.d.) as $N(z, 1)$. This implies:

Furthermore, because $X$ and $Y$ are conditionally independent given $Z$, their conditional covariance is zero:

Step 2: Evaluating Marginal Independence via the Law of Total Covariance

To determine if $X$ and $Y$ are independent, we evaluate their marginal covariance. If they are independent, their covariance must be exactly $0$. We apply the Law of Total Covariance:

Substitute the conditional expectations and covariance we established in Step 1:

The covariance of a variable with itself is simply its variance:

Since $\text{Cov}(X, Y) = 1/12 \neq 0$, the random variables $X$ and $Y$ are not marginally independent.

Step 3: Testing for Bivariate Normality for Part (b)

If the joint distribution $(X,Y)$ follows a bivariate normal distribution, then a necessary (though not sufficient on its own) condition is that the marginal distributions of $X$ and $Y$ must be univariate normal distributions. Let us find the marginal distribution of $X$ using its Moment Generating Function (MGF), $M_X(t)$.

By the Law of Iterated Expectations, the marginal MGF is the expected value of the conditional MGF:

Conditional on $Z$, $X \sim N(Z, 1)$. The MGF of a Normal distribution $N(\mu, \sigma^2)$ is $e^{\mu t + \frac{1}{2}\sigma^2 t^2}$. Substituting $\mu = Z$ and $\sigma^2 = 1$:

Now, we take the expectation over the uniform distribution of $Z$:

Step 4: Evaluating the MGF of Z

Since $Z \sim U(0,1)$, its MGF is given by the integral:

Substituting this back into the marginal MGF for $X$ gives:

The MGF of a normal distribution must be strictly of the form $\exp(\mu t + \frac{1}{2}\sigma^2 t^2)$. Because of the additional $\frac{e^t - 1}{t}$ term, $M_X(t)$ is clearly not the MGF of a normal distribution. Therefore, the marginal distribution of $X$ is not normal (it is a continuous mixture of normals).

📌 Q7 Bivariate Normal: Sufficiency and Ancillarity (12 Marks)

🧠 Approach & Key Concepts

This problem assesses your mastery of statistical inference, specifically data reduction techniques using Sufficient and Ancillary statistics.

• For part (a): To find the minimal sufficient statistic, we will construct the joint likelihood function and apply the Lehmann-Scheffé Theorem (also known as the ratio test for minimal sufficiency). By examining the ratio of likelihoods for two sample points, we will identify the coarsest partition of the sample space that makes the ratio parameter-free.
• For part (b): A statistic is ancillary if its probability distribution does not depend on the parameter $\theta$. We will evaluate the marginal distribution of $Q_1$ to test for individual ancillarity. For the joint statistic $Q = (Q_1, Q_2)$, we will derive the cross-moment $\mathbb{E}[Q_1 Q_2]$ to demonstrate that their joint distribution structurally depends on the correlation parameter $\theta$.

✍️ Step-by-Step Proof / Derivation

Step 1: Constructing the Joint Likelihood Function for Part (a)

We are given a random sample $(X_1, Y_1), \dots, (X_n, Y_n)$ from a bivariate normal distribution with means $\mu_X = \mu_Y = 0$, variances $\sigma_X^2 = \sigma_Y^2 = 1$, and correlation $\rho = \theta$. The joint density (likelihood function) $L(\theta)$ is the product of the marginal densities:

Simplifying this by bringing the product into the exponent as a sum yields:

Step 2: Applying the Lehmann-Scheffé Theorem

Let $(\mathbf{x}, \mathbf{y})$ and $(\mathbf{x}', \mathbf{y}')$ be two distinct sample points. We construct the likelihood ratio:

For this ratio to be independent of the parameter $\theta$ for all $\theta \in (-1, 1)$, the terms multiplied by the functionally independent parameter components, namely $-\frac{1}{2(1-\theta^2)}$ and $\frac{\theta}{1-\theta^2}$, must perfectly vanish. This requires:

By the Lehmann-Scheffé theorem, the minimal sufficient statistic is the vector $T$ that defines this equivalence class:

Step 3: Evaluating Ancillarity of $Q_1$ for Part (b)

A statistic is ancillary if its distribution does not depend on $\theta$. Let's examine $Q_1 = \sum_{i=1}^n X_i^2$.

From the joint bivariate normal density $f_\theta(x, y)$, integrating out $y$ gives the marginal distribution of $X$. Since $\mu_X = 0$ and $\sigma_X^2 = 1$, the marginal distribution is standard normal:

Because the $X_i$ are independent standard normal variables, the sum of their squares follows a Chi-square distribution with $n$ degrees of freedom:

Since the $\chi^2_{(n)}$ distribution has no dependence on $\theta$, we conclude that $Q_1$ is strictly ancillary for $\theta$.

Step 4: Evaluating Ancillarity of the Joint Statistic $Q = (Q_1, Q_2)$

By symmetry, the marginal distribution of $Y_i$ is also $N(0,1)$, meaning $Q_2 = \sum_{i=1}^n Y_i^2 \sim \chi^2_{(n)}$. Thus, $Q_2$ is individually ancillary. However, for the joint vector $Q = (Q_1, Q_2)$ to be ancillary, its joint distribution must be independent of $\theta$.

To check this, let us evaluate the covariance between $Q_1$ and $Q_2$:

We compute $\text{Cov}(X_i^2, Y_i^2) = \mathbb{E}[X_i^2 Y_i^2] - \mathbb{E}[X_i^2]\mathbb{E}[Y_i^2]$. Since $X_i, Y_i \sim N(0,1)$, we know $\mathbb{E}[X_i^2] = 1$ and $\mathbb{E}[Y_i^2] = 1$.

Using the properties of conditional expectation for bivariate normals, $Y_i | X_i \sim N(\theta X_i, 1-\theta^2)$:

Recall that the 4th moment of a standard normal is $3$. Thus:

Returning to the covariance:

Because the covariance of $Q_1$ and $Q_2$ is a direct function of $\theta$, their joint distribution fundamentally depends on $\theta$. Therefore, the vector $Q$ cannot be ancillary.

📌 Q8 Hypothesis Testing with Geometric Constraints (12 Marks)

🧠 Approach & Key Concepts

This problem falls under the framework of Linear Models with Constraints. The fundamental constraint is geometric: the sum of the true angles of a planar triangle is exactly $180^\circ$ (or $\pi$ radians). The strategy involves:

• Defining a suitable unbiased estimator for the difference $\theta_1 - \theta_2$.
• Finding an independent estimator for the unknown error variance $\sigma^2$ using the geometric constraint $\sum \theta_i = \pi$.
• Constructing a Student's t-statistic (or equivalent F-statistic) by taking the ratio of the standard normal variable to the square root of the normalized chi-square variable.

✍️ Step-by-Step Proof / Derivation

Step 1: Formulating the Model and the Constraint

We are given that the measurements $Y_i$ are subject to independent $N(0, \sigma^2)$ errors. Thus, the measurement model is:

Since the angles belong to a triangle, their true sum is a known constant. Let $S = 180^\circ$ (or $\pi$). Thus, we have the deterministic constraint:

Step 2: Constructing the Test Statistic Numerator

We are testing $H_0: \theta_1 - \theta_2 = 0$. A natural, unbiased estimator for $\theta_1 - \theta_2$ is the difference in their measurements. Let $U = Y_1 - Y_2$. By the properties of normal combinations:

Under the null hypothesis $H_0$, $U \sim N(0, 2\sigma^2)$. Standardizing this gives us a standard normal variable:

Step 3: Estimating the Unknown Variance $\sigma^2$

Because $\sigma^2$ is unknown, we need to estimate it using the remaining degrees of freedom. Consider the sum of the measurements, $V = Y_1 + Y_2 + Y_3$.

Thus, $V \sim N(S, 3\sigma^2)$. We can construct a $\chi^2$ random variable with 1 degree of freedom:

Step 4: Proving Independence of $U$ and $V$

For the t-statistic to be valid, the numerator ($Z$) and the denominator (function of $X$) must be independent. Since they are linear combinations of jointly normal variables, we only need to show their covariance is zero:

Since $\text{Cov}(U, V) = 0$, $U$ and $V$ are strictly independent. Consequently, $Z$ and $X$ are independent.

Step 5: Constructing the Final t-Test

We define the test statistic $T$ as the ratio of the standard normal variable $Z$ to the square root of the chi-square variable $X$ divided by its degrees of freedom ($\nu = 1$):

The unknown parameter $\sigma$ perfectly cancels out, leaving a completely computable statistic:

Under $H_0$, this statistic $T$ follows a Student's t-distribution with $1$ degree of freedom (which is equivalent to a standard Cauchy distribution).

📌 Q9 Most Powerful Test and Neyman-Pearson Lemma (12 Marks)

🧠 Approach & Key Concepts

This problem asks for the Most Powerful (MP) test for a simple null versus a simple alternative hypothesis. We will use the Neyman-Pearson Lemma (NPL). The key steps are to find the joint likelihood function, identify the sufficient statistic (which will be the first order statistic $X_{(1)}$ due to the indicator function), formulate the Likelihood Ratio $\Lambda$, and determine the critical region based on the size $\alpha$. Because the support of the distribution depends on the parameter $\theta$, the likelihood ratio will be piecewise, leading to different randomized test structures depending on the condition $e^{2n(\theta_0 - \theta_1)} \lessgtr \alpha$.

✍️ Step-by-Step Proof / Derivation

Step 1: Likelihood Function and Sufficient Statistic

The joint probability density function (likelihood) of the sample is:

where $X_{(1)} = \min(X_1, \dots, X_n)$. By the Factorization Theorem, $X_{(1)}$ is a minimal sufficient statistic for $\theta$. We must find the distribution of $X_{(1)}$ under $H_0$ to evaluate probabilities. For $y \geq \theta_0$:

Step 2: Constructing the Likelihood Ratio ($\Lambda$)

We are testing $H_0: \theta = \theta_0$ against $H_1: \theta = \theta_1$ (with $\theta_1 > \theta_0$). The likelihood ratio is:

Since $\theta_1 > \theta_0$, we analyze $\Lambda$ piecewise over the support of $H_0$ ($X_{(1)} \geq \theta_0$):

• If $\theta_0 \leq X_{(1)} < \theta_1$, the numerator is $0$, so $\Lambda = 0$.
• If $X_{(1)} \geq \theta_1$, both indicators are $1$, so $\Lambda = e^{2n(\theta_1 - \theta_0)} = k^*$.

By the Neyman-Pearson Lemma, the Most Powerful test function $\phi(X_{(1)})$ has the form:

where $c$ and $\gamma$ are chosen such that $\mathbb{E}_{\theta_0}[\phi(X_{(1)})] = \alpha$. Note that $P_{\theta_0}(\Lambda = k^*) = P_{\theta_0}(X_{(1)} \geq \theta_1) = e^{-2n(\theta_1 - \theta_0)} = e^{2n(\theta_0 - \theta_1)}$.

Step 3: Deriving the Test for Case (a) $e^{2n(\theta_0 - \theta_1)} < \alpha$

In this case, the probability of obtaining the maximum likelihood ratio $k^*$ under $H_0$ is strictly less than the allowed size $\alpha$:

If we set the threshold $c = k^*$, our test size would be too small. Therefore, we must lower the threshold to $c = 0$. The test rule becomes:

• Reject $H_0$ with probability $1$ if $\Lambda > 0$ (i.e., $X_{(1)} \geq \theta_1$).
• Reject $H_0$ with probability $\gamma_1$ if $\Lambda = 0$ (i.e., $\theta_0 \leq X_{(1)} < \theta_1$).

We solve for $\gamma_1$ using the size constraint:

Step 4: Deriving the Test for Case (b) $e^{2n(\theta_0 - \theta_1)} > \alpha$

In this case, the probability of obtaining the maximum likelihood ratio $k^*$ under $H_0$ exceeds $\alpha$:

If we always rejected $H_0$ when $X_{(1)} \geq \theta_1$, the Type I error would exceed $\alpha$. Thus, we must set $c = k^* = e^{2n(\theta_1 - \theta_0)}$ and introduce randomization at this boundary. The test rule becomes:

• Reject $H_0$ with probability $\gamma_2$ if $\Lambda = k^*$ (i.e., $X_{(1)} \geq \theta_1$).
• Reject $H_0$ with probability $0$ if $\Lambda < k^*$ (i.e., $\theta_0 \leq X_{(1)} < \theta_1$).

We solve for $\gamma_2$ using the size constraint:

📌 Q10 Multivariate Normal Distribution (12 Marks)

🧠 Approach & Key Concepts

This problem evaluates your structural understanding of the Multivariate Normal Distribution.

• For part (a): Direct integration is impossible. Instead, we exploit the symmetry of the multivariate normal distribution. We look for a distribution-preserving transformation that flips the sign of the target expression, proving it is symmetric around $0$.
• For part (b): The Multiple Correlation Coefficient $R_{1 \cdot 234}$ measures the maximum correlation between $X_1$ and a linear combination of $(X_2, X_3, X_4)$. It is derived from partitioning the covariance matrix.
• For part (c): We compute the Partial Correlation Coefficient by finding the conditional covariance matrix. A critical property of the multivariate normal is that uncorrelated blocks are strictly independent.

✍️ Step-by-Step Proof / Derivation

Step 1: Analyzing Independence and Symmetry for Part (a)

Let $W = X_1 - X_2 X_3 X_4$. We want to find $\mathbb{P}(W > 0)$. Notice the zero entries in the covariance matrix $\boldsymbol{\Sigma}$:

• $\text{Cov}(X_1, X_2) = 0$, $\text{Cov}(X_1, X_3) = 0$
• $\text{Cov}(X_4, X_2) = 0$, $\text{Cov}(X_4, X_3) = 0$

Because the variables are jointly normal, zero covariance implies strict independence. Thus, the random vector $(X_1, X_4)$ is completely independent of the random vector $(X_2, X_3)$.

Furthermore, since the mean vector is $\mathbf{0}$, the distribution of $(X_1, X_4)$ is symmetric. Specifically, if we apply the transformation $(X_1, X_4) \mapsto (-X_1, -X_4)$, the mean remains $(0,0)$ and the covariance matrix remains unchanged because $\text{Cov}(-X_1, -X_4) = (-1)(-1)\text{Cov}(X_1, X_4) = 1$.

Therefore, $(-X_1, X_2, X_3, -X_4)$ has the exact same joint distribution as $(X_1, X_2, X_3, X_4)$.

Step 2: Evaluating the Probability $\mathbb{P}(W > 0)$

Let us apply this transformation to our variable $W$:

Since the joint distribution is invariant under this transformation, $W$ and $-W$ are identically distributed ($W \stackrel{d}{=} -W$). This means the distribution of $W$ is perfectly symmetric around $0$.

Because $W$ is a continuous random variable (a polynomial of continuous normals), $\mathbb{P}(W = 0) = 0$. Therefore:

Thus, $\mathbb{P}(X_1 > X_2 X_3 X_4) = \mathbb{P}(X_1 - X_2 X_3 X_4 > 0) = 0.5$.

Step 3: Partitioning the Covariance Matrix for Part (b)

The squared multiple correlation coefficient of $X_1$ with $\mathbf{X}_{-1} = (X_2, X_3, X_4)^\top$ is given by:

From the given $\boldsymbol{\Sigma}$, we extract the sub-matrices:

Step 4: Computing $R_{1 \cdot 234}$

Notice that $\boldsymbol{\Sigma}_{-1, -1}$ is block-diagonal. The vector $\boldsymbol{\Sigma}_{1, -1}$ only has a non-zero entry in the 3rd position (corresponding to $X_4$). Therefore, we only need the bottom-right entry of the inverse matrix $\boldsymbol{\Sigma}_{-1, -1}^{-1}$, which is simply $1/4$.

Now, divide by $\boldsymbol{\Sigma}_{11}$:

Taking the positive square root (since $R \geq 0$):

Step 5: Computing the Partial Correlation for Part (c)

We need the partial correlation $\rho_{23 \cdot 14}$, which is derived from the conditional covariance matrix of $(X_2, X_3)$ given $(X_1, X_4)$:

We inspect the cross-covariance matrix $\boldsymbol{\Sigma}_{23, 14}$ between the block $(X_2, X_3)$ and the block $(X_1, X_4)$. From $\boldsymbol{\Sigma}$, all these cross-terms are zero.

Because the cross-covariance is a zero matrix (confirming our observation in Step 1 that the blocks are independent), conditioning on $(X_1, X_4)$ provides zero information about $(X_2, X_3)$. Thus, the conditional covariance is exactly equal to the marginal covariance:

The partial correlation is therefore identical to the marginal Pearson correlation between $X_2$ and $X_3$:

📚 Paper Summary & Key Focus Areas

🎯 Core Concepts Tested in This Paper

• Real Analysis & Topology (Q1, Q2): Mastery of the Monotone Convergence Theorem, fixed-point limits, and recognizing topological traps (e.g., knowing when to apply Bolzano-Weierstrass and when to provide counterexamples for unbounded sets).
• Matrix Theory & Linear Algebra (Q3): Exploiting structural properties like idempotency and trace-rank equivalence for projection matrices, and applying the Cauchy-Schwarz inequality for PSD verification.
• Combinatorics & Probability (Q4, Q5, Q6): Identifying structural isomorphisms (like mapping pillars to a $C_4$ graph) to use Chromatic Polynomials. Distinguishing between conditional and marginal independence using the Law of Total Covariance, and applying Jensen's Inequality to bound moments.
• Advanced Statistical Inference (Q7, Q9): Deriving minimal sufficient statistics via the Lehmann-Scheffé (ratio) test, proving ancillarity, and rigorously constructing randomized Most Powerful (MP) tests using the Neyman-Pearson Lemma under different boundary size ($\alpha$) conditions.
• Multivariate Normal & Linear Models (Q8, Q10): Partitioning covariance matrices to compute multiple and partial correlation coefficients. Using zero-covariance to establish strict block independence, and constructing valid t-statistics under geometric constraints.