Machine Learning Foundations · End Term · Jan 2026 · FN

**1. Understand the given random variable and transformation:** We are given that \(X \sim U[1, 3]\), meaning \(X\) is uniformly distributed on the interval \([1, 3]\) with probability density function \(f_X(x) = \frac{1}{3 - 1} = \frac{1}{2}\) for \(1 \le x \le 3\). We define the linear transformation: \[Y = 2X + 4\] **2. Determine the support and distribution of \(Y\):** A non-degenerate linear transformation \(Y = aX + b\) (with \(a > 0\)) of a continuous uniform random variable on \([u, v]\) is also uniformly distributed on the transformed interval \([au + b, av + b]\). Substituting \(u = 1, v = 3\) and \(a = 2, b = 4\): - Lower endpoint: \(y_{\min} = 2(1) + 4 = 6\) - Upper endpoint: \(y_{\max} = 2(3) + 4 = 10\) **3. Verify using the PDF transformation method:** Since \(x = \frac{y - 4}{2}\), the Jacobian is \(\left|\frac{dx}{dy}\right| = \frac{1}{2}\). Thus, for \(y \in [6, 10]\): \[f_Y(y) = f_X\left(\frac{y-4}{2}\right) \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = \frac{1}{10 - 6}\] This confirms that \(Y \sim U[6, 10]\). **Answer: A — U[6, 10].** A: **Correct:** Mapping the endpoints \(x = 1\) and \(x = 3\) through \(Y = 2X + 4\) gives \(Y \sim U[6, 10]\). B: **Incorrect:** A linear transformation of a uniform random variable remains uniformly distributed, not Normal. C: **Incorrect:** Transforming \([1, 3]\) by \(2X + 4\) shifts and scales the interval to \([6, 10]\); it does not stay on \([1, 3]\). D: **Incorrect:** Multiplying \(X \in [1, 3]\) by \(2\) gives \([2, 6]\), and adding \(4\) shifts it to \([6, 10]\), not \([5, 7]\).

**1. Recall the second-derivative test for convexity:** A twice-differentiable function \(h(x)\) on an open interval \((0, \infty)\) is convex if and only if its second derivative satisfies \(h''(x) \ge 0\) for all \(x \in (0, \infty)\). **2. Check the convexity of \(f(x) = x \ln x\) on \((0, \infty)\):** Using the product rule to differentiate \(f(x) = x \ln x\): \[f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\] Differentiating a second time: \[f''(x) = \frac{1}{x}\] Since \(x > 0\) on \((0, \infty)\), we have \(f''(x) = \frac{1}{x} > 0\) everywhere on \((0, \infty)\). Thus, \(f(x)\) is strictly convex. **3. Check the convexity of \(g(x) = x^2 - \ln x\) on \((0, \infty)\):** Differentiating \(g(x) = x^2 - \ln x\) twice: \[g'(x) = 2x - \frac{1}{x}\] \[g''(x) = 2 + \frac{1}{x^2}\] For all \(x > 0\), both \(2 > 0\) and \(\frac{1}{x^2} > 0\), so \(g''(x) > 0\) on \((0, \infty)\). (Equivalently, \(x^2\) is convex and \(-\ln x\) is convex on \((0, \infty)\), and the sum of two convex functions is convex.) Thus, both \(f\) and \(g\) are convex. **Answer: A — Both f and g are convex.** A: **Correct:** For all \(x > 0\), \(f''(x) = \frac{1}{x} > 0\) and \(g''(x) = 2 + \frac{1}{x^2} > 0\), so both functions are convex on \((0, \infty)\). B: **Incorrect:** Not only \(f\), but also \(g(x) = x^2 - \ln x\) is convex because \(g''(x) = 2 + 1/x^2 > 0\) for all \(x > 0\). C: **Incorrect:** Not only \(g\), but also \(f(x) = x \ln x\) is convex because \(f''(x) = 1/x > 0\) for all \(x > 0\). D: **Incorrect:** Both functions have strictly positive second derivatives on \((0, \infty)\) and are therefore convex.

**1. Write out the general quadratic form \(x^T A x\) in \(\mathbb{R}^2\):** Let \(x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) and \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\). Expanding \(x^T A x\) gives: \[x^T A x = \begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = a_{11} x_1^2 + (a_{12} + a_{21}) x_1 x_2 + a_{22} x_2^2\] **2. Match coefficients with the given polynomial:** We are given: \[f(x) = x_1^2 - 4x_1 x_2 + 2x_2^2\] Comparing coefficients term by term, a matrix \(A\) represents \(f(x)\) if and only if: - \(a_{11} = 1\) - \(a_{22} = 2\) - \(a_{12} + a_{21} = -4\) **3. Check each option:** - Option A: \(\begin{bmatrix} 1 & -2 \\ -2 & 2 \end{bmatrix}\) has \(a_{11}=1, a_{22}=2\), and \(a_{12}+a_{21} = -2 + (-2) = -4\). (Valid!) - Option B: \(\begin{bmatrix} 1 & -4 \\ 0 & 2 \end{bmatrix}\) has \(a_{11}=1, a_{22}=2\), and \(a_{12}+a_{21} = -4 + 0 = -4\). (Valid!) - Option C: \(\begin{bmatrix} 1 & -4 \\ -4 & 2 \end{bmatrix}\) has \(a_{12}+a_{21} = -4 + (-4) = -8 \neq -4\). (Invalid.) - Option D: \(\begin{bmatrix} 2 & -2 \\ -2 & 1 \end{bmatrix}\) has \(a_{11}=2 \neq 1\) and \(a_{22}=1 \neq 2\). (Invalid.) **Answer: A, B — [[1, -2], [-2, 2]] and [[1, -4], [0, 2]].** A: **Correct:** Expanding \(x^T \begin{bmatrix} 1 & -2 \\ -2 & 2 \end{bmatrix} x\) gives \(x_1^2 + (-2 - 2)x_1 x_2 + 2x_2^2 = x_1^2 - 4x_1 x_2 + 2x_2^2\). B: **Correct:** Expanding \(x^T \begin{bmatrix} 1 & -4 \\ 0 & 2 \end{bmatrix} x\) gives \(x_1^2 + (-4 + 0)x_1 x_2 + 2x_2^2 = x_1^2 - 4x_1 x_2 + 2x_2^2\). C: **Incorrect:** The off-diagonal entries sum to \(-4 + (-4) = -8\), which produces \(x_1^2 - 8x_1 x_2 + 2x_2^2\). D: **Incorrect:** The diagonal entries are swapped, producing \(2x_1^2 - 4x_1 x_2 + x_2^2\) instead of \(x_1^2 - 4x_1 x_2 + 2x_2^2\).

**1. Define a Hermitian matrix:** A square complex matrix \(M\) is called Hermitian (or self-adjoint) if it equals its own conjugate transpose: \[M^* = M\] Recall the properties of the conjugate transpose: \((X + Y)^* = X^* + Y^*\), \((XY)^* = Y^* X^*\), and \((X^*)^* = X\). Note also that \(M\) must be a square matrix for \(M^* = M\) to even be defined. **2. Check each candidate matrix:** - **Option A:** \(M = A^* + A\), where \(A\) is a complex square matrix. Taking the conjugate transpose: \[M^* = (A^* + A)^* = (A^*)^* + A^* = A + A^* = M\] Thus, \(A^* + A\) is always Hermitian. - **Option B:** \(M = A^* - A\), where \(A\) is a complex square matrix. Taking the conjugate transpose: \[M^* = (A^* - A)^* = A - A^* = -(A^* - A) = -M\] This is skew-Hermitian, not Hermitian (unless \(A = A^*\)). - **Option C:** \(M = A^* A\), where \(A\) is any complex \(m \times n\) matrix. First, \(A^*\) is \(n \times m\) and \(A\) is \(m \times n\), so \(A^* A\) is an \(n \times n\) square matrix. Taking its conjugate transpose: \[M^* = (A^* A)^* = A^* (A^*)^* = A^* A = M\] Thus, \(A^* A\) is always Hermitian! - **Option D:** \(A^* + A\) where \(A\) is a non-square \(m \times n\) matrix (\(m \neq n\)). Since \(A^*\) has size \(n \times m\) and \(A\) has size \(m \times n\), their sum \(A^* + A\) is not even defined. **Answer: A, C — A^* + A (for square A) and A^* A (for any m x n complex matrix A).** A: **Correct:** For any square matrix \(A\), \((A^* + A)^* = (A^*)^* + A^* = A + A^* = A^* + A\), so it is Hermitian. B: **Incorrect:** Since \((A^* - A)^* = A - A^* = -(A^* - A)\), this matrix is skew-Hermitian rather than Hermitian. C: **Correct:** For any \(m \times n\) matrix \(A\), \(A^* A\) is an \(n \times n\) square matrix satisfying \((A^* A)^* = A^* (A^*)^* = A^* A\). D: **Incorrect:** When \(m \neq n\), \(A^*\) is \(n \times m\) and \(A\) is \(m \times n\), so the matrix addition \(A^* + A\) is undefined.

**1. Analyze each statement about convex sets and convex functions:** - **Statement A:** The convex hull of a set \(S\), denoted \(\text{conv}(S)\), is the smallest convex set containing \(S\). If \(S\) is already convex, it contains all convex combinations of its points, so \(\text{conv}(S) = S\). Conversely, if \(\text{conv}(S) = S\), then \(S\) equals a convex set and is therefore convex. Thus, Statement A is **true**. - **Statement B:** Suppose \(S \subseteq \mathbb{R}\) is a non-empty convex set with a finite number of elements. If \(S\) contained two distinct real numbers \(a < b\), then by convexity \(S\) would have to contain the entire continuous interval \([a, b]\), which has infinitely many points. Therefore, a finite non-empty convex subset of \(\mathbb{R}\) can only contain a single point (at most—and hence exactly—one element). Thus, Statement B is **true**. - **Statement C:** If \(f\) is a convex function, then \(-f\) is a *concave* function, not convex (unless \(f\) is affine). For example, \(f(x) = x^2\) is convex, while \(-f(x) = -x^2\) is strictly concave. Thus, Statement C is **false**. - **Statement D:** A function \(f\) is convex *if and only if* its epigraph \(\text{epi}(f) = \{(x, t) : t \ge f(x)\}\) is a convex set. For a non-convex function (like \(f(x) = -x^2\)), the epigraph is not convex. Thus, Statement D is **false**. **Answer: A, B — A set S is convex iff convex-hull(S) = S, and any finite non-empty convex set in R has exactly one element.** A: **Correct:** A set contains all convex combinations of its elements if and only if it equals its own convex hull. B: **Correct:** Any convex set containing two distinct points must contain the infinite line segment between them; hence a finite non-empty convex set in \(\mathbb{R}\) has exactly one element. C: **Incorrect:** Negating a convex function reverses Jensen's inequality, making \(-f\) concave rather than convex (unless \(f\) is linear/affine). D: **Incorrect:** The epigraph of \(f\) is a convex set if and only if \(f\) is a convex function, not for an arbitrary function.

**1. Understand feasible directions at a boundary point:** We have a constrained region defined by \(g(x, y) = 3x^2 - 2xy + 3y^2 \le 1\), and a point \(p = \left(0, \frac{1}{\sqrt{3}}\right)\). First, check that \(p\) lies on the boundary \(g(x, y) = 1\): \[g\left(0, \frac{1}{\sqrt{3}}\right) = 3(0)^2 - 2(0)\left(\frac{1}{\sqrt{3}}\right) + 3\left(\frac{1}{\sqrt{3}}\right)^2 = 3\left(\frac{1}{3}\right) = 1\] For a smooth convex constraint \(g(x, y) \le 1\) active at \(p\), a non-zero vector \(d = (d_1, d_2)\) is a feasible direction (meaning \(p + \alpha d\) stays inside \(\{g \le 1\}\) for small \(\alpha > 0\)) if it points into the interior of the region, i.e.: \[\nabla g(p)^T d < 0\] **2. Compute the gradient \(\nabla g(p)\):** Taking the partial derivatives of \(g(x, y) = 3x^2 - 2xy + 3y^2\): \[\frac{\partial g}{\partial x} = 6x - 2y, \quad \frac{\partial g}{\partial y} = -2x + 6y\] Evaluating at \(p = \left(0, \frac{1}{\sqrt{3}}\right)\): \[\nabla g(p) = \left(-\frac{2}{\sqrt{3}},\; \frac{6}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}(-1, 3)\] **3. Test each direction \(d = (d_1, d_2)\) using \(\nabla g(p)^T d < 0 \iff -d_1 + 3d_2 < 0 \iff d_1 > 3d_2\):** - **Option A:** \(d = (1, 0) \implies -1(1) + 3(0) = -1 < 0\). (Feasible!) - **Option B:** \(d = (3, -1) \implies -1(3) + 3(-1) = -6 < 0\). (Feasible!) - **Option C:** \(d = (-1, 0) \implies -1(-1) + 3(0) = +1 > 0\). (Infeasible.) - **Option D:** \(d = (-5, 2) \implies -1(-5) + 3(2) = 5 + 6 = +11 > 0\). (Infeasible.) **Answer: A, B — (1, 0) and (3, -1).** A: **Correct:** At \(p = (0, 1/\sqrt{3})\), the directional derivative along \(d = (1, 0)\) is \(\nabla g(p)^T d = -2/\sqrt{3} < 0\), so moving in direction \((1, 0)\) decreases \(g\) and keeps the point feasible. B: **Correct:** Along \(d = (3, -1)\), \(\nabla g(p)^T d = \frac{2}{\sqrt{3}}(-3 - 3) = -\frac{12}{\sqrt{3}} < 0\), so \((3, -1)\) is a feasible direction. C: **Incorrect:** Along \(d = (-1, 0)\), \(\nabla g(p)^T d = +2/\sqrt{3} > 0\), which immediately increases \(g(x, y)\) above \(1\). D: **Incorrect:** Along \(d = (-5, 2)\), \(\nabla g(p)^T d = \frac{2}{\sqrt{3}}(5 + 6) = \frac{22}{\sqrt{3}} > 0\), moving outside the feasible region.

**1. State Markov's inequality:** For any non-negative random variable \(X \ge 0\) and any threshold \(a > 0\), Markov's inequality provides the upper bound: \[P(X \ge a) \le \frac{E[X]}{a}\] **2. Identify the given values:** - \(X\) is the number of automobiles sold weekly (so \(X \ge 0\)). - Expected value \(E[X] = 16\). - Threshold \(a = 20\). **3. Compute the upper bound:** Substituting \(E[X] = 16\) and \(a = 20\) into Markov's inequality: \[P(X \ge 20) \le \frac{16}{20} = \frac{4}{5} = 0.8\] **Answer: 0.8.**

**1. Understand the classifier and loss function:** We are given a binary classifier \(h(x)\) for \(x = (x_1, x_2, x_3)^T\): \[h(x) = \begin{cases} 1, & \text{if } x_1 - 2x_2 + x_3 \ge 0 \\ -1, & \text{otherwise} \end{cases}\] The 0-1 loss for each sample is \(\ell(h(x_i), y_i) = 0\) if \(h(x_i) = y_i\) and \(1\) if \(h(x_i) \neq y_i\). The empirical risk (average loss) over \(N = 4\) samples is: \[R_{\text{emp}}(h) = \frac{1}{4} \sum_{i=1}^{4} \ell(h(x_i), y_i)\] **2. Evaluate \(h(x_i)\) on each of the four data points:** - **Point 1:** \(x^{(1)} = (1, 0, 1)^T, y_1 = 1\). Compute \(1 - 2(0) + 1 = 2 \ge 0 \implies h(x^{(1)}) = 1 = y_1\). Loss \(= 0\). - **Point 2:** \(x^{(2)} = (0, 1, 1)^T, y_2 = -1\). Compute \(0 - 2(1) + 1 = -1 < 0 \implies h(x^{(2)}) = -1 = y_2\). Loss \(= 0\). - **Point 3:** \(x^{(3)} = (1, 1, 1)^T, y_3 = 1\). Compute \(1 - 2(1) + 1 = 0 \ge 0 \implies h(x^{(3)}) = 1 = y_3\). Loss \(= 0\). - **Point 4:** \(x^{(4)} = (0, 1, 0)^T, y_4 = 1\). Compute \(0 - 2(1) + 0 = -2 < 0 \implies h(x^{(4)}) = -1 \neq y_4\). Loss \(= 1\). **3. Compute the average loss:** Out of \(4\) samples, exactly \(1\) sample is misclassified: \[R_{\text{emp}}(h) = \frac{0 + 0 + 0 + 1}{4} = \frac{1}{4} = 0.25\] **Answer: 0.25.**

**1. Recall how shifting a matrix by \(cI\) affects its eigenvalues:** If \(\lambda\) is an eigenvalue of a square matrix \(A\) with eigenvector \(v\) (so \(Av = \lambda v\)), then for the shifted matrix \(B = A - 3I\): \[(A - 3I)v = Av - 3Iv = \lambda v - 3v = (\lambda - 3)v\] Thus, each eigenvalue \(\lambda_i\) of \(A\) shifts to \(\lambda_i - 3\) for \(A - 3I\). **2. Compute the eigenvalues of \(B = A - 3I\):** We are given that the eigenvalues of \(A\) are \(\lambda_1 = 1\), \(\lambda_2 = 3\), and \(\lambda_3 = 7\). Subtracting \(3\) from each eigenvalue gives the eigenvalues of \(B\): - \(\mu_1 = 1 - 3 = -2\) - \(\mu_2 = 3 - 3 = 0\) - \(\mu_3 = 7 - 3 = 4\) **3. Sum the eigenvalues of \(B\):** \[\mu_1 + \mu_2 + \mu_3 = (-2) + 0 + 4 = 2\] **Answer: 2.**

**1. Determine the rate parameter \(\lambda\) of the Exponential distribution:** For an exponentially distributed random variable \(X \sim \text{Exponential}(\lambda)\), the mean is \(E[X] = \frac{1}{\lambda}\). We are given that the mean is \(\frac{1}{2}\), so: \[\frac{1}{\lambda} = \frac{1}{2} \implies \lambda = 2\] **2. Set up the tail probability equation:** For \(c > 0\), the survival probability is \(P(X > c) = e^{-\lambda c} = e^{-2c}\). We are given that \(P(X > c) = 0.25 = \frac{1}{4}\): \[e^{-2c} = \frac{1}{4} \implies e^{2c} = 4\] **3. Solve for \(c\):** Taking the natural logarithm on both sides: \[2c = \ln(4) = 2\ln(2) \implies c = \ln(2) \approx 0.6931\] Rounding to two decimal places gives \(0.69\). **Answer: 0.69.**

**1. Understand the given joint density function:** We are given the joint PDF of \(X\) and \(Y\): \[f(x, y) = \begin{cases} 2e^{-x}e^{-2y}, & x > 0,\; y > 0 \\ 0, & \text{otherwise} \end{cases}\] Notice that \(f(x, y) = (e^{-x})(2e^{-2y})\), which means \(X \sim \text{Exponential}(1)\) and \(Y \sim \text{Exponential}(2)\) are independent random variables. **2. Set up the double integral for \(P(X < Y)\):** To find \(P(X < Y)\), we integrate \(x\) from \(0\) to \(y\), and \(y\) from \(0\) to \(\infty\): \[P(X < Y) = \int_{0}^{\infty} \left( \int_{0}^{y} e^{-x}\,dx \right) 2e^{-2y}\,dy\] **3. Evaluate the inner and outer integrals:** First, evaluate the inner integral with respect to \(x\): \[\int_{0}^{y} e^{-x}\,dx = \left[-e^{-x}\right]_{0}^{y} = 1 - e^{-y}\] Now substitute this into the outer integral: \[P(X < Y) = \int_{0}^{\infty} 2e^{-2y}(1 - e^{-y})\,dy = \int_{0}^{\infty} \left(2e^{-2y} - 2e^{-3y}\right) dy\] Since \(\int_0^\infty ke^{-ky}\,dy = 1\), we get: \[P(X < Y) = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.3333\] Rounding to two decimal places gives \(0.33\). **Answer: 0.33.**

**1. Write down the likelihood function for the dataset:** We are given the observed sample \(D = \{1, 2, 4, 5, 8\}\) of size \(n = 5\) drawn independently from the PDF: \[f(x; \theta) = \begin{cases} \dfrac{2x}{\theta^2}, & 0 < x \le \theta \\ 0, & \text{otherwise} \end{cases}\] For the likelihood \(L(\theta) = \prod_{i=1}^5 f(x_i; \theta)\) to be non-zero, every observed sample \(x_i\) must lie inside the support \((0, \theta]\). That requires \(\theta \ge \max(x_1, x_2, x_3, x_4, x_5) = 8\). **2. Analyze how \(L(\theta)\) behaves for \(\theta \ge 8\):** For \(\theta \ge 8\), the likelihood function is: \[L(\theta) = \prod_{i=1}^{5} \frac{2x_i}{\theta^2} = \frac{2^5 (1 \cdot 2 \cdot 4 \cdot 5 \cdot 8)}{\theta^{10}} = \frac{10240}{\theta^{10}}\] Because \(\theta^{10}\) is in the denominator and \(\theta > 0\), \(L(\theta)\) is a strictly decreasing function of \(\theta\) on the interval \([8, \infty)\). **3. Find the Maximum Likelihood Estimate (MLE):** To maximize \(L(\theta) = \frac{10240}{\theta^{10}}\) subject to \(\theta \ge 8\), we must choose the smallest permissible value of \(\theta\), which is: \[\hat{\theta}_{\text{MLE}} = \max_{i} x_i = 8\] **Answer: 8.**

**1. Understand the given probability density function:** The interview duration \(X\) (in hours) has the PDF: \[f_X(x) = \begin{cases} \dfrac{1}{18}(6 - x), & 0 \le x \le 6 \\ 0, & \text{otherwise} \end{cases}\] We need to find the expected time \(E[X]\). **2. Set up the integral for the expected value \(E[X]\):** By the definition of mathematical expectation for a continuous random variable: \[E[X] = \int_{-\infty}^{\infty} x f_X(x)\,dx = \int_{0}^{6} x \cdot \frac{1}{18}(6 - x)\,dx = \frac{1}{18} \int_{0}^{6} (6x - x^2)\,dx\] **3. Evaluate the integral step by step:** Integrating term by term: \[E[X] = \frac{1}{18} \left[ 3x^2 - \frac{x^3}{3} \right]_{0}^{6}\] Substitute the upper limit \(x = 6\): \[E[X] = \frac{1}{18} \left( 3(36) - \frac{216}{3} \right) = \frac{1}{18} (108 - 72) = \frac{36}{18} = 2\] **Answer: 2.**

**1. Set up the Lagrangian function:** We want to minimize \(f(x, y) = 8x^2 - 2y\) subject to the constraint \(g(x, y) = x^2 + y^2 - 1 = 0\). The Lagrangian is: \[\mathcal{L}(x, y, \lambda) = 8x^2 - 2y + \lambda(x^2 + y^2 - 1)\] **2. Find the stationary points:** Setting the partial derivatives with respect to \(x\), \(y\), and \(\lambda\) to zero: 1. \(\frac{\partial \mathcal{L}}{\partial x} = 16x + 2\lambda x = 2x(8 + \lambda) = 0 \implies x = 0\) or \(\lambda = -8\). 2. \(\frac{\partial \mathcal{L}}{\partial y} = -2 + 2\lambda y = 0 \implies \lambda y = 1\). 3. \(x^2 + y^2 = 1\). **3. Evaluate the candidate points:** - **Case 1 (\(x = 0\)):** From \(x^2 + y^2 = 1\), we get \(y^2 = 1 \implies y = 1\) or \(y = -1\). - At \((0, 1)\): \(f(0, 1) = 8(0)^2 - 2(1) = -2\). - At \((0, -1)\): \(f(0, -1) = 8(0)^2 - 2(-1) = +2\). - **Case 2 (\(\lambda = -8\)):** From \(\lambda y = 1\), we get \(y = -\frac{1}{8}\), and \(x^2 = 1 - \frac{1}{64} = \frac{63}{64}\). - Here, \(f(x, y) = 8\left(\frac{63}{64}\right) - 2\left(-\frac{1}{8}\right) = \frac{63}{8} + \frac{2}{8} = \frac{65}{8} = 8.125\). Comparing all candidate values (\(-2\), \(2\), and \(8.125\)), the global minimum value of \(8x^2 - 2y\) on the unit circle is \(-2\) (achieved at \(x = 0, y = 1\)). (Alternatively, since \(x^2 \ge 0\) and \(y \le 1\) on \(x^2 + y^2 = 1\), we immediately have \(8x^2 - 2y \ge 0 - 2(1) = -2\).) **Answer: -2.**