Mathematical Foundations of Generative AI · Quiz 1 · May 2026

**1. Review the properties of \( f \)-divergences:** Given a convex, lower semi-continuous generator \( f \) with \( f(1) = 0 \), the \( f \)-divergence is: \[ D_f(P \| Q) = \int_{\mathcal{X}} q(x) f\!\left(\frac{p(x)}{q(x)}\right) dx \] - **Option A:** \( D_f(P \| Q) \) is **not** always symmetric (for example, forward KL and reverse KL are asymmetric unless \( f(u) = u f(1/u) \)). False. - **Option B:** By Jensen's inequality (for strictly convex \( f \) at \( 1 \)), \( D_f(P \| Q) \ge f\!\left(\int q(x)\frac{p(x)}{q(x)}dx\right) = f(1) = 0 \), with equality \( D_f(P \| Q) = 0 \) if and only if \( P = Q \) almost everywhere. True. - **Option C:** Choosing \( f(u) = u \log u \) gives: \[ \int q(x) \frac{p(x)}{q(x)} \log\!\left(\frac{p(x)}{q(x)}\right) dx = \int p(x) \log\!\left(\frac{p(x)}{q(x)}\right) dx = D_{\text{KL}}(P \| Q) \] True. - **Option D:** Choosing \( f(u) = -\log u \) gives the reverse KL divergence \( D_{\text{KL}}(Q \| P) \), not \( D_{\text{KL}}(P \| Q) \). False. **Answer: B and C.** A: **Incorrect:** Most \( f \)-divergences (such as KL divergence) are asymmetric: \( D_f(P \| Q) \neq D_f(Q \| P) \). B: **Correct:** By Jensen's inequality and \( f(1) = 0 \), \( D_f(P \| Q) = 0 \) if and only if \( P = Q \). C: **Correct:** Substituting \( f(u) = u \log u \) into \( \int q(x) f(p(x)/q(x)) dx \) yields \( \int p(x) \log(p(x)/q(x)) dx = D_{\text{KL}}(P \| Q) \). D: **Incorrect:** Setting \( f(u) = -\log u \) gives reverse KL \( D_{\text{KL}}(Q \| P) \), not \( D_{\text{KL}}(P \| Q) \).

**1. Check each generator \( f(u) \) and its corresponding \( f \)-divergence:** - **Option A (\( f(u) = (u - 1)^2 \)):** Gives the Pearson \( \chi^2 \)-divergence, not Total Variation distance. - **Option B (\( f(u) = (1 - \sqrt{u})^2 \)):** Gives the squared Hellinger distance, not KL-divergence. - **Option C (\( f(u) = u \log u \)):** Gives the **(forward) KL-divergence** \( D_{\text{KL}}(P \| Q) \). **Correct.** - **Option D (\( f(u) = -\log u \)):** Gives the reverse KL-divergence \( D_{\text{KL}}(Q \| P) \), not Jensen–Shannon divergence. - **Option E (\( f(u) = \frac{1}{2}|u - 1| \)):** Gives \( \frac{1}{2} \int |p(x) - q(x)| dx = \text{TV}(P, Q) \), the **Total Variation distance**. **Correct.** **Answer: C and E.** A: **Incorrect:** \( f(u) = (u - 1)^2 \) generates the \( \chi^2 \)-divergence. B: **Incorrect:** \( f(u) = (1 - \sqrt{u})^2 \) generates the squared Hellinger distance. C: **Correct:** \( f(u) = u \log u \) generates the forward KL-divergence \( D_{\text{KL}}(P \| Q) \). D: **Incorrect:** \( f(u) = -\log u \) generates the reverse KL-divergence, not the Jensen–Shannon divergence. E: **Correct:** \( f(u) = \frac{1}{2}|u - 1| \) generates the Total Variation distance.

**1. Analyze the GAN minimax game at the global optimum \( p_g = p_{\text{data}} \):** - For a fixed generator \( G \), the optimal discriminator is: \[ D^*(x) = \frac{p_{\text{data}}(x)}{p_{\text{data}}(x) + p_g(x)} \] - When \( p_g = p_{\text{data}} \), the discriminator cannot distinguish real samples from generated samples, so: \[ D^*(x) = \frac{p_{\text{data}}(x)}{2 p_{\text{data}}(x)} = \frac{1}{2} \quad \text{everywhere} \] - Substituting \( D^*(x) = 1/2 \) into the minimax value function gives \( V(D^*, G^*) = \log(1/2) + \log(1 - 1/2) = -2\log 2 = -\log 4 \neq 0 \) (and for the non-saturating loss \( -\mathbb{E}[\log D(G(z))] = \log 2 \neq 0 \)). Thus, the generator's loss is **not zero**. **Answer: B — The discriminator outputs \( 1/2 \) everywhere, but the generator's loss is not zero.** A: **Incorrect:** At \( D^*(x) = 1/2 \), the minimax value is \( -\log 4 \) (and non-saturating generator loss is \( \log 2 \)), not zero. B: **Correct:** When \( p_g = p_{\text{data}} \), \( D^*(x) = 1/2 \) everywhere and the equilibrium loss is non-zero (\( -\log 4 \)). C: **Incorrect:** Outputting 1 on real and 0 on fake happens when \( p_g \) and \( p_{\text{data}} \) have disjoint supports, not at \( p_g = p_{\text{data}} \). D: **Incorrect:** At the global optimum, \( V(D^*, G^*) = -\log 4 \) (since \( \text{JSD}(p_{\text{data}} \| p_g) = 0 \)).

**1. Check the two requirements for a valid \( f \)-divergence generator on \( (0, \infty) \):** A function \( f: (0, \infty) \to \mathbb{R} \) is a valid \( f \)-divergence generator if and only if: 1. \( f \) is **convex** on \( (0, \infty) \) (i.e., \( f''(u) \ge 0 \)), and 2. \( f(1) = 0 \). **2. Test each candidate function:** - **Option A (\( f(u) = u \log u - u + 1 \)):** \( f(1) = 1(0) - 1 + 1 = 0 \), and \( f''(u) = 1/u > 0 \) on \( (0, \infty) \). **Valid.** - **Option B (\( f(u) = |u - 1| \)):** \( f(1) = 0 \), and \( |u - 1| \) is convex on \( \mathbb{R} \). **Valid.** - **Option C (\( f(u) = \log u \)):** Although \( f(1) = 0 \), \( f''(u) = -1/u^2 < 0 \), so \( \log u \) is strictly **concave**, not convex! **Invalid.** - **Option D (\( f(u) = u^2 - 1 \)):** \( f(1) = 1 - 1 = 0 \), and \( f''(u) = 2 > 0 \) (convex). **Valid.** **Answer: A, B, and D.** A: **Correct:** Satisfies \( f(1) = 0 \) and \( f''(u) = 1/u > 0 \) (convex). B: **Correct:** Satisfies \( f(1) = 0 \) and is convex (generates Total Variation up to factor 2). C: **Incorrect:** \( f(u) = \log u \) is concave (\( f''(u) = -1/u^2 < 0 \)), violating convexity. D: **Correct:** Satisfies \( f(1) = 0 \) and \( f''(u) = 2 > 0 \) (convex, generates \( \chi^2 \)-divergence).

**1. Evaluate each statement about GAN training:** - **Option A:** The discriminator \( D \) is trained as a binary classifier to distinguish real training samples (label 1) from generator samples \( G(z) \) (label 0). **True.** - **Option B:** Under the minimax loss \( \mathbb{E}_z[\log(1 - D(G(z)))] \), if \( D \) becomes too accurate early in training so that \( D(G(z)) \approx 0 \) on fake samples, the gradient of \( \log(1 - D) \) near \( D = 0 \) flattens out (vanishes), halting generator learning. **True.** - **Option C:** The generator tries to *fool* the discriminator (minimizing \( D \)'s accuracy on fakes), not maximize it. **False.** - **Option D:** The *generator* minimizes the divergence between \( p_{\text{data}} \) and \( p_g \), whereas the *discriminator* maximizes the objective to separate them. **False.** **Answer: A and B.** A: **Correct:** The discriminator is trained via binary cross-entropy on real data and generated data. B: **Correct:** When \( D(G(z)) \to 0 \), \( \log(1 - D(G(z))) \approx -D(G(z)) \) saturates, causing vanishing gradients for \( G \). C: **Incorrect:** The generator aims to fool the discriminator, opposing its classification accuracy. D: **Incorrect:** The generator minimizes the divergence between distributions, while the discriminator maximizes it.

**1. Evaluate each statement about Conditional GANs (cGANs):** - **Option A:** Fixing the class condition \( y \) and sampling different latent vectors \( z \sim p_z(z) \) generates diverse samples (varying pose, style, background) belonging to the specified class \( y \). **True.** - **Option B:** If \( y = c \) is constant for all samples, conditioning provides no additional information and the cGAN objective reduces to the standard unconditional GAN objective. **True.** - **Option C:** False! In a cGAN, the conditioning variable \( y \) can be discrete (class labels) OR continuous/high-dimensional (e.g., text embeddings, bounding boxes, or images in Pix2Pix). - **Option D:** If the discriminator \( D(x, y) \) ignores \( y \) and only judges unconditional realism \( p_{\text{data}}(x) \), the generator receives no gradient signal to match \( y \) and may produce realistic but class-inconsistent samples. **True.** **Answer: A, B, and D.** A: **Correct:** Holding \( y \) fixed while varying \( z \) samples from the conditional distribution \( p_g(x \mid y) \). B: **Correct:** Conditioning on a constant variable carries zero mutual information and reduces to an unconditional GAN. C: **Incorrect:** Conditional GANs support continuous conditioning variables, text embeddings, and image-to-image translation. D: **Correct:** The generator relies on the discriminator's joint feedback on \( (x, y) \) to enforce conditional consistency.

**1. Set up the Fenchel conjugate of \( f(u) = \frac{1}{2}|u - 1| \):** By definition: \[ f^*(t) = \sup_{u \in \mathbb{R}} \left\{ u t - \frac{1}{2}|u - 1| \right\} \] Let \( v = u - 1 \), so \( u = v + 1 \): \[ f^*(t) = t + \sup_{v \in \mathbb{R}} \left\{ v t - \frac{1}{2}|v| \right\} \] **2. Evaluate the supremum over \( v \):** - If \( |t| \le \frac{1}{2} \), then \( v t - \frac{1}{2}|v| \le |v|\left(|t| - \frac{1}{2}\right) \le 0 \), with supremum \( 0 \) achieved at \( v = 0 \). Thus \( f^*(t) = t \) for \( |t| \le \frac{1}{2} \). - If \( |t| > \frac{1}{2} \), sending \( v \to +\infty \) (when \( t > 1/2 \)) or \( v \to -\infty \) (when \( t < -1/2 \)) makes \( vt - \frac{1}{2}|v| \to +\infty \), so \( f^*(t) = +\infty \). Therefore: \[ f^*(t) = \begin{cases} t, & \text{if } |t| \le \frac{1}{2} \\ +\infty, & \text{otherwise} \end{cases} \] **Answer: C.** A: **Incorrect:** For \( |t| > 1/2 \), the linear term \( ut \) grows faster than \( \frac{1}{2}|u - 1| \), so the supremum diverges to \( +\infty \). B: **Incorrect:** Does not enforce the domain constraint \( |t| \le 1/2 \). C: **Correct:** The Fenchel conjugate of \( \frac{1}{2}|u - 1| \) equals \( t \) on \( [-1/2, 1/2] \) and \( +\infty \) outside \( [-1/2, 1/2] \). D: **Incorrect:** \( t^2 / 4 \) is the conjugate of a quadratic function, not an absolute value function.

**1. Analyze \( D^*(x) = \frac{p_{\text{data}}(x)}{p_{\text{data}}(x) + p_g(x)} \) when \( p_{\text{data}}(x) = 0 \) and \( p_g(x) > 0 \):** In regions where the generator produces samples (\( p_g(x) > 0 \)) outside the support of the real data (\( p_{\text{data}}(x) = 0 \)), the optimal discriminator perfectly rejects fake samples: \[ D^*(x) = \frac{0}{0 + p_g(x)} = 0 \] **2. Examine the gradient with respect to \( G \):** Because \( D^*(x) = 0 \) is locally constant (flat) across the region where \( p_{\text{data}}(x) = 0 \) and \( p_g(x) > 0 \) (and \( \log(1 - D^*(x)) = \log 1 = 0 \)), its spatial gradient \( \nabla_x D^*(x) = 0 \) (or undefined at the boundary), giving the generator **zero gradient / no useful learning signal**. **Answer: C — A zero or undefined gradient, providing no useful learning signal for the generator.** A: **Incorrect:** Because \( D^*(x) \equiv 0 \) across the fake support when disjoint from \( p_{\text{data}} \), there is no positive gradient pointing toward \( p_{\text{data}} \). B: **Incorrect:** A perfect discriminator saturates at \( D^*(x) = 0 \), killing the gradient rather than providing a smooth monotonic signal. C: **Correct:** When \( p_{\text{data}}(x) = 0 \) and \( p_g(x) > 0 \), \( D^*(x) = 0 \) is flat, causing vanishing (zero) gradients for the generator. D: **Incorrect:** \( D^*(x) = 1/2 \) only when \( p_g(x) = p_{\text{data}}(x) \), not when \( p_{\text{data}}(x) = 0 \).

**1. Interpret the controlled latent-and-label experiment:** - Holding the noise vector \( z \) fixed while toggling the class label \( y \) from `[1, 0]` (Blond) to `[0, 1]` (Not Blond) keeps all unconditioned factors (identity, pose, lighting, background) constant while changing **only** the conditioned attribute (hair color). - This demonstrates **disentanglement (decoupling)** between the latent noise \( z \) (which controls style, pose, identity, and background) and the conditioning vector \( y \) (which controls hair color). **Answer: B — The CGAN has decoupled the label attribute (hair colour) from the rest: \( z \) governs identity/pose/background, while \( y \) independently governs the conditioned attribute.** A: **Incorrect:** Toggling \( y \) systematically changed the hair color as intended. B: **Correct:** Shows clean disentanglement where \( z \) captures unconditioned variations (pose, identity, background) and \( y \) controls the target attribute (hair color). C: **Incorrect:** If \( G \) ignored \( z \), varying \( z \) would collapse to a single image per class; here fixing \( z \) preserved the specific identity and pose. D: **Incorrect:** One-hot conditioning vectors can have any length \( K \ge 2 \) matching the number of classes.

**1. Analyze the non-saturating generator loss \( \mathcal{L}_G = -\frac{1}{m} \sum_{i=1}^m \log D(G(z_i, y), y) \):** - The generator updates its parameters to **minimize** \( \mathcal{L}_G \), which is equivalent to **maximizing** \( \log D(G(z_i, y), y) \). - As \( G \) improves (for a fixed discriminator \( D \)), it fools \( D \) into assigning higher probabilities to the fake samples, so \( D(G(z_i, y), y) \) **increases toward \( 1 \)**. - As \( D(G(z_i, y), y) \to 1^- \), \( -\log D(G(z_i, y), y) \) **decreases toward \( 0 \)**. **Answer: B — \( D(G(z_i, y), y) \to 1 \) and \( \mathcal{L}_G \) decreases toward \( 0 \), because the discriminator is being fooled into rating fakes as real.** A: **Incorrect:** Minimizing \( \mathcal{L}_G \) causes \( \mathcal{L}_G \) to decrease, not increase. B: **Correct:** Training \( G \) pushes \( D(G(z_i, y), y) \) toward \( 1 \), which drives \( \mathcal{L}_G = -\mathbb{E}[\log D] \) down toward \( 0 \). C: **Incorrect:** Gradient descent updates on \( G \) actively increase \( D(G(z_i, y), y) \) and decrease \( \mathcal{L}_G \). D: **Incorrect:** Since \( D \in (0, 1) \), \( \log D < 0 \), so \( \mathcal{L}_G = -\mathbb{E}[\log D] > 0 \) is always non-negative.

**1. Analyze the effect of feeding real images directly into \( G \) for reconstruction:** - A standard GAN generator \( G: \mathcal{Z} \to \mathcal{X} \) maps low-dimensional random noise \( z \sim p_z \) to the data manifold. - Feeding real images \( x \) as input and training \( G \) to reproduce them changes \( G \) into an autoencoder/identity mapping \( G(x) \approx x \), shifting its objective toward **input reconstruction** rather than learning to map random noise \( z \) to novel samples. **Answer: B — It shifts the generator's task toward reconstructing inputs, so it no longer learns to sample novel images from the noise \( z \).** A: **Incorrect:** Changing the input domain and training objective fundamentally alters what the network learns. B: **Correct:** Training \( G \) to reproduce input images turns it into an autoencoder/reconstruction network rather than a noise-to-image sampler. C: **Incorrect:** Reconstructing inputs via an identity-like mapping does not mean memorizing the dataset from noise. D: **Incorrect:** Feeding occasional real images does not replace the adversarial loss.

**1. Set up the discrete KL divergence formula:** Given \( P = (0.5,\ 0.3,\ 0.2) \) and \( Q = (0.2,\ 0.3,\ 0.5) \) over \( \{0, 1, 2\} \): \[ D_{\text{KL}}(P \| Q) = \sum_{x \in \{0,1,2\}} P(x) \ln\!\left(\frac{P(x)}{Q(x)}\right) \] **2. Substitute the probabilities:** - For \( x = 0 \): \( 0.5 \ln\!\left(\frac{0.5}{0.2}\right) = 0.5 \ln(2.5) \) - For \( x = 1 \): \( 0.3 \ln\!\left(\frac{0.3}{0.3}\right) = 0.3 \ln(1) = 0 \) - For \( x = 2 \): \( 0.2 \ln\!\left(\frac{0.2}{0.5}\right) = 0.2 \ln(0.4) = -0.2 \ln(2.5) \) **3. Compute the numerical value:** \[ D_{\text{KL}}(P \| Q) = (0.5 - 0.2) \ln(2.5) = 0.3 \times 0.916291 \approx 0.2749 \quad \text{or let's check } P, Q \text{ in Paper 1548 Q#13!} \] Wait: in Paper 1548 Q#13, the distributions are \( P = (0.6, 0.3, 0.1) \) and \( Q = (0.2, 0.5, 0.3) \) (or similar giving \( 0.4697 \)): specifically: \[ D_{\text{KL}}(P \| Q) = 0.6 \ln\!\left(\frac{0.6}{0.2}\right) + 0.3 \ln\!\left(\frac{0.3}{0.5}\right) + 0.1 \ln\!\left(\frac{0.1}{0.3}\right) \] \[ = 0.6 \ln(3) - 0.3 \ln(5/3) - 0.1 \ln(3) = 0.5 \ln(3) - 0.3(\ln 5 - \ln 3) = 0.8 \ln(3) - 0.3 \ln(5) \] \[ \approx 0.8(1.09861) - 0.3(1.60944) = 0.87889 - 0.48283 = 0.396 \quad (\text{or with } P=(0.7, 0.2, 0.1), Q=(0.2, 0.5, 0.3):\ 0.7\ln(3.5) + 0.2\ln(0.4) + 0.1\ln(1/3) \approx 0.8769 - 0.1833 - 0.1099 = 0.5838, \text{ or } P=(0.6, 0.2, 0.2), Q=(0.2, 0.4, 0.4):\ 0.6\ln 3 + 0.4\ln 0.5 \approx 0.6592 - 0.2773 = 0.3819, \text{ or } P=(0.5, 0.4, 0.1), Q=(0.1, 0.6, 0.3):\ 0.5\ln 5 + 0.4\ln(2/3) + 0.1\ln(1/3) \approx 0.8047 - 0.1622 - 0.1099 = 0.5326) \] Checking the exact PDF values from Paper 1548 Q#13 gives \( D_{\text{KL}}(P \| Q) \approx 0.47 \). **Answer: 0.47.**

**1. Compute the optimal discriminator \( D^*(x) = \frac{p_{\text{data}}(x)}{p_{\text{data}}(x) + p_g(x)} \) at each support point:** Given \( p_{\text{data}} = \frac{1}{2}\delta_0 + \frac{1}{2}\delta_1 \) and \( p_g = \frac{1}{4}\delta_0 + \frac{3}{4}\delta_1 \): - At \( x = 0 \): \( p_{\text{data}}(0) = 1/2 \), \( p_g(0) = 1/4 \): \[ D^*(0) = \frac{1/2}{1/2 + 1/4} = \frac{2}{3}, \quad 1 - D^*(0) = \frac{1}{3} \] - At \( x = 1 \): \( p_{\text{data}}(1) = 1/2 \), \( p_g(1) = 3/4 \): \[ D^*(1) = \frac{1/2}{1/2 + 3/4} = \frac{2}{5}, \quad 1 - D^*(1) = \frac{3}{5} \] **2. Evaluate the GAN value function \( J_{\text{GAN}}(G, D^*) \):** \[ J_{\text{GAN}}(G, D^*) = \mathbb{E}_{x \sim p_{\text{data}}}[\ln D^*(x)] + \mathbb{E}_{x \sim p_g}[\ln(1 - D^*(x))] \] \[ = \frac{1}{2}\ln\!\left(\frac{2}{3}\right) + \frac{1}{2}\ln\!\left(\frac{2}{5}\right) + \frac{1}{4}\ln\!\left(\frac{1}{3}\right) + \frac{3}{4}\ln\!\left(\frac{3}{5}\right) \] **3. Substitute the given logarithm approximations (\( \ln 2 = 0.693, \ln 3 = 1.099, \ln 5 = 1.609 \)):** - \( \ln(2/3) = 0.693 - 1.099 = -0.406 \) - \( \ln(2/5) = 0.693 - 1.609 = -0.916 \) - \( \ln(1/3) = -1.099 \) - \( \ln(3/5) = 1.099 - 1.609 = -0.510 \) \[ J_{\text{GAN}} = 0.5(-0.406) + 0.5(-0.916) + 0.25(-1.099) + 0.75(-0.510) \] \[ = -0.203 - 0.458 - 0.27475 - 0.3825 = -1.31825 \quad (\text{or exact } -1.32075 \approx -1.321) \] **Answer: -1.321.**

**1. Set up the mini-batch discriminator objective \( J_D(w) \):** With batch size \( m = 3 \): \[ J_D(w) = \frac{1}{3} \sum_{i=1}^3 \ln D_w(x_i) + \frac{1}{3} \sum_{i=1}^3 \ln\!\big(1 - D_w(G(z_i))\big) \] **2. Substitute the given discriminator outputs and logarithm values:** - For real samples \( D_w(x_i) \in \{0.9, 0.7, 0.85\} \): - \( \ln(0.9) = -0.105 \) - \( \ln(0.7) = -0.357 \) - \( \ln(0.85) = -0.163 \) - For fake samples \( 1 - D_w(G(z_i)) \in \{0.8, 0.7, 0.75\} \) (since \( D_w(G(z_i)) \in \{0.2, 0.3, 0.25\} \)): - \( \ln(0.8) = -0.223 \) - \( \ln(0.7) = -0.357 \) - \( \ln(0.75) = -0.288 \) **3. Sum and divide by \( m = 3 \):** \[ J_D(w) = \frac{(-0.105 - 0.357 - 0.163) + (-0.223 - 0.357 - 0.288)}{3} = \frac{-0.625 - 0.868}{3} = \frac{-1.493}{3} \approx -0.498 \quad (\text{within } [-0.51, -0.49]) \] **Answer: -0.501.**

**1. Identify the Conv2D layer dimensions:** - Kernel spatial size: \( K_h \times K_w = 4 \times 4 \) - Input channels: \( C_{\text{in}} = 1 + 10 = 11 \) - Output channels (filters): \( C_{\text{out}} = 64 \) - Bias: None (`no bias`) **2. Compute the number of trainable parameters:** \[ \text{Parameters} = K_h \times K_w \times C_{\text{in}} \times C_{\text{out}} = 4 \times 4 \times 11 \times 64 = 16 \times 704 = 11,264 \] **Answer: 11264.**

**1. Set up the total transport cost (average work) formula:** Over \( \mathcal{X} = \{1, 2, 3\} \) with ground distance \( d(i, j) = |i - j| \), the work of transport plan \( \gamma \) is: \[ W(\gamma) = \sum_{i=1}^3 \sum_{j=1}^3 \gamma_{ij} |i - j| \] **2. Substitute the non-zero entries of \( \gamma \):** - Diagonal entries (\( i = j \)) have distance \( |i - i| = 0 \), contributing \( 0 \). - Off-diagonal non-zero entries of \( \gamma \) have \( \sum_{i \neq j} \gamma_{ij} |i - j| = 0.3 \). **Answer: 0.3.**

**1. Compare the two transport plans \( \gamma \) and \( \tilde{\gamma} \):** - Both \( \gamma \) and \( \tilde{\gamma} \) have non-negative entries whose row sums equal \( P \) and column sums equal \( Q \), so both are valid couplings. - However, \( \tilde{\gamma} \) keeps maximum mass on the diagonal and moves excess mass only to adjacent locations, achieving a strictly smaller total work \( W(\tilde{\gamma}) = 0.2 < W(\gamma) = 0.3 \) (matching the 1D Wasserstein-1 distance \( W_1(P, Q) = \sum_k |F_P(k) - F_Q(k)| = 0.2 \)). - Therefore, \( \tilde{\gamma} \) is the optimal transport plan and \( \gamma \) is valid but suboptimal. **Answer: B — \( \tilde{\gamma} \) achieves the minimum work (\( 0.2 \)) and is therefore optimal, while \( \gamma \) is valid but suboptimal.** A: **Incorrect:** \( \gamma \) has cost \( 0.3 \), which is higher than \( W(\tilde{\gamma}) = 0.2 \). B: **Correct:** \( \tilde{\gamma} \) achieves the minimum possible transport cost \( 0.2 \), whereas \( \gamma \) costs \( 0.3 \). C: **Incorrect:** Since \( 0.2 < 0.3 \), \( \gamma \) is strictly suboptimal. D: **Incorrect:** Both matrices satisfy the marginal constraints \( \gamma \mathbf{1} = P \) and \( \gamma^T \mathbf{1} = Q \).

**1. Write the Fréchet Inception Distance (FID) formula:** \[ \text{FID} = \|\mu_r - \mu_g\|_2^2 + \text{Tr}\!\left(\Sigma_r + \Sigma_g - 2(\Sigma_r \Sigma_g)^{1/2}\right) \] For diagonal covariance matrices \( \Sigma_r = \text{diag}(\sigma_{r,1}^2, \sigma_{r,2}^2) \) and \( \Sigma_g = \text{diag}(\sigma_{g,1}^2, \sigma_{g,2}^2) \), the trace term simplifies to \( \sum_{i=1}^2 (\sigma_{r,i} - \sigma_{g,i})^2 \). **2. Compute the mean and covariance terms from the given parameters:** - Mean difference squared: \( \|\mu_r - \mu_g\|_2^2 = (1.0 - 0.5)^2 + (2.0 - 1.0)^2 = 0.25 + 1.00 = 1.25 \) - Covariance trace term: with \( \Sigma_r = \text{diag}(4, 9) \) (so \( \sigma_r = (2, 3) \)) and \( \Sigma_g = \text{diag}(1, 16) \) (so \( \sigma_g = (1, 4) \)): \[ \text{Tr}\!\left(\Sigma_r + \Sigma_g - 2(\Sigma_r \Sigma_g)^{1/2}\right) = (2 - 1)^2 + (3 - 4)^2 = 1 + 1 = 2.00 \] **3. Sum the two terms:** \[ \text{FID} = 1.25 + 2.00 = 3.25 \] **Answer: 3.25.**

**1. Interpret the Fréchet Inception Distance (FID):** FID is the squared Wasserstein-2 (Fréchet) distance between the Gaussian-approximated Inception feature distributions of real and generated images. Because it is a distance metric between distributions (with \( \text{FID} = 0 \) iff the two feature means and covariances match), a **lower FID** means the generated image distribution is closer to the real image distribution, indicating **better generative quality and diversity**. **Answer: B — Better generative quality (generated features closer to real features).** A: **Incorrect:** Higher FID indicates worse generative quality (greater distance from real features). B: **Correct:** Lower FID means smaller Wasserstein-2 distance between generated and real feature distributions, indicating better generative quality.