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Question 2 MCQ · 2.0 marks
Consider two discrete probability distributions
and
, defined over a sample space
. The probability mass functions are given as follows:
Compute
. Use
for the calculation.
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1
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0
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Published solution
**1. State discrete KL:** \(D_{KL}(p\|q)=\sum_{x:p(x)>0}p(x)\log_2[p(x)/q(x)]\).
**2. Check support:** At x1, p(x1)=0.5 but q(x1)=0. A distribution q assigns no probability to an event that p says can occur.
**3. Apply the extended-value definition:** That term is \(0.5\log_2(0.5/0)=+\infty\). The shared x3 term is zero, but cannot remove the infinite term. Therefore KL is positive infinity, regardless of log base.
**Answer: C.**
A: **Incorrect:** A finite value misses the support mismatch.
B: **Incorrect:** Zero would require identical distributions on their mass.
C: **Correct:** Positive p mass at a zero-q point makes this KL infinite.
D: **Incorrect:** KL is nonnegative and cannot be -1.
A standard GAN cannot reconstruct a specific input sample because it does
not learn an explicit encoder mapping inputs to the latent space.
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The blurriness in VAE-generated samples is primarily due to stochastic
sampling in the latent space.
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GANs can generate realistic samples despite not learning an encoder, as they
sample from a structured latent distribution.
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The regularization term in VAEs improves the sharpness of generated samples
by enforcing consistency in reconstructions.
Published solution
**1. Compare mappings:** A standard GAN trains a generator from a latent prior to data, but no explicit inverse encoder. It can therefore generate samples without providing a direct input-to-code reconstruction operation.
**2. Explain image quality:** Adversarial training can favor realistic generated samples. VAE blur is more closely associated with likelihood/reconstruction averaging, decoder limitations and the modeling tradeoff than with stochastic latent sampling alone.
**3. Interpret reconstruction carefully:** A’s intended meaning is that a basic GAN does not supply direct reconstruction through a learned encoder. A particular input may still be approximately reconstructed through a separate GAN-inversion optimization; absence of an encoder is not a mathematical impossibility of every reconstruction.
**Answer: A, C — Selected: A standard GAN cannot reconstruct a specific input sample because it does not learn an explicit encoder mapping inputs to the latent space.; GANs can generate realistic samples despite not learning an encoder, as they sample from a structured latent distribution.**
A: **Correct:** The basic GAN has no explicit inference encoder; this concerns direct reconstruction, not the impossibility of later optimization-based inversion.
B: **Incorrect:** Stochastic sampling alone is not the primary general explanation for VAE blur.
C: **Correct:** A trained generator can produce realistic data from its latent prior without learning an encoder.
D: **Incorrect:** KL regularization shapes the latent distribution and does not by itself guarantee sharper images.
In a Generative Adversarial Network, suppose for a data point
, the real data distribution assigns
probability
, while the generator assigns probability
. If
is the
optimal discriminator, what probability does it assign to
being generated? Enter the answer
correct to two decimal places.
Published solution
**1. Identify real versus generated probability:** Under equal real/fake priors, \(D^*(x)=p_{data}(x)/(p_{data}(x)+p_G(x))\) is the probability of being real.
**2. Substitute:** \(D^*=0.25/(0.25+0.75)=0.25\).
**3. Answer the question’s direction:** It asks for being generated, so use \(1-D^*=0.75\). Equivalently, divide the generated density by the total density.
**Answer: 0.75.**
, calculate the responsibility (posterior probability) of
component
for this point, i.e.,
. Provide the answer to three decimal places.
Published solution
**1. Use Bayes responsibilities:** \[\gamma_1(x)=\frac{\pi_1\mathcal N(x;\mu_1,\sigma_1^2)}{\pi_1\mathcal N(x;\mu_1,\sigma_1^2)+\pi_2\mathcal N(x;\mu_2,\sigma_2^2)}.\]
**2. Evaluate unnormalized weights:** At x=7, component 1 has mean 5, standard deviation 1 and weight 0.6; component 2 has mean 10, standard deviation 2 and weight 0.4. Cancel the common \(1/\sqrt{2\pi}\) factor, leaving \(u_1=0.6e^{-2}\approx0.081201\) and \(u_2=(0.4/2)e^{-9/8}\approx0.064930\).
**3. Normalize:** \(\gamma_1=u_1/(u_1+u_2)\approx0.55566\), rounded to 0.556. The second Gaussian’s 1/2 density normalization must be retained.
**Answer: 0.556.**
**1. Count one affine projection:** Mapping D input features to D outputs requires D-squared weights and D biases.
**2. Count separate query, key and value projections:** \[N_{params}=3(D^2+D)=3D^2+3D.\]
**3. Explain sequence-length independence:** The same projection weights are applied at every token, so the N input vectors do not multiply the parameter count. An output projection is not included in this particular question.
**Answer: B.**
A: **Incorrect:** This counts weights but omits all three bias vectors.
B: **Correct:** Three D-by-D weight matrices and three D-length biases give this count.
C: **Incorrect:** This counts only one of the three projections.
D: **Incorrect:** Six D-squared terms do not describe the specified three affine projections.
In a self-attention mechanism, each input vector has dimension
. The queries, keys, and values
are computed using separate linear transformations, each mapping from
. Suppose
there are
input vectors.
Based on the above data, answer the given subquestions.
How many attention weights
are computed in total if we have
and
?
Published solution
**1. Count query-key pairs:** Every one of N query vectors attends to every one of N key vectors in the stated single-head setting.
**2. Form the attention matrix:** Its shape is N-by-N, so it contains N-squared weights.
**3. Substitute N=100:** \[100^2=10{,}000.\] Feature width D=16 affects dot-product computation per score, not the number of score entries.
**Answer: 10000.**
Suppose the value function values for different states are
The sampled log-policy gradient scalars
are given as
Based on the above data, answer the given subquestions.
Compute the full discounted return
. Enter the answer correct to
two decimal places.
Published solution
**1. Use the supplied discounting:** Rewards are (3,1,-2,5), and gamma=0.8.
**2. Compute discounted terms:** \[R=3+0.8(1)+0.8^2(-2)+0.8^3(5).\]
The terms are 3,0.8,-1.28 and 2.56.
**3. Add:** \(R=3+0.8-1.28+2.56=5.08\). Negative rewards remain negative when discounted.
**Answer: 5.08.**
Suppose the value function values for different states are
The sampled log-policy gradient scalars
are given as
Based on the above data, answer the given subquestions.
Compute the gradient estimate
where
is the advantage estimate given by
Enter the answer correct to
three decimal places.
Published solution
**1. Compute returns from each step:** Working backward with gamma=0.8 gives R3=5, R2=-2+0.8(5)=2, R1=1+0.8(2)=2.6 and R0=3+0.8(2.6)=5.08.
**2. Subtract the value baselines:** Values are (4.5,2.0,1.5,5.0), so the sampled advantages are (0.58,0.6,0.5,0).
**3. Form the stated scalar gradient estimate:** \[\hat g=0.4(0.58)+0.3(0.6)-0.2(0.5)+0.1(0)=0.232+0.18-0.10=0.312.\] This uses the question’s sum of g_t times advantage without adding an extra discount factor outside those terms.
**Answer: 0.312.**
represents a token. A single-head attention module uses
Hence
Each row of
and
correspond to the representation of a single input token. Use scaled dot-
product attention with
and scale factor
.
For the first query only, use the approximations
Based on the above data, answer the given subquestions.
Which token receive the lowest attention score from the query
before scaling? (For token 1,
enter your answer as 1, and so on.)
Published solution
**1. Compute the first query:** The first token row is (1,2,0). Multiplying by the supplied WQ gives q1=(1,2).
**2. Compute keys:** From K=XWK, the three rows are k1=(1,3), k2=(1,1), k3=(3,2).
**3. Compare raw dot products:** \(q_1\cdot k_1=7\), \(q_1\cdot k_2=3\), \(q_1\cdot k_3=7\). Token 2 has the lowest score. Scaling by the same positive square root does not change the order.
**Answer: 2.**
represents a token. A single-head attention module uses
Hence
Each row of
and
correspond to the representation of a single input token. Use scaled dot-
product attention with
and scale factor
.
For the first query only, use the approximations
Based on the above data, answer the given subquestions.
Compute the attention weight on token 3 for query 1. Enter the answer correct to three decimal
places.
Published solution
**1. Find scaled scores:** The first-query dot products are (7,3,7); dividing by \(\sqrt2\) gives approximately (4.950,2.121,4.950).
**2. Apply softmax with the supplied approximations:** Exponentials are (141.175,8.339,141.175), totaling 290.689.
**3. Normalize token 3:** \[a_{13}=141.175/290.689\approx0.48566.\] Rounded to three decimals, its weight is 0.486. Tokens 1 and 3 receive equal weights.
**Answer: 0.486.**
represents a token. A single-head attention module uses
Hence
Each row of
and
correspond to the representation of a single input token. Use scaled dot-
product attention with
and scale factor
.
For the first query only, use the approximations
Based on the above data, answer the given subquestions.
Compute the sum of the first and the second coordinate of the attention output for query 1. Enter
the answer correct to two decimal places.
Published solution
**1. Compute value vectors:** Multiplying X by WV gives v1=(3,2), v2=(1,2), v3=(2,1).
**2. Compute first-query weights:** The raw scores are (7,3,7), and scaled softmax using the provided exponentials gives weights (141.175,8.339,141.175)/290.689.
**3. Sum output coordinates efficiently:** The sums of the value-vector coordinates are 5,3 and 3. Thus
\[o_{11}+o_{12}=\frac{5(141.175)+3(8.339)+3(141.175)}{290.689}=3.9713\ldots.\]
Rounded to two decimals, the requested sum is 3.97.
**Answer: 3.97.**
Causal masking is applied to the scores exactly as in autoregressive transformers. The masking
uses
for the mask matrix
. Use
Based on the above data, answer the given subquestions.
For token 3, compute the attention probability assigned to position 4.
Published solution
**1. Apply autoregressive visibility:** Query position 3 may attend to positions 1,2 and 3. Position 4 is a future position.
**2. Mask the score:** Its additive mask entry is negative infinity, regardless of the original finite score.
**3. Apply softmax:** \(e^{-\infty}=0\), while allowed positions have a positive denominator. Hence \(a_{34}=0\).
**Answer: 0.**
Causal masking is applied to the scores exactly as in autoregressive transformers. The masking
uses
for the mask matrix
. Use
Based on the above data, answer the given subquestions.
Which position receives the highest attention weight for token 4? Enter your answer as either 1 or
2 or 3 or 4.
Published solution
**1. Inspect row 4:** Its supplied scaled scores are (0,1,2,5).
**2. Check the causal mask:** The final query position may attend to every one of positions 1 through 4, so none of these four entries is masked.
**3. Compare:** Softmax preserves order. Score 5 is largest and belongs to position 4, so position 4 receives the highest attention probability.
**Answer: 4.**
Causal masking is applied to the scores exactly as in autoregressive transformers. The masking
uses
for the mask matrix
. Use
Based on the above data, answer the given subquestions.
Compute the ratio
. Use the masked attention weights. Enter
the answer correct to three decimal places.
Published solution
**1. Normalize the token-3 self weight:** After masking, row 3 has scores (1,2,4), so \[a_{33}=e^4/(e^1+e^2+e^4)=54.598/(2.718+7.389+54.598).\]
**2. Normalize token 2’s weight on position 1:** Its allowed scores are (2,3), giving \(a_{21}=e^2/(e^2+e^3)=7.389/(7.389+20.085)\).
**3. Form the ratio:** \[\frac{a_{33}}{a_{21}}=\frac{54.598/64.705}{7.389/27.474}\approx3.137.\] Each row needs its own softmax denominator; a single raw-score ratio would be incorrect.
**Answer: 3.137.**
**1. Build four positional coordinates:** At position j=3 with model width four, the sinusoidal encoding is \((\sin3,\cos3,\sin0.03,\cos0.03)\), approximately (0.141,-0.990,0.030,0.9996).
**2. Add to the base embedding:** The source gives x=(2,1,0,2), so \[\tilde x=(2.141,0.010,0.030,2.9996).\]
**3. Match the rounded vector:** It is approximately (2.14,0.01,0.03,3.00), option A. Only two sine/cosine pairs are used for a four-dimensional vector; the PDF’s broad i-range notation is understood within those four coordinates.
**Answer: A.**
A: **Correct:** This matches coordinate-wise addition and rounding.
B: **Incorrect:** The second coordinate should add a negative cosine and become 0.01, not 1.99.
C: **Incorrect:** The third coordinate should be 0.03, not 0.30.
D: **Incorrect:** The first coordinate must include sin(3), producing 2.14.
Based on the above data, answer the given subquestions.
The sinusoidal positional encoding is added to the token embedding
to form
. Which of the
following statements is true?
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It increases the dimensionality of the embedding vector to encode position.
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It adds position information to the embedding while preserving embedding
dimensionality.
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It replaces the semantic information in the embedding with positional
information.
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It ensures that embeddings are normalized across positions.
Published solution
**1. Match vector shapes:** The token embedding and positional encoding each have the same model dimension.
**2. Add coordinate-wise:** \(\tilde x=x+PE(j)\) combines content and position without concatenating extra coordinates.
**3. Interpret the effect:** The representation gains positional information while preserving its dimension. Addition does not remove semantic information by definition and does not enforce unit norm or another normalization across positions.
**Answer: B — Selected: It adds position information to the embedding while preserving embedding dimensionality.**
A: **Incorrect:** Concatenation would increase width; addition does not.
B: **Correct:** Equal-dimensional vector addition preserves width and adds position information.
C: **Incorrect:** The original content vector remains one component of the sum.
D: **Incorrect:** Adding sine/cosine values does not guarantee normalized embeddings.
For two sampled states, the old and new full action distributions over two actions are:
State 1:
State 2:
Assume the sampled actions were the first action in both states.The advantages are
Use PPO clipping with
. Use natural logarithms with
Based on the above data, answer the given subquestions.
Compute the sampled-action ratio for state 1. Enter the answer correct to one decimal place.
Published solution
**1. Define the importance ratio:** For the sampled action, \(r=\pi_{new}(a\mid s)/\pi_{old}(a\mid s)\).
**2. Read state 1’s first-action probabilities:** They are 0.36 for the new policy and 0.30 for the old policy.
**3. Divide:** \(r_1=0.36/0.30=1.2\). Use the sampled action’s probability, not a ratio of entire distributions.
**Answer: 1.2.**
For two sampled states, the old and new full action distributions over two actions are:
State 1:
State 2:
Assume the sampled actions were the first action in both states.The advantages are
Use PPO clipping with
. Use natural logarithms with
Based on the above data, answer the given subquestions.
Compute the clipped PPO contribution for state 1. Enter the answer correct to two decimal places.
Published solution
**1. Write the clipped objective term:** \(L=\min(rA,\operatorname{clip}(r,1-\epsilon,1+\epsilon)A)\).
**2. Substitute state 1:** r=0.36/0.30=1.2, A=1.8 and epsilon=0.2, so clipping bounds are [0.8,1.2]. The ratio already equals the upper endpoint.
**3. Compare the two products:** Both are \(1.2(1.8)=2.16\). Their minimum is 2.16.
**Answer: 2.16.**
For two sampled states, the old and new full action distributions over two actions are:
State 1:
State 2:
Assume the sampled actions were the first action in both states.The advantages are
Use PPO clipping with
. Use natural logarithms with
Based on the above data, answer the given subquestions.
Compute the sampled-action ratio for state 2. Enter the answer correct to one decimal place.
Published solution
**1. Read the sampled action:** The first action is sampled in state 2, with old probability 0.60 and new probability 0.48.
**2. Compute the ratio:** \[r_2=0.48/0.60=0.8.\]
**3. Interpret:** The new policy assigns this action 80% of its old probability. This also equals the lower clipping endpoint for epsilon=0.2.
**Answer: 0.8.**
For two sampled states, the old and new full action distributions over two actions are:
State 1:
State 2:
Assume the sampled actions were the first action in both states.The advantages are
Use PPO clipping with
. Use natural logarithms with
Based on the above data, answer the given subquestions.
Compute the clipped PPO contribution for state 2. Enter the answer correct to one decimal place.
Published solution
**1. Compute the state-2 ratio:** \(r_2=0.48/0.60=0.8\). Its advantage is A2=-1.5.
**2. Apply the clipping interval:** With epsilon=0.2, [0.8,1.2] leaves this ratio unchanged.
**3. Evaluate the objective:** Both candidate products equal \(0.8(-1.5)=-1.2\), so \(\min(-1.2,-1.2)=-1.2\). For negative advantages the min operation still applies to the signed products, not to their magnitudes.
**Answer: -1.2.**
For two sampled states, the old and new full action distributions over two actions are:
State 1:
State 2:
Assume the sampled actions were the first action in both states.The advantages are
Use PPO clipping with
. Use natural logarithms with
Based on the above data, answer the given subquestions.
Compute
for state 1. Enter the answer correct to four decimal places.
Published solution
**1. Use full-distribution KL:** The question asks \(D_{KL}(\pi_{old}\|\pi_{new})\), so include both actions in state 1:
\[D=0.30\ln(0.30/0.36)+0.70\ln(0.70/0.64).\]
**2. Use the supplied logarithm approximations:** \(D\approx0.30(-0.182)+0.70(0.090)=-0.0546+0.0630=0.0084\).
**3. Report precision:** The four-decimal answer under the supplied approximations is 0.0084. Using unrounded natural logs gives approximately 0.0080, also within the exported accepted range; the discrepancy is from the intentionally coarse log approximations.
**Answer: 0.0084.**
