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📗 [4 points] Consider a classification problem with \(n\) = classes \(y \in \left\{1, 2, ..., n\right\}\), and two binary features \(x_{1}, x_{2} \in \left\{0, 1\right\}\). Suppose \(\mathbb{P}\left\{Y = y\right\}\) = , \(\mathbb{P}\left\{X_{1} = 1 | Y = y\right\}\) = , \(\mathbb{P}\left\{X_{2} = 1 | Y = y\right\}\) = . Which class will naive Bayes classifier produce on a test item with \(X_{1}\) = and \(X_{2}\) = .
Hint
See Fall 2016 Final Q18, Fall 2011 Midterm Q20. Use the Bayes rule: \(\mathbb{P}\left\{Y = y | X_{1} = x_{1}, X_{2} = x_{2}\right\} = \dfrac{\mathbb{P}\left\{X_{1} = x_{1}, X_{2} = x_{2} | Y = y\right\} \mathbb{P}\left\{Y = y\right\}}{\displaystyle\sum_{y'=1}^{n} \mathbb{P}\left\{X_{1} = x_{1}, X_{2} = x_{2} | Y = y'\right\} \mathbb{P}\left\{Y = y'\right\}}\), which is equal to \(\dfrac{\mathbb{P}\left\{X_{1} = x_{1} | Y = y\right\} \mathbb{P}\left\{X_{2} = x_{2} | Y = y\right\} \mathbb{P}\left\{Y = y\right\}}{\displaystyle\sum_{y'=1}^{n} \mathbb{P}\left\{X_{1} = x_{1} | Y = y'\right\} \mathbb{P}\left\{X_{2} = x_{2} | Y = y'\right\} \mathbb{P}\left\{Y = y'\right\}}\), due to the independence assumption of Naive Bayes. For Bayesian network that are not Naive, the second equality is not true. Naive Bayes classifier selects the \(y\) that maximizes \(\mathbb{P}\left\{Y = y | X_{1} = x_{1}, X_{2} = x_{2}\right\}\): since the denominators for these probabilities are the same, and the prior probability is constant, the classifier is effectively selecting the \(y\) that maximizes \(\mathbb{P}\left\{X_{1} = x_{1} | Y = y\right\} \mathbb{P}\left\{X_{2} = x_{2} | Y = y\right\}\) which is a function in \(y\). You can try different values of \(y\) to find the maximizer or use the first derivative condition if the number of classes is large (i.e. compare the integers near the places where the first derivative is zero and the end points).
📗 Answer: .
📗 [4 points] Consider the problem of detecting if an email message contains a virus. Say we use four random variables to model this problem: Boolean (binary) class variable \(V\) indicates if the message contains a virus or not, and three Boolean feature variables: \(A, B, C\). We decide to use a Naive Bayes Classifier to solve this problem so we create a Bayesian network with arcs from \(V\) to each of \(A, B, C\). Their associated CPTs (Conditional Probability Table) are created from the following data: \(\mathbb{P}\left\{V = 1\right\}\) = , \(\mathbb{P}\left\{A = 1 | V = 1\right\}\) = , \(\mathbb{P}\left\{A = 1 | V = 0\right\}\) = , \(\mathbb{P}\left\{B = 1 | V = 1\right\}\) = , \(\mathbb{P}\left\{B = 1 | V = 0\right\}\) = , \(\mathbb{P}\left\{C = 1 | V = 1\right\}\) = , \(\mathbb{P}\left\{C = 1 | V = 0\right\}\) = . Compute \(\mathbb{P}\){ \(A\) = , \(B\) = , \(C\) = }.
Hint
See Spring 2017 Final Q7. Naive Bayes is a special simple Bayesian Network, so the way to compute the joint probabilities is the same (product of conditional probabilities given the parents): \(\mathbb{P}\left\{A = a, B = b, C = c\right\} = \mathbb{P}\left\{A = a, B = b, C = c, V = 0\right\} + \mathbb{P}\left\{A = a, B = b, C = c, V = 1\right\}\) and \(\mathbb{P}\left\{A = a, B = b, C = c, V = v\right\} = \mathbb{P}\left\{A = a | V = v\right\} \mathbb{P}\left\{B = b | V = v\right\} \mathbb{P}\left\{C = c | V = v\right\} \mathbb{P}\left\{V = v\right\}\).
📗 Answer: .
📗 [3 points] Consider the following directed graphical model over binary variables: \(A \to B \leftarrow C\). Given the CPTs (Conditional Probability Table):
Variable
Probability
Variable
Probability
\(\mathbb{P}\left\{A = 1\right\}\)
\(\mathbb{P}\left\{C = 1\right\}\)
\(\mathbb{P}\left\{B = 1 | A = C = 1\right\}\)
\(\mathbb{P}\left\{B = 1 | A = 0, C = 1\right\}\)
\(\mathbb{P}\left\{B = 1 | A = 1, C = 0\right\}\)
\(\mathbb{P}\left\{B = 1 | A = C = 0\right\}\)
What is the probability that \(\mathbb{P}\){ \(A\) = , \(B\) = , \(C\) = }? Hint
See Fall 2019 Final Q22 Q23 Q24 Q25, Spring 2018 Final Q24 Q25, Fall 2014 Final Q9, Fall 2006 Final Q20, Fall 2005 Final Q20. For any Bayes net, the joint probability can always be computed as the product of the conditional probabilities (conditioned on the parent node variable). For a causal chain \(A \to B \to C\), \(\mathbb{P}\left\{A = a, B = b, C = c\right\} = \mathbb{P}\left\{A = a\right\} \mathbb{P}\left\{B = b | A = a\right\} \mathbb{P}\left\{C = c | B = b\right\}\). For a common cause \(A \leftarrow B \to C\), \(\mathbb{P}\left\{A = a, B = b, C = c\right\} = \mathbb{P}\left\{A = a | B = b\right\} \mathbb{P}\left\{B = b\right\} \mathbb{P}\left\{C = c | B = b\right\}\). For a common effect \(A \to B \leftarrow C\), \(\mathbb{P}\left\{A = a, B = b, C = c\right\} = \mathbb{P}\left\{A = a\right\} \mathbb{P}\left\{B = b | A = a, C = c\right\} \mathbb{P}\left\{C = c\right\}\).
📗 Answer: .
📗 [5 points] Consider the following directed graphical model over binary variables: \(A \to B \to C\). Given the CPTs (Conditional Probability Table):
Variable
Probability
Variable
Probability
\(\mathbb{P}\left\{A = 1\right\}\)
\(\mathbb{P}\left\{B = 1 | A = 1\right\}\)
\(\mathbb{P}\left\{B = 1 | A = 0\right\}\)
\(\mathbb{P}\left\{C = 1 | B = 1\right\}\)
\(\mathbb{P}\left\{C = 1 | B = 0\right\}\)
What is the probability that \(\mathbb{P}\){ \(A\) = \(|\) \(C\) = }? Hint
See Fall 2019 Final Q22 Q23 Q24 Q25, Spring 2018 Final Q24 Q25, Fall 2014 Final Q9, Fall 2006 Final Q20, Fall 2005 Final Q20. For any type of network, one way (brute force, not really efficient) is to use the marginal distributions: \(\mathbb{P}\left\{A = a | C = c\right\} = \dfrac{\mathbb{P}\left\{A = a, C = c\right\}}{\mathbb{P}\left\{C = c\right\}} = \dfrac{\displaystyle\sum_{b'} \mathbb{P}\left\{A = a, B = b', C = c\right\}}{\displaystyle\sum_{a', b'} \mathbb{P}\left\{A = a', B = b', C = c\right\}}\). The joint probabilities can be calculated the same way as the previous question.
📗 Answer: .
📗 [5 points] Consider the following directed graphical model over binary variables: \(A \leftarrow B \to C\). Given the CPTs (Conditional Probability Table):
Variable
Probability
Variable
Probability
\(\mathbb{P}\left\{B = 1\right\}\)
\(\mathbb{P}\left\{C = 1 | B = 1\right\}\)
\(\mathbb{P}\left\{C = 1 | B = 0\right\}\)
\(\mathbb{P}\left\{A = 1 | B = 1\right\}\)
\(\mathbb{P}\left\{A = 1 | B = 0\right\}\)
What is the probability that \(\mathbb{P}\){ \(A\) = \(|\) \(C\) = }? Hint
See Fall 2019 Final Q22 Q23 Q24 Q25, Spring 2018 Final Q24 Q25, Fall 2014 Final Q9, Fall 2006 Final Q20, Fall 2005 Final Q20. For any type of network, one way (brute force, not really efficient) is to use the marginal distributions: \(\mathbb{P}\left\{A = a | C = c\right\} = \dfrac{\mathbb{P}\left\{A = a, C = c\right\}}{\mathbb{P}\left\{C = c\right\}} = \dfrac{\displaystyle\sum_{b'} \mathbb{P}\left\{A = a, B = b', C = c\right\}}{\displaystyle\sum_{a', b'} \mathbb{P}\left\{A = a', B = b', C = c\right\}}\). The joint probabilities can be calculated the same way as the previous question.
📗 Answer: .
📗 [4 points] Consider the following Bayesian Network containing 5 Boolean random variables. How many numbers must be stored in total in all CPTs (Conditional Probability Table) associated with this network (excluding the numbers that can be calculated from other numbers)?
Hint
See Fall 2019 Final Q21, Spring 2017 Final Q8, Fall 2011 Midterm Q15. A node with one parent \(\mathbb{P}\left\{B | A\right\}\) requires storing \(\left(n_{B} - 1\right) n_{A}\) probabilities; a node with no parent \(\mathbb{P}\left\{A\right\}\) requires storing \(n_{A} - 1\) probabilities, and a node with two parents \(\mathbb{P}\left\{C | A, B\right\}\) requires storing \(\left(n_{C} - 1\right) n_{A} n_{B}\) probabilities, ...
📗 Answer: .
📗 [2 points] An n-gram language model computes the probability \(\mathbb{P}\left\{w_{n} | w_{1}, w_{2}, ..., w_{n-1}\right\}\). How many parameters need to be estimated for a -gram language model given a vocabulary size of ?
Hint
See Fall 2019 Final Q8. The idea is similar to computing the number of conditional probabilities that are required to store for a Bayesian network: \(\mathbb{P}\left\{A | B, C, D, ...\right\}\) requires storing \(\left(n_{A} - 1\right) n_{B} n_{C} n_{D} ...\) probabilities. The reason that there is a \(-1\) in the first term is because \(\mathbb{P}\left\{A = n_{A} | B = b, C = c, D = d, ...\right\} = 1 - \displaystyle\sum_{i=1}^{n_{A} - 1} \mathbb{P}\left\{A = i | B = b, C = c, D = d, ...\right\}\) can be computed from the other \(n_{A} - 1\) probabilities. Note that this is true for any combination of \(b, c, d, ...\) and there are \(n_{B} n_{C} n_{D} ...\) such combinations.
📗 Answer: .
📗 [4 points] Given the following transition matrix for a bigram model with words "", "" and "": . Row \(i\) column \(j\) is \(\mathbb{P}\left\{w_{t} = j | w_{t-1} = i\right\}\). What is the probability that the third word is "" given the first word is ""?
Hint
See Fall 2019 Final Q30. Sum over all possible values of the second word: \(\mathbb{P}\left\{w_{3} = j | w_{1} = i\right\} = \displaystyle\sum_{k=1}^{3} \mathbb{P}\left\{w_{3} = j | w_{2} = k\right\} \mathbb{P}\left\{w_{2} = k | w_{1} = i\right\}\), where the \(\mathbb{P}\left\{w_{t} | w_{t-1}\right\}\) probabilities are given by the transition matrix.
📗 Answer: .
📗 [3 points] Welcome to the Terrible-Three-Day-Tour! We will visit New York on Day 1. The rules for Day 2 and Day 3 are:
(a) If we were at New York the day before, with probability we will stay in New York, and with probability we will go to Baltimore.
(b) If we were at Baltimore the day before, with probability we will stay in Baltimore, and with probability we will go to Washington D.C.
On average, before you start the tour, what is your chance to visit (at least on one of the two days)? Hint
See Fall 2009 Final Q13. W is visited only if the sequence is NBW, and B is not visited only if the sequence is NNN. Compute the sequence probabilities \(\mathbb{P}\left\{w_{3} = X, w_{2} = Y, w_{1} = Z\right\} = \mathbb{P}\left\{w_{3} = X | w_{2} = Y\right\} \mathbb{P}\left\{w_{2} = Y | w_{1} = Z\right\} \mathbb{P}\left\{w_{1} = Z\right\}\) using the transition probabilities.
📗 Answer: .
📗 [5 points] Andy is a three-month old baby. He can be happy (state 0), hungry (state 1), or having a wet diaper (state 2). Initially when he wakes up from his nap at 1pm, he is happy. If he is happy, there is a chance that he will remain happy one hour later, a chance to be hungry by then, and a chance to have a wet diaper. Similarly, if he is hungry, one hour later he will be happy with chance, hungry with chance, and wet diaper with chance. If he has a wet diaper, one hour later he will be happy with chance, hungry with chance, and wet diaper with chance. He can smile (observation 0) or cry (observation 1). When he is happy, he smiles of the time and cries of the time; when he is hungry, he smiles of the time and cries of the time; when he has a wet diaper, he smiles of the time and cries of the time.
What is the probability that the particular observed sequence (or \(Y_{1}, Y_{2}\) = ) happens (in the first two periods)?
Note: if the weights are not shown clearly, you could move the nodes around with mouse or touch. Hint
See Fall 2019 Final Q28 Q29, Spring 2018 Final Q28 Q29, Spring 2017 Final Q9, Fall 2008 Final Q4. The question is asking for \(\mathbb{P}\left\{Y_{1} = y_{1}, Y_{2} = y_{2}\right\}\) which is equal to \(\displaystyle\sum_{x_{1}, x_{2} \in \left\{s_{0}, s_{1}, s_{2}\right\}} \mathbb{P}\left\{Y_{1} = y_{1}, Y_{2} = y_{2}, X_{1} = x_{1}, X_{2} = x_{2}\right\}\) using the definition of the marginal distribution, which is equal to \(\displaystyle\sum_{x_{1}, x_{2} \in \left\{s_{0}, s_{1}, s_{2}\right\}} \mathbb{P}\left\{Y_{1} = y_{1} | X_{1} = x_{1}\right\} \mathbb{P}\left\{Y_{2} = y_{2} | X_{2} = x_{2}\right\} \mathbb{P}\left\{X_{2} = x_{2} | X_{1} = x_{1}\right\} \mathbb{P}\left\{ X_{1} = x_{1}\right\}\) using the Bayesian network joint distribution formula (i.e. the joint probability is the product of the conditional probabilities given the parents), and all of these conditional probabilities are given in the diagram (or in the description of the question).
📗 Answer: .
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