<font color=blue>Solutions</font> to Third Midterm Exam</h1> <h4>David Mix Barrington</h4> <h4>17 April 2009</h4> <p>Question text is in black, solutions in <font color=blue>blue</font>. <h3>Directions:</h3> <p><ul> <li> Answer the problems on the exam pages. <li> There are six problems for 100 total points. Actual scale is A=90, C=62. <li> If you need extra space use the back of a page. <li> No books, notes, calculators, or collaboration. <li> The first four questions are true/false, with five points for the correct boolean answer and up to five for a correct justification. <li> When the answer to a question is a number, you may give your answer in the form of an expression using arithmetic operations, powers, falling powers, or the factorial function. Probabilities may be given as either fractions or decimals. </ul> <p><pre> Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 30 points Q6: 30 points Total: 100 points </pre> <p><ul> <p><li><b>Question 1 (10):</b> <i>True or false with justification:</i> If I believe that there is a 42% chance that the Red Sox will win tonight's game, and I can bet on their winning at three-to-two odds (with no service charge), my expected return (expected money won minus expected money lost) is positive. <p><font color=blue> TRUE. 3-2 odds correspond to a probability of (3/2)/(1 + 3/2) = 3/5 that the <i>offerer</i> of the odds will win. But their winning probability is only 58%, less than that. We can also calculate the expected value directly -- for each dollar bet, I have expected winnings of (0.42)(3/2) = 0.63 and expected losses of (0.58)(1) = 0.58, for expected net winnings of 0.05, which is positive. </font> <p><li><b>Question 2 (10):</b> <i>True or false with justification:</i> I think that there is a 1/3 probability that it will rain tomorrow. I consult my local TV station, which predicts that it will rain, but I know that their predictions are correct only 60% of the time. After considering their prediction, I still believe that the probability of rain is less than 50%. <p><font color=blue> TRUE. Let R be the event of rain and let T be the event that the TV station predicts rain. We have prior odds O(R) = 1/2 (computed as Pr(R)/Pr(¬R)) and likelihood ratio L(T|R) = Pr(R|T)/Pr(R|¬T) = (0.60)/(0.40) = 3/2. Thus the posterior odds O(R|T) = (1/2)(3/2) = 3/4 and the coresponding probability is (3/4)/(1 + 3/4) = 3/7, less than 50%. </font> <p><li><b>Question 3 (10):</b> <i>True or false with justification:</i> Let <code>PR</code> be an object that produces a boolean value whenever its <code>next</code> method is run. Suppose that any even number of consecutive booleans produced by <code>PR</code> consist of exactly half <code>true</code> and exactly half <code>false</code>. Then <code>PR</code> may be used in place of a true generator of uniform random booleans in any application. <p><font color=blue> FALSE. The even-numbered results are all determined by the results that precede them, so they could be predicted if the earlier results were known. In fact the string of results could be as simple as 010101... and still meet the conditions. For many applications, predictability makes the generator useless. For example, if you used the sequence as a one-time pad in a cryptographic application, an adversary would have an advantage in figuring out the even-numbered bits and could decipher those bits of the message if successful in doing so. </font> <p><li><b>Question 4 (10):</b> <i>True or false with justification:</i> A particular test for Disease C has a false positive rate of 10% and a false negative rate of 30%. If a patient tests positive with this test, I can conclude that there is a 60% chance that she has Disease C. <p><font color=blue> FALSE. From the given data we can compute L(T|C) = (0.7)/(0.1) = 7, where T is the event of a positive test and C is the event that the patient has Disease C. Thus the posterior odds are seven times the prior odds. But the prior odds are not given, and we can conclude nothing about Pr(C) without them. </font> <p><li><b>Question 5 (30):</b> I am going to throw two fair, independent, ordinary six-sided dice. Let M be the <i>maximum</i> of the two numbers thrown, and let T be the <i>total</i> of the two numbers thrown. Let D be the event that the numbers on the two dice are the same ("doubles"). <ul> <li> (a,5) For each number i in the set {1, 2, 3, 4, 5, 6}, compute Pr(M = i). <p><font color=blue> There are 36 equally likely ways the two dice can come up. One of these (1-1) has maximum 1, so Pr(M = 1) = 1/36. Similarly, by inspecting the cases, we get that Pr(M = 2) = 3/36, Pr(M = 3) = 5/36, Pr(M = 4) = 7/36, Pr(M = 5) = 9/36, and Pr(M = 6) = 11/36. Another way to solve this besides direct inspection is that Pr(M = i) = Pr(both dice are ≤ i) - Pr(both dice are ≤ i-1) = (i/6)<sup>2</sup> - ((i-1)/6)<sup>2</sup> = (2i - 1)/36. </font> <li> (b,5) Compute the conditional probabilities Pr(D | M = 6) and Pr(D | M ≠ 6). <p><font color=blue> Pr(D | M = 6) = Pr(D ∧ (M = 6))/Pr(M = 6) = (1/36)/(11/36) = 1/11. <p>Pr(D | M ≠ 6) = Pr( D ∧ (M ≠ 6))/Pr(M ≠ 6) = (5/36)/(25/36) = 1/5. </font> <li> (c,5) Compute the prior odds O(M = 6) and the posterior odds O(M = 6 | D). <p><font color=blue> O(M = 6) = Pr(M = 6)/Pr(M ≠ 6) = (11/36)/(25/36) = 11/25. <p>L(D | M = 6) = Pr(D | M = 6)/Pr(D | M ≠ 6) = (1/11)/(1/5) = 5/11. <p>O(M = 6 | D) = O(M = 6)L(D | M = 6) = (11/25)(5/11) = 1/5. <p>The posterior odds can also be calculated directly from the conditional probabilities, but we will need the likelihood ratio for part (f). </font> <li> (d,5) Compute the conditional probabilities Pr(T = 7 | M = 6) and Pr(T = 7 | M ≠ 6). <p><font color=blue> Pr(T = 7 | M = 6) = Pr((T = 7) ∧ (M = 6))/Pr(M = 6) = (2/36)/(11/36) = 2/11. <p>Pr(T = 7 | M ≠ 6) = Pr((T = 7) ∧ (M ≠ 6))/Pr(M ≠ 6) = (4/36)/(25/36) = 4/25. </font> <li> (e,5) Compute the posterior odds O(M = 6 | T = 7). <p><font color=blue> L(T = 7 | M = 6) = Pr(T = 7 | M = 6)/Pr(T = 7 | M ≠ 6) = (2/11)/(4/25) = 25/22. <p>O(M = 6 | T = 7) = O(M = 6)L(T = 7 | M = 6) = (11/25)/(25/22) = 1/2, so the posterior probability is 1/3. </font> <li> (f,5) Multiply the prior odds O(M = 6) by <i>both</i> the likelihood ratios you used in parts (c) and (e). Is the result the posterior odds O(M = 6 | D ∧ (T = 7))? Explain your answer. <p><font color=blue> O(M = 6)L(D | M = 6)L(T = 7 | M = 6) = (11/25)(5/11)(25/22) = 5/22. <p><i>If</i> the events D and T = 7 were conditionally independent with respect to M = 6, then this calculation would give the posterior odds of M = 6 given both pieces of evidence. But in fact the two events D and T = 7 are disjoint and thus not conditionally independent. The posterior odds with respect to the event D ∧ (T = 7) are <i>not defined</i>, because the probability of this event is 0 and thus we cannot define conditional probabilities with respect to it. </font> </ul> <p><li><b>Question 6 (30):</b> Last year I saw 25 movies in the theatre, 15 of which I liked. Three professional critics, named Paula, Randy, and Simon, also rated each of these movies "thumbs up" or "thumbs down". I would like to use a Naive Bayes Classifier to help me use these critics' opinions of future movies to predict whether I will like them. Summarizing their opinions: <ul> <li> Paula liked 12 of the 15 movies that I liked, and liked 8 of the 10 movies that I did not like. <li> Randy liked 10 of the movies that I liked, and liked 4 of the movies that I did not like. <li> Simon liked 3 of the movies that I liked, and liked 5 of the movies that I did not like. </ul> <p>Here are your questions: <ul> <li> (a,5) Suppose I am considering a new movie that comes from the same distribution as the 25 movies from last year. What should be my prior odds and prior probability that I will like the movie? <p><font color=blue> Let M be the event that I like a movie. The maximum likelihood estimator for M is the fraction of observed movies that I liked, which is 15/25 = 3/5. The prior odds that I like a new movie are thus (3/5)/(1 - 3/5) = 3/2. </font> <li> (b,10) What should be my posterior odds and probability that I will like the new movie if all three critics like it? What if all three critics hate it? <p><font color=blue> Let P, R, and S be the events that Paula, Randy, and Simon (respectively) like a given movie. We can compute the likelihood ratios for the six possible critical opinions with respect to M: <p>L(P|M) = Pr(P|M)/Pr(P|¬M) = (12/15)/(8/10) = 1. <p>L(R|M) = Pr(R|M)/PR(R|¬M) = (10/15)/(4/10) = 5/3. <p>L(S|M) = Pr(S|M)/PR(S|¬M) = (3/15)/(5/10) = 2/5. <p>Thus O(M|P∧R∧S) = (3/2)(1)(5/3)(2/5) = 1, so the probability that I like a movie that all three critics liked is 50%. <p>L(¬P|M) = Pr(¬P|M)/Pr(¬P|¬M) = (3/15)/(2/10) = 1. <p>L(¬R|M) = Pr(¬R|M)/PR(¬R|¬M) = (5/15)/(6/10) = 5/9. <p>L(¬S|M) = Pr(¬S|M)/PR(¬S|¬M) = (12/15)/(5/10) = 8/5. <p>Thus O(M|¬P∧¬R∧¬S) = (3/2)(1)(5/9)(8/5) = 4/3, so the probability that I like a movie that all three critics hated is (4/3/1 + 4/3) = 4/7. </font> <li> (c,10) Given the three critics, there are eight possible sets of their opinions on a new movie. What set or sets of critics' opinions make it most likely that I will like the new movie, and what are the odds and probability in that case? What set or sets will make is least likely, and what are the odds and probability in that case? <p><font color=blue> Paula's opinions are independent of mine, so her rating of a movie has no effect on the posterior odds of my liking it. Randy's liking a movie increases the odds that I will, and Simon's liking it decreases those odds. So the movies I am most likely to like at those that Randy likes and Simon hates, whatever Paula thinks of them. The posterior odds for such a movie are (3/2)(1)(5/3)(8/5) = 4, so there is a probability of 4/(1 + 4) = 4/5 that I will like it. <p>Similarly, the movies I am least likely to like are ones that Randy hates and Simon likes, whatever Paula thinks of them. The posterior odds for such a movie are (3/2)(1)(5/9)(2/5) = 1/3, and thus the probability that I will like such a movie is (1/3)/(1 + 1/3) = 1/4. </font> <li> (d,5) I would classify 10 of the 25 movies from last year as science fiction, and I liked 8 of those 10. When I evaluate the probability that I will like a new movie, I would like to take into account whether it is a science fiction movie. I could either add this feature to my Naive Bayes Classifier, or I could run two separate Naive Bayes Classifiers, one for the SF movies and one for the non-SF movies. What are some reasons why one of these ideas would be better? Do you need information that I have not given you? <p><font color=blue> Since I appear to like SF movies on average more often than non-SF movies, we should be able to get information from knowing whether a new movie is SF. <i>If</i> being SF is conditionally independent of the three critics' opinions, with respect to my liking the movie, adding SF as a new feature to the NBC is a good idea. But if, for example, Simon hates all SF movies, conditional independence does not hold, and we would do better to run two separate NBC's so that opinions about one type of movie do not influence the other. (If Simon hates all SF movies whether I like them or not, his likelihood ratio for those movies will be 1, or close to it because of smoothing. But his opinions about the non-SF movies might be helpful if separated out.) <p>We can get an idea about the conditional independence by comparing the fraction of SF and non-SF movies each critic liked (information I have not yet given you) and comparing these fractions to the fraction for all movies. We could also test the assumption that the three critics' opinions are conditionally independent of <i>each other</i> with respect to my opinion, but this assumption is inherent in building an NBC. </font> </ul> </ul> <p><small>Last modified 23 April 2009</small> </body> </html>