HW3 Solutions. Answer: Let X be the random variable which denotes the number of packets lost.

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1 HW3 Solutions 1. (20 pts.) Packets Over the Internet n packets are sent over the Internet (n even). Consider the following probability models for the process: (a) Each packet is routed over a different path and is lost independently with probability p. (b) All n packets are routed along the same path and with probability p, one of the links along the path fails and all n packets are lost. Otherwise all packets are received. (c) The n packets are divided into 2 groups of n/2 packets, and each group is routed along a different path and lost with probability p. Losses of different groups are independent events. In each of the three models, compute the distribution of the number of packet losses. For n 6 and p 0.3, plot the distribution in each of the three cases. Does the distribution depend on the probability model? Which of the three routing protocols do you prefer? Answer: Let X be the random variable which denotes the number of packets lost. (a) (6 pts.) Since each of the n packets is independently lost with probability p, we know that X Bin(n, p). Hence, for k {0, 1,..., n}, ( ) n P(X k) p k (1 p) n k k (b) (6 pts.) In this case, either all or none of the packets are lost. So P(X 0) 1 p and P(X n) p. 1

2 (c) (6 pts.) In this case, either no packets, exactly n/2 packets, or all n packets are lost. When n/2 packets are lost, it could be because path 1 failed while path 2 worked, or because path 2 failed while path 1 worked. Let Y i represent whether path i fails (Y i 1 if it fails, Y i 0 if it does not). Then X (n/2)(y 1 + Y 2 ) and Y i Bern(p). Hence, we get P(X 0) (1 p) 2 P(X n/2) 2p(1 p) P(X n) p 2 (2 pts.) As can be seen from the probability distribution tables and plots, the distributions depend on the probability models of packet loss. Which protocol is most preferable is quite dependent on the context, and it s possible to construct situations where each of the protocols performs best: 2

3 If the packets were being transmitted without using an error-correcting code, then all of them are needed at the recipient s end to reconstruct the whole message. In this case, protocol (a) succeeds with probability (1 p) n, protocol (b) with probability 1 p and protocol (c) with probability (1 p) 2. Therefore, protocol (b), which routes all packets on the same path, would work best in this situation. Suppose you used an error-correcting code that can tolerate the loss of up to 35% of packets, and suppose n 100 and p 0.3. Then protocol (a) succeeds with probability 0.884, protocol (b) with probability 0.7, and protocol (c) with probability Therefore, protocol (a) works best in this situation. Suppose you used an error-correcting code that can tolerate the loss of up to 50% of packets, and suppose n 3 and p 0.5. Then protocol (a) and (b) succeed with probability 0.5, while protocol (c) succeeds with probability In this case, protocol (c) works best. Finally, suppose that each packet that makes it to the recipient is worth the same amount; the total value of the communication is proportional to the number of packets received. Then it might not matter which protocol you choose: in the long run, they will deliver the same value, on average. Any of these sorts of answers would be acceptable, as long as you give a coherent justification for which protocol you said is preferable. Comment: There is a particularly important and common scenario where protocol (a) performs best by a wide margin. Suppose we know the packet loss probability p. Then if the number of packets n is large, it turns out that the actual number of packets lost will be pretty close to np (it will almost always be between n(p ɛ) and n(p + ɛ), for some small constant ɛ). Hence, in such a case, it makes sense to use an error-correcting code which is designed to correct up to a p + ɛ of lost packets. By doing so, protocol (a) almost always succeeds in delivering enough packets that the error-correcting code can make up for the ones that were lost. This success probability can be made as close to 1 as desired (by increasing n). On the other hand, protocol (b) will drop all the packets with probability p and protocol (c) with probability p 2, so they are considerably worse in this situation. 2. (20 pts.) Coupon Collection Consider the coupon collection problem with n distinct baseball cards and m cereal boxes bought. Compute the distribution, mean and variance of the number of Babe Ruth cards acquired. (Hint: Think and write in terms of random variables.) Answer: Each box of cereal is analogous to a biased coin flip. We either get Babe Ruth (coin comes up 1 heads) w.p. n or we don t w.p. 1 1 n (coin comes up tails). Since each box of cereal is independent of all others then X Bin(m, 1/n) ( ) ( ) m 1 n 1 m k P(X k) k n k n where X is the random variable denoting the number of cards Babe Ruth cards acquired. 3. (20 pts.) Family Planning Mr and Mrs Brown decide to continue having children until they either have their first girl or until they have five children. Assume that each child is equally likely to be a boy or a girl, independent of all other children, and that there are no multiple births. Let B and G denote the numbers of boys and girls respectively that the Browns have. (a) (10 pts.) Write down the sample space together with the probability of each outcome 3

4 Answer: Ω {g, bg, bbg, bbbg, bbbbg, bbbbb}, where b represents boy and g represents girl in the strings in Ω. For any ω Ω, let l(ω) represent the length of the string (i.e. the number of children born to the Browns). Then, P(ω) ( 1 2 )l(ω), since the gender of all children are independent and each gender is equally likely for every child. (b) (10 pts.) Compute and plot the distributions of the random variables B and G. Answer: P(G 0) P(G 1) The distribution of B is as follows: P(B 0) 1 2 P(B 1) P(B 2) P(B 3) P(B 4) P(B 5)

5 4. (20 pts.) Conditional Independence and Medical Diagnosis There is a disease which affects 2% of the population. A medical test is available which gives a false positive rate of 5% and a mis-detection rate of 3%. (a) (10 pts.) Compute the false discovery rate and the false omission rate of using this test. Answer: Define the random variables H and T as in the lecture notes (H 0 if the patient is healthy, H 1 if affected, T 0 if the test is negative, T 1 if positive). P(H 1) Misdetection rate is P(T 0 H 1) False positive rate is P(T 1 H 0) To compute the false discovery rate, by Bayes rule, P(H 0)P(T 1 H 0) P(H 0 T 1) P(H 0)P(T 1 H 0) + P(H 1)P(T 1 H 1) (1 0.02)0.05 (1 0.02) (1 0.03) To compute the false omission rate, by Bayes rule, P(H 1)P(T 0 H 1) P(H 1 T 0) P(H 1)P(T 0 H 1) + P(H 0)P(T 0 H 0) (1 0.02)(1 0.05) (b) (10 pts.) The doctors complain that the false discovery rate for this test is too high. The medical test company responds with a new test that has a false positive rate of 6% and a mis-detection rate of 4%. Although this new test has worse false positive rate and mis-detection rate compared to the old test, the company claims that when used in conjunction with the old test, will give a lower false discovery 5

6 rate because the results of the two tests are conditionally independent given the disease state of the patient. More specifically, the company recommends the doctors diagnose a patient to have a disease if and only if both the old and the new tests are positive. Do you agree that the company s claim? Justify your answer. Answer: Let the result of the old test be T 1, and that of the new test be T 2. To compute the false discovery rate of the combined test, P(H 0 T 1 1, T 2 1) P(H 0)P(T 1 1, T 2 1 H 0) P(H 0)P(T 1 1, T 2 1 H 0) + P(H 1)P(T 1 1, T 2 1 H 1) P(H 0)P(T 1 1 H 0)P(T 2 1 H 0) P(H 0)P(T 1 1 H 0)P(T 2 1 H 0) + P(H 1)P(T 1 1 H 1)P(T 2 1 H 1) (1 0.02) (1 0.02) (1 0.03)(1 0.04) The false discovery rate is indeed lower. 5. (20 pts.) Typos Assume a 10-page scientific journal paper has 10 typos in it. (a) (10 pts.) If the typos are randomly distributed over the 10 pages, find the probability of having a single typo on each page. Answer: Each typo can be in any of the 10 pages, thus there are 10 different positions for each typo. Since there are 10 pages, the total number of ways we can distribute the typos to the pages is Having a single typo on each of the 10 pages implies sampling the typos 10 times without replacement, since no more than a single typo can be on any page. Thus, there are 10! ways of having a single typo on each page. Hence, the probability of this event is 10!/ (b) (10 pts.) If 8 of the typos are spelling errors and the other two are mathematical errors, find the probability of having 4 spelling errors and a single mathematical error on the first five pages of the paper. Answer: Given a set of 4 spelling errors and 1 mathematical errors, there are 5 5 ways of arranging them over the first five pages, and there are 5 5 ways of arranging the remaining 5 errors over the last 5 pages, leading to a total of 5 10 arrangements. Since there are ( 8 2 4)( 1) different sets of 4 spelling and 1 mathematical errors, the total number of ways we can arrange the errors is ( ( 8 4) 2 1) Hence, the probability of this event is ( ( ( 8 4) 2 1) 5 10 )/