HW Solution 7 Due: Oct 24

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1 ECS 315: Probability and Random Processes 2014/1 HW Solution 7 Due: Oct 24 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra question at the end is optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1 (M2010). The random variable V has pmf { 1 + c, v { 2, 2, 3} p V (v) = v 2 0, otherwise. (a) Find the value of the constant c. (b) Let W = V 2 V + 1. Find the pmf of W. (c) Find EV (d) Find E [V 2 ] (e) Find Var V (f) Find σ V (g) Find EW Solution: 7-1

2 ECS 315 HW Solution 7 Due: Oct /1 (a) The pmf must sum to 1. Hence, The value of c must be Note that this gives 1 ( 2) 2 + c + 1 (2) 2 + c + 1 (3) 2 + c = 1. c = 1 3 ( ) = p V ( 2) = p V (2) = and p V (3) = (b) W = V 2 V + 1. So, when V = 2, 2, 3, we have W = 7, 3, 7, respectively. Hence, W takes only two values, 7 and 3. the corresponding probabilities are P [W = 7] = p V ( 2) + p V (3) = and P [W = 3] = p V (2) = Hence, the pmf of W is given by 41, w = 3, p W (w) =, w = 7, 108 0, otherwise. 0.38, w = 3, 0.62, w = 7, 0, otherwise. (c) We apply the formula EV = v vp V (v). Here, EV = ( 2) p V ( 2) + (2) p V (2) + (3) p V (3) = ( 2) (2) + (3) = (d) We apply the formula E [V 2 ] = v v 2 p V (v). Here, E [ V 2] = ( 2) 2 p V ( 2) 2 + (2) p V (2) + (3) 2 p V (3) = =

3 ECS 315 HW Solution 7 Due: Oct /1 (e) Var V = EV 2 (EV ) 2 = (f) σ V = Var V (g) We show two techniques to calculate EW. Method 1: EW = w wp W (w) = = = Method 2: EW = E [V 2 V + 1] = E [V 2 ] E [V ]+1 = = Problem 2. Suppose X is a uniform discrete random variable on { 3, 2, 1, 0, 1, 2, 3, 4}. Find (a) EX (b) E [X 2 ] (c) Var X (d) σ X Solution: All of the calculations in this question are simply plugging in numbers into appropriate formulas. (a) EX = 0.5 (b) E [X 2 ] = 5.5 (c) Var X = 5.25 (d) σ X = Alternatively, we can find a formula for the general case of uniform random variable X on the sets of integers from a to b. Note that there are n = b a + 1 values that the random variable can take. Hence, all of them has probability 1 n. (a) EX = b i=a i 1 = 1 b n n i=a i = 1 n(a+b) = a+b. n

4 ECS 315 HW Solution 7 Due: Oct /1 (b) First, note that b b ( ) (i + 1) (i 2) i (i 1) = i (i 1) 3 i=a ( b ) = 1 b (i + 1) i (i 1) i (i 1) (i 2) 3 i=a i=a i=a = 1 ((b + 1) b (b 1) a (a 1) (a 2)) 3 where the last equality comes from the fact that there are many terms in the first sum that is repeated in the second sum and hence many cancellations. Now, Therefore, b i 2 = i=a b i=a = 1 3 b (i (i 1) + i) = i=a b i (i 1) + i=a b i i=a ((b + 1) b (b 1) a (a 1) (a 2)) + n (a + b) 2 i 2 1 n = 1 a + b ((b + 1) b (b 1) a (a 1) (a 2)) + 3n 2 = 1 3 a2 1 6 a ab b b2 (c) Var X = E [X 2 ] (EX) 2 = 1 12 (d) σ X = n Var X = (b a) (b a + 2) = 1 12 (n 1)(n + 1) = n Problem 3. (Expectation + pmf + Gambling + Effect of miscalculation of probability) In the eighteenth century, a famous French mathematician Jean Le Rond d Alembert, author of several works on probability, analyzed the toss of two coins. He reasoned that because this experiment has THREE outcomes, (the number of heads that turns up in those two tosses can be 0, 1, or 2), the chances of each must be 1 in 3. In other words, if we let N be the number of heads that shows up, Alembert would say that [Mlodinow, 2008, p 50 51] p N (n) = 1/3 for N = 0, 1, 2. (7.1) 7-4

5 ECS 315 HW Solution 7 Due: Oct /1 We know that Alembert s conclusion was wrong. His three outcomes are not equally likely and hence classical probability formula can not be applied directly. The key is to realize that there are FOUR outcomes which are equally likely. We should not consider 0, 1, or 2 heads as the possible outcomes. There are in fact four equally likely outcomes: (heads, heads), (heads, tails), (tails, heads), and (tails, tails). These are the 4 possibilities that make up the sample space. The actual pmf for N is 1/4, n = 0, 2, p N (n) = 1/2 n = 1, 0 otherwise. Suppose you travel back in time and meet Alembert. You could make the following bet with Alembert to gain some easy money. The bet is that if the result of a toss of two coins contains exactly one head, then he would pay you $150. Otherwise, you would pay him $100. Let R be Alembert s profit from this bet and Y be the your profit from this bet. (a) Then, R = 150 if you win and R = +100 otherwise. Use Alembert s miscalculated probabilities from (7.1) to determine the pmf of R (from Alembert s belief). (b) Use Alembert s miscalculated probabilities from (7.1) (or the corresponding (miscalculated) pmf found in part (a)) to calculate ER, the expected profit for Alembert. Remark: You should find that ER > 0 and hence Alembert will be quite happy to accept your bet. (c) Use the actual probabilities, to determine the pmf of R. (d) Use the actual pmf, to determine ER. Remark: You should find that ER < 0 and hence Alembert should not accept your bet if he calculates the probabilities correctly. (e) Note that Y = +150 if you win and Y = 100 otherwise. Use the actual probabilities to determine the pmf of Y. (f) Use the actual probabilities, to determine EY. Remark: You should find that EY > 0. This is the amount of money that you expect to gain each time that you play with Alembert. Of course, Alembert, who still believes that his calculation is correct, will ask you to play this bet again and again believing that he will make profit in the long run. By miscalculating probabilities, one can make wrong decisions (and lose a lot of money)! Solution: 7-5

6 ECS 315 HW Solution 7 Due: Oct /1 (a) P [R = 150] = P [N = 1] and P [R = +100] = P [N 1] = P [N = 0] + P [N = 2]. So, p N (1), r = 150, p R (r) = p N (0) + p N (2), r = +100, 0, otherwise. Using Alembert s miscalculated pmf, 1/3, r = 150, p R (r) = 2/3, r = +100, 0, otherwise (b) From p R (r) in part (a), we have ER = r p R(r) = 1 ( 150) = (c) Again, Using the actual pmf, 1, r = 150, 2 1 p R (r) = + 1, r = +100, 4 4 0, otherwise p N (1), r = 150, p R (r) = p N (0) + p N (2), r = +100, 0, otherwise = { 1, 2 r = 150 or + 100, 0, otherwise. (d) From p R (r) in part (c), we have ER = r p R(r) = 1 ( 150) = (e) Observe that Y = R. Hence, using the answer from part (d), we have p R (r) = { 1, 2 r = +150 or 100, 0, otherwise. (f) Observe that Y = R. Hence, EY = ER. Using the actual probabilities, ER = 25 from part (d). Hence, EY = +25. Extra Questions Here are some questions for those who want extra practice. 7-6

7 ECS 315 HW Solution 7 Due: Oct /1 Problem 4. A random variables X has support containing only two numbers. Its expected value is EX = 5. Its variance is Var X = 3. Give an example of the pmf of such a random variable. Solution: We first find σ X = Var X = 3. Recall that this is the average deviation from the mean. Because X takes only two values, we can make them at exactly ± 3 from the mean; that is x 1 = 5 3 and x 2 = In which case, we automatically have EX = 5 and Var X = 3. Hence, one example of such pmf is { 1 p X (x) =, x = 5 ± 3 2 0, otherwise We can also try to find a general formula for x 1 and x 2. If we let p = P [X = x 2 ], then q = 1 p = P [X = x 1 ]. Given p, the values of x 1 and x 2 must satisfy two conditions: EX = m and Var X = σ 2. (In our case, m = 5 and σ 2 = 3.) From EX = m, we must have that is x 1 q + x 2 p = m; (7.2) x 1 = m q x p 2 q. From Var X = σ 2, we have E [X 2 ] = Var X + EX 2 = σ 2 + m 2 and hence we must have Substituting x 1 from (7.2) into (7.3), we have x 2 1q + x 2 2p = σ 2 + m 2. (7.3) x 2 2p 2x 2 mp + ( pm 2 qσ 2) = 0 whose solutions are Using (7.2), we have x 2 = 2mp ± 4m 2 p 2 4p (pm 2 qσ 2 ) 2p = 2mp ± 2σ pq 2p x 1 = m ) q p p (m q ± σ p q = m σ q. Therefore, for any given p, there are two pmfs: 1 p, x = m σ p X (x) = p, x = m + σ 0, otherwise, p 1 p 1 p p q = m ± σ p. 7-7

8 ECS 315 HW Solution 7 Due: Oct /1 or 1 p, x = m + σ p p X (x) = p, x = m σ p 0, otherwise. 1 p 1 p Problem 5. For each of the following families of random variable X, find the value(s) of x which maximize p X (x). (This can be interpreted as the mode of X.) (a) P(α) (b) Binomial(n, p) (c) G 0 (β) (d) G 1 (β) Remark [Y&G, p. 66]: For statisticians, the mode is the most common number in the collection of observations. There are as many or more numbers with that value than any other value. If there are two or more numbers with this property, the collection of observations is called multimodal. In probability theory, a mode of random variable X is a number x mode satisfying p X (x mode ) p X (x) for all x. For statisticians, the median is a number in the middle of the set of numbers, in the sense that an equal number of members of the set are below the median and above the median. In probability theory, a median, X median, of random variable X is a number that satisfies P [X < X median ] = P [X > X median ]. Neither the mode nor the median of a random variable X need be unique. A random variable can have several modes or medians. Solution: We first note that when α > 0, p (0, 1), n N, and β (0, 1), the above pmf s will be strictly positive for some values of x. Hence, we can discard those x at which p X (x) = 0. The remaining points are all integers. To compare them, we will evaluate p X(i+1) p X (i). (a) For Poisson pmf, we have p X (i + 1) p X (i) = e α α i+1 (i+1)! e α α i i! = α i + 1. Notice that 7-8

9 ECS 315 HW Solution 7 Due: Oct /1 p X(i+1) p X (i) > 1 if and only if i < α 1. p X(i+1) p X (i) = 1 if and only if i = α 1. p X(i+1) p X (i) < 1 if and only if i > α 1. Let τ = α 1. This implies that τ is the place where things change. Moving from i to i + 1, the probability strictly increases if i < τ. When i > τ, the next probability value (at i + 1) will decrease. (i) Suppose α (0, 1), then α 1 < 0 and hence i > α 1 for all i. (Note that i are are nonnegative integers.) This implies that the pmf is a strictly decreasing function and hence the maximum occurs at the first i which is i = 0. (ii) Suppose α N. Then, the pmf will be strictly increasing until we reaches i = α 1. At which point, the next probability value is the same. Then, as we further increase i, the pmf is strictly decreasing. Therefore, the maximum occurs at α 1 and α. (iii) Suppose α / N and α 1. Then we will have have any i = α 1. The pmf will be strictly increasing where the last increase is from i = α 1 to i + 1 = α = α. After this, the pmf is strictly decreasing. Hence, the maximum occurs at α. To summarize, arg max p X (x) = x 0, α (0, 1), α 1 and α, α is an integer, α, α > 1 is not an integer. (b) For binomial pmf, we have p X (i + 1) p X (i) = n! (i+1)!(n i 1)! pi+1 (1 p) n i 1 n! i!(n i)! pi (1 p) n i = (n i) p (i + 1) (1 p). Notice that p X(i+1) p X (i) > 1 if and only if i < np 1 + p = (n + 1)p 1. p X(i+1) p X (i) = 1 if and only if i = (n + 1)p 1. p X(i+1) p X (i) < 1 if and only if i > (n + 1)p

10 ECS 315 HW Solution 7 Due: Oct /1 Let τ = (n+1)p 1. This implies that τ is the place where things change. Moving from i to i + 1, the probability strictly increases if i < τ. When i > τ, the next probability value (at i + 1) will decrease. (i) Suppose (n+1)p is an integer. The pmf will strictly increase as a function of i, and then stays at the same value at i = τ = (n + 1)p 1 and i + 1 = (n + 1)p = (n + 1)p. Then, it will strictly decrease. So, the maximum occurs at (n + 1)p 1 and (n + 1)p. (ii) Suppose (n + 1)p is not an integer. Then, there will not be any i that is = τ. Therefore, we only have the pmf strictly increases where the last increase occurs when we goes from i = τ to i + 1 = τ + 1. After this, the probability is strictly decreasing. Hence, the maximum is unique and occur at τ + 1 = (n + 1)p = (n + 1)p. To summarize, arg max p X (x) = x { (n + 1)p 1 and (n + 1)p, (n + 1)p is an integer, (n + 1)p, (n + 1)p is not an integer. (c) p X(i+1) p X (i) = β < 1. Hence, p X (i) is strictly decreasing. The maximum occurs at the smallest value of i which is 0. (d) p X(i+1) p X (i) = β < 1. Hence, p X (i) is strictly decreasing. The maximum occurs at the smallest value of i which is 1. Problem 6. An article in Information Security Technical Report [ Malicious Software Past, Present and Future (2004, Vol. 9, pp. 618)] provided the data (shown in Figure 7.1) on the top ten malicious software instances for The clear leader in the number of registered incidences for the year 2002 was the Internet worm Klez. This virus was first detected on 26 October 2001, and it has held the top spot among malicious software for the longest period in the history of virology. Suppose that 20 malicious software instances are reported. Assume that the malicious sources can be assumed to be independent. (a) What is the probability that at least one instance is Klez? (b) What is the probability that three or more instances are Klez? (c) What are the expected value and standard deviation of the number of Klez instances among the 20 reported? 7-10

11 (b) What is the probability of no hits? [ Malicious Software Past, Present and Future (2004, Vol. 9, pp. 6 18)] provided the following data on the top ten malicious (c) What are the mean and variance of software instances for The clear leader in the num A statistical process control ch ber of registered incidences for the year 2002 was the Internet worm Klez, and it is still one of the most widespread threats. This virus was first detected on 26 October 2001, and it has of 20 parts from a metal punching pro hour. Typically, 1% of the parts require the number of parts in the sample of 20 ECS 315 held the HW top spot Solution among malicious 7 Due: software Oct for 24 the longest process problem 2014/1 is suspected if X exc period in the history of virology. Place Name % Instances 1 I-Worm.Klez 61.22% 2 I-Worm.Lentin 20.52% 3 I-Worm.Tanatos 2.09% 4 I-Worm.BadtransII 1.31% 5 Macro.Word97.Thus 1.19% 6 I-Worm.Hybris 0.60% 7 I-Worm.Bridex 0.32% 8 I-Worm.Magistr 0.30% 9 Win95.CIH 0.27% 10 I-Worm.Sircam 0.24% The 10 most widespread malicious programs for 2002 (Source Kaspersky Labs). than three standard deviations. (a) If the percentage of parts that req 1%, what is the probability that X more than three standard deviation (b) If the rework percentage increas probability that X exceeds 1? (c) If the rework percentage increase probability that X exceeds 1 in at le hours of samples? Because not all airline passen reserved seat, an airline sells 125 ticket only 120 passengers. The probability th show up is 0.10, and the passengers be (a) What is the probability that every up can take the flight? (b) What is the probability that the flig seats? Figure 7.1: The 10 most widespread malicious programs for 2002 (Source Kaspersky Labs) This exercise illustrates that p Suppose that 20 malicious software instances are reported. schedules and costs. A manufacturing pr Solution: Let N Assume be thethat number the malicious of instances sources can (among be assumed theto 20) be independent. that are orders Klez. to fill. Then, Each order requires one N binomial(n, p) where n = 20 and p = purchased from a supplier. However, ty (a) What is the probability that at least one instance is Klez? ponents are identified as defective, and (a) P [N 1] = 1 P (b)[n What < 1] is the = 1 P probability [N = that 0] = three 1 p or more instances are assumed to be independent. N (0) = 1 ( ) 20 Klez? (a) If the manufacturer stocks 100 co (c) What are the mean and standard deviation of the number probability that the 100 orders of Klez instances among the 20 reported? reordering components? (b) Heart failure is due to either natural occurrences (b) If the manufacturer stocks 102 co P [N 3] = 1 P [N < 3] = 1 (P [N = 0] + P [N = 1] + P [N (87%) or outside factors (13%). Outside factors are related to probability = 2]) that the 100 orders induced substances 2 ( ) 20 or foreign objects. Natural occurrences are reordering components? caused = 1 by arterial blockage, (0.6122) disease, k (0.3878) and infection. 20 k Suppose (c) If the manufacturer stocks 105 co k that 20 patients k=0 will visit an emergency room with heart failure. probability that the 100 orders Assume that causes of heart failure between individuals are reordering components? (c) EN = np = 20 independent = Consider the lengths of stay at σ N = Var N = (a) What np(1is the p) probability = 20 that three individuals have conditions caused by outside factors? pendently arrive for service. department in Exercise Assume (b) What is the probability that three or more individuals have (a) What is the probability that the le conditions caused by outside factors? one person is less than or equal to 4 (c) What are the mean and standard deviation of the number (b) What is the probability that exactly of individuals with conditions caused by outside factors? than 4 hours? 7-11

12 ECS 315: Probability and Random Processes 2014/1 HW 8 Due: Nov 6 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra question at the end is optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1 (Yates and Goodman, 2005, Q3.1.3). The CDF of a random variable W is 0, w < 5, (w + 5)/8, 5 w < 3, F W (w) = 1/4, 3 w < 3, 1/4 + 3 (w 3) /8, 3 w < 5, 1, w 5. (a) What is P [W 4]? (b) What is P [ 2 < W 2]? (c) What is P [W > 0]? (d) What is the value of a such that P [W a] = 1/2? Problem 2 (Yates and Goodman, 2005, Q3.2.1). The random variable X has probability density function { cx 0 x 2, f X (x) = 0, otherwise. Use the pdf to find 8-1

13 ECS 315 HW 8 Due: Nov /1 (a) the constant c, (b) P [0 X 1], (c) P [ 1/2 X 1/2], (d) the cdf F X (x). Problem 3 (Yates and Goodman, 2005, Q3.2.3). The CDF of random variable W is 0, w < 5, (w + 5)/8, 5 w < 3, F W (w) = 1/4, 3 w < 3, 1/4 + 3 (w 3) /8, 3 w < 5, 1, w 5. Find its pdf f W (w). Problem 4 (Yates and Goodman, 2005, Q3.3.4). The pdf of random variable Y is { y/2 0 y < 2, f Y (y) = 0, otherwise. What are E [Y ] and Var Y? Problem 5 (Yates and Goodman, 2005, Q3.3.6). The cdf of random variable V is 0 v < 5, F V (v) = (v + 5) 2 /144, 5 v < 7, 1 v 7. (a) What is E [V ]? (b) What is Var[V ]? (c) What is E [V 3 ]? 8-2

14 ECS HW8 Page 1 Y&G Q3.1.3 Wednesday, October 03, :14 PM

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16 ECS HW8 Page 3 Y&G Q3.2.1 Wednesday, October 03, :50 PM

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18 ECS HW8 Page 5 Y&G Q3.2.3 Wednesday, October 03, :18 PM

19 ECS HW8 Page 6 Y&G Q3.3.4 Tuesday, September 07, :08 PM

20 ECS HW8 Page 7 Y&G Q3.3.6 Wednesday, October 03, :37 PM

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22 ECS 315: Probability and Random Processes 2014/1 HW Solution 9 Due: Nov 14 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra question at the end is optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1 (Randomly Phased Sinusoid). Suppose Θ is a uniform random variable on the interval (0, 2π). (a) Consider another random variable X defined by where t is some constant. Find E [X]. X = 5 cos(7t + Θ) (b) Consider another random variable Y defined by Y = 5 cos(7t 1 + Θ) 5 cos(7t 2 + Θ) where t 1 and t 2 are some constants. Find E [Y ]. Solution: First, because Θ is a uniform random variable on the interval (0, 2π), we know that f Θ (θ) = 1 1 2π (0,2π)(t). Therefore, for any function g, we have E [g(θ)] = g(θ)f Θ (θ)dθ. 9-1

23 ECS 315 HW Solution 9 Due: Nov /1 (a) X is a function of Θ. E [X] = 5E [cos(7t + Θ)] = 5 2π 1 cos(7t + θ)dθ. Now, we know 0 2π that integration over a cycle of a sinusoid gives 0. So, E [X] = 0. (b) Y is another function of Θ. E [Y ] = E [5 cos(7t 1 + Θ) 5 cos(7t 2 + Θ)] = 2π 0 = 25 2π 2π 0 cos(7t 1 + θ) cos(7t 2 + θ)dθ. Recall 1 the cosine identity Therefore, cos(a) cos(b) = 1 (cos (a + b) + cos (a b)). 2 EY = 25 4π 2π = 25 4π 1 2π 5 cos(7t 1 + θ) 5 cos(7t 2 + θ)dθ 0 cos (14t + 2θ) + cos (7 (t 1 t 2 )) dθ ( 2π 0 cos (14t + 2θ) dθ + 2π 0 cos (7 (t 1 t 2 )) dθ ). The first integral gives 0 because it is an integration over two period of a sinusoid. The second integrand in the second integral is a constant. So, EY = 25 4π cos (7 (t 1 t 2 )) 2π 0 dθ = 25 4π cos (7 (t 1 t 2 )) 2π = 25 2 cos (7 (t 1 t 2 )). Problem 2 (Yates and Goodman, 2005, Q3.4.5). X is a continuous uniform RV on the interval ( 5, 5). (a) What is its pdf f X (x)? (b) What is its cdf F X (x)? 1 This identity could be derived easily via the Euler s identity: cos(a) cos(b) = eja + e ja ejb + e jb = 1 ( e ja e jb + e ja e jb + e ja e jb + e ja e jb) = 1 ( e ja e jb + e ja e jb + e ja e jb + e ja e jb ) = 1 (cos (a + b) + cos (a b))

24 ECS 315 HW Solution 9 Due: Nov /1 (c) What is E [X]? (d) What is E [X 5 ]? (e) What is E [ e X]? Solution: For a uniform random variable X on the interval (a, b), we know that { 0, x < a or x > b, f X (x) = a x b 1, b a and { 0, x < a or x > b, F X (x) = x a, a x b. b a In this problem, we have a = 5 and b = 5. (a) f X (x) = { 0, x < 5 or x > 5, x 5 (b) F X (x) = (c) EX = In general, EX = { 0, x < a or x > b, x+5 10 a x b. xf X (x) dx = b a 5 5 x 1 10 dx = 1 10 x x 1 b a dx = 1 b xdx = 1 b a b a a With a = 5 and b = 5, we have EX = 0. (d) E [X 5 ] = In general, x 5 f X (x) dx = E [ X 5] = b a 5 5 x dx = 1 10 x 6 6 ( = ( 5) 2) = x 2 2 x 5 1 b a dx = 1 b x 5 dx = 1 b a b a a With a = 5 and b = 5, we have E [X 5 ] = b a = 1 b 2 a 2 b a 2 ( = ( 5) 6) = 0. x 6 6 b a = 1 b a = a + b 2. b 6 a 6. 2

25 ECS 315 HW Solution 9 Due: Nov /1 (e) In general, E [ e X] = b a e x 1 b a dx = 1 b e x dx = 1 b a b a ex b a = eb e a b a. a With a = 5 and b = 5, we have E [ e X] = e5 e Problem 3. A random variable X is a Gaussian random variable if its pdf is given by f X (x) = 1 2πσ e 1 2( x m σ ) 2, for some constants m and σ. Furthermore, when a Gaussian random variable has m = 0 and σ = 1, we say that it is a standard Gaussian random variable. There is no closed-form expression for the cdf of the standard Gaussian random variable. The cdf itself is denoted by Φ and its values (or its complementary values Q( ) = 1 Φ( )) are traditionally provided by a table. We refer to this kind of table as the Φ table. Examples of such tables are Table 3.1 and Table 3.2 in [Y&G]. Suppose Z is a standard Gaussian random variable. (a) Use the Φ table to find the following probabilities: (i) P [Z < 1.52] (ii) P [Z < 1.52] (iii) P [Z > 1.52] (iv) P [Z > 1.52] (v) P [ 1.36 < Z < 1.52] (b) Use the Φ table to find the value of c that satisfies each of the following relation. (i) P [Z > c] = 0.14 (ii) P [ c < Z < c] = 0.95 Solution: (a) (i) P [Z < 1.52] = Φ(1.52) = (ii) P [Z < 1.52] = Φ( 1.52) = 1 Φ(1.52) = =

26 ECS 315 HW Solution 9 Due: Nov /1 (b) (iii) P [Z > 1.52] = 1 P [Z < 1.52] = = (iv) It is straightforward to see that the area of P [Z > 1.52] is the same as P [Z < 1.52] = Φ(1.52). So, P [Z > 1.52] = Alternatively, P [Z > 1.52] = 1 P [Z 1.52] = 1 Φ( 1.52) = 1 (1 Φ(1.52)) = Φ(1.52). (v) P [ 1.36 < Z < 1.52] = Φ(1.52) Φ( 1.36) = Φ(1.52) (1 Φ(1.36)) = Φ(1.52)+ Φ(1.36) 1 = = (i) P [Z > c] = 1 P [Z c] = 1 Φ(c). So, we need 1 Φ(c) = 0.14 or Φ(c) = = In the Φ table, we do not have exactly 0.86, but we have and Because 0.86 is closer to , we answer the value of c whose φ(c) = Therefore, c (ii) P [ c < Z < c] = Φ(c) Φ( c) = Φ(c) (1 Φ(c)) = 2Φ(c) 1. So, we need 2Φ(c) 1 = 0.95 or Φ(c) = From the Φ table, we have c Problem 4. The peak temperature T, as measured in degrees Fahrenheit, on a July day in New Jersey is a N (85, 100) random variable. Remark: Do not forget that, for our class, the second parameter in N (, ) is the variance (not the standard deviation). (a) Express the cdf of T in terms of the Φ function Hint: Recall that the cdf of a random variable T is given by F T (t) = P [T t]. For T N (m, σ 2 ), F T (t) = t 1 e 1 2( x m σ ) 2 dx. 2πσ The Φ function, which is the cdf of the standard Gaussian random variable, is given by Φ (z) = z 1 2π e 1 2 x2 dx. (b) Express each of the following probabilities in terms of the Φ function(s). Make sure that the arguments of the Φ functions are positive. (Positivity is required so that we can directly use the Φ/Q tables to evaluate the probabilities.) (i) P [T > 100] 9-5

27 ECS 315 HW Solution 9 Due: Nov /1 (ii) P [T < 60] (iii) P [70 T 100] (c) Express each of the probabilities in part (b) in terms of the Q function(s). Again, make sure that the arguments of the Q functions are positive. (d) Evaluate each of the probabilities in part (b) using the Φ/Q tables. (e) Observe that Table 3.1 stops at z = 2.99 and Table 3.2 starts at z = Why is it better to give a table for Q(z) instead of Φ(z) when z is large? Solution: (a) Continue from the hint. We perform a change of variable using y = x m σ to get F T (t) = x m σ 1 2π e 1 2 y2 dy = Φ ( x m ( ) t 85 Here, T N (85, 10 2 ). Therefore, F T (t) = Φ. 10 σ ). (b) (i) P [T > 100] = 1 P [T 100] = 1 F T (100) = 1 Φ ( ) = 1 Φ(1.5) (ii) P [T < 60] = P [T 60] because T is a continuous random variable and hence (iii) P [T = 60] = 0. Now, P [T 60] = F T (60) = Φ ( ) = Φ ( 2.5) = 1 Φ (2.5). Note that, for the last equality, we use the fact that Φ( x) = 1 Φ(x). ( ) ( ) P [70 T 100] = F T (100) F T (70) = Φ Φ = Φ (1.5) Φ ( 1.5) = Φ (1.5) (1 Φ (1.5)) = 2Φ (1.5) 1. (c) In this question, we use the fact that Q(x) = 1 Φ(x). (i) 1 Φ(1.5) = Q(1.5). (ii) 1 Φ (2.5) = Q(2.5). 9-6

28 ECS 315 HW Solution 9 Due: Nov /1 (d) (iii) 2Φ (1.5) 1 = 2(1 Q(1.5)) 1 = 2 2Q(1.5) 1 = 1 2Q(1.5). (i) 1 Φ(1.5) = = (ii) 1 Φ (2.5) = = (iii) 2Φ (1.5) 1 = 2 (0.9332) 1 = (e) When z is large, Φ(z) will start with The first few significant digits will all be the same and hence not quite useful to be there. Extra Questions Here are optional questions for those who want more practice. Problem 5. Let X be a uniform random variable on the interval [0, 1]. Set [ A = 0, 1 ) [, B = 0, 1 ) [ , 3 ) [, and C = 0, 1 ) [ , 3 ) [ 1 8 2, 5 ) 8 Are the events [X A], [X B], and [X C] independent? Solution: Note that P [X A] = P [X B] = P [X C] = dx = 1 2, dx + dx dx = 1 2, and dx dx dx = 1 2. [ 3 4, 7 )

29 ECS 315 HW Solution 9 Due: Nov /1 Now, for pairs of events, we have P ([X A] [X B]) = 1 4 dx = 1 4 = P [X A] P [X B], (9.1) P ([X A] [X C]) = P ([X B] [X C]) = dx + dx dx = 1 4 dx = 1 4 = P [X A] P [X C], and (9.2) = P [X B] P [X C]. (9.3) Finally, P ([X A] [X B] [X C]) = 1 8 dx = 1 8 = P [X A] P [X B] P [X C]. (9.4) 0 From (9.1), (9.2), (9.3) and (9.4), we can conclude that the events [X A], [X B], and [X C] are independent. Problem 6 (Q3.5.6). Solve this question using Table 3.1 and/or Table 3.2 from [Yates & Goodman, 2005]: A professor pays 25 cents for each blackboard error made in lecture to the student who points out the error. In a career of n years filled with blackboard errors, the total amount in dollars paid can be approximated by a Gaussian random variable Y n with expected value 40n and variance 100n. (a) What is the probability that Y 20 exceeds 1000? (b) How many years n must the professor teach in order that P [Y n > 1000] > 0.99? Solution: We are given 2 that Y n N (40n, 100n). Recall that when X N (m, σ 2 ), ( ) x m F X (x) = Φ. (9.5) σ 2 Note that the expected value and the variance in this question are proportional to n. This naturally occurs when we consider the sum of i.i.d. random variables. The approximation by Gaussian random variable is a result of the central limit theorem (CLT). 9-8

30 ECS 315 HW Solution 9 Due: Nov /1 (a) Here n = 20. So, we have Y n N (40 20, ) = N (800, 2000). For this random variable m = 800 and σ = We want to find P [Y 20 > 1000] which is the same as 1 P [Y ]. Expressing this quantity using cdf, we have Apply (9.5) to get P [Y 20 > 1000] = 1 Φ P [Y 20 > 1000] = 1 F Y20 (1000). ( ) = 1 Φ(4.472) Q(4.47) (b) Here, the value of n is what we want. So, we will need to keep the formula in the general form. Again, from (9.5), for Y n N (40n, 100n), we have ( ) ( ) n 100 4n P [Y n > 1000] = 1 F Yn (1000) = 1 Φ 10 = 1 Φ. n n To find the value of n such that P [Y n > 1000] > 0.99, we will first find the value of z which make 1 Φ (z) > (9.6) At this point, we may try to solve for the value of Z by noting that (9.6) is the same as Φ (z) < (9.7) Unfortunately, the tables that we have start with Φ(0) = 0.5 and increase to something close to 1 when the argument of the Φ function is large. This means we can t directly find 0.01 in the table. Of course, 0.99 is in there and therefore we will need to solve (9.6) via another approach. To do this, we use another property of the Φ function. Recall that 1 Φ(z) = Φ( z). Therefore, (9.6) is the same as Φ ( z) > (9.8) From our table, we can then conclude that (9.7) (which is the same as (9.8)) will happen when z > (If you have MATLAB, then you can get a more accurate answer of ) Now, plugging in z = 100 4n n, we have 4n 100 n > To solve for n, we first let x = n. In which case, we have 4x2 100 > 2.33 or, equivalently, 4x x 100 > 0. The two x roots are x = and x > 5.3. So, We need x < or x > 5.3. Note that x = n and therefore can not be negative. So, we only have one case; that is, we need x > 5.3. Because n = x 2, we then conclude that we need n > 28.1 years. 9-9

31 ECS 315: Probability and Random Processes 2014/1 HW Solution 10 Due: Nov 20 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra question at the end is optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1. Consider each random variable X defined below. Let Y = 1 + 2X. (i) Find and sketch the pdf of Y and (ii) Does Y belong to any of the (popular) families discussed in class? If so, state the name of the family and find the corresponding parameters. (a) X U(0, 1) (b) X E(1) (c) X N (0, 1) Solution: See handwritten solution Problem 2. Consider each random variable X defined below. Let Y = 1 2X. (i) Find and sketch the pdf of Y and (ii) Does Y belong to any of the (popular) families discussed in class? If so, state the name of the family and find the corresponding parameters. (a) X U(0, 1) (b) X E(1) (c) X N (0, 1) Solution: See handwritten solution 10-1

32 ECS 315 HW Solution 10 Due: Nov /1 Problem 3. Let X E(5) and Y = 2/X. Find (a) F Y (y). (b) f Y (y). (c) EY Hint: Because d 10 e y = 10 e 10 dy y 2 y > 0 for y 0. We know that e 10 y is an increasing function on our range of integration. In particular, consider y > 10/ ln(2). Then, e 10 y > 1. Hence, y e y dy > 10/ ln y e y dy > 10/ ln y 2 dy = 10/ ln 2 5 y dy. Remark: To be technically correct, we should be a little more careful when writing Y = 2 because it is undefined when X = 0. Of course, this happens with 0 probability; X so it won t create any serious problem. However, to avoid the problem, we may define Y by { 2/X, X 0, Y = 0, X = 0. Solution: (a) We consider two cases: y 0 and y > 0. (i) Because X > 0, we know that Y = 2 must be > 0 and hence, F X Y (y) = 0 for y 0. Note that Y can not be = 0. We need X = or to make Y = 0. However, ± are not real numbers therefore they are not possible X values. (ii) For y > 0, F Y (y) = P [Y y] = P [ ] [ 2 X y = P X 2 ]. y Note that, for the last equality, we can freely move X and y without worrying about flipping the inequality or division by zero because both X and y considered here are strictly positive. Now, for X E(λ) and x > 0, we have P [X x] = λe λt dt = e λt x = e λx Therefore, x F Y (y) = e 5( 2 y) = e 10 y. 10-2

33 ECS 315 HW Solution 10 Due: Nov /1 Combining the two cases above we have F Y (y) = { e 10 y, y > 0 0, y 0 (b) We show two methods that can be applied here to find f Y (y). Method 1: Because we have already derive the cdf in the previous part, we can find the pdf via the cdf by f Y (y) = d F dy Y (y). This gives f Y at all points except at y = 0 which we will set f Y to be 0 there. Hence, f Y (y) = { 10e y 2 y, y > 0 0, y 0. Method 2: See handwritten solution. (c) We can find EY from f Y (y) found in the previous part or we can even use f X (x) Method 1: EY = From the hint, we have yf Y (y) = 0 y y 2 e y dy = y e y dy EY > 10/ ln y e y dy > = 5 ln y 10/ ln 2 =. 10/ ln y 2 dy = 10/ ln 2 5 y dy Therefore, EY =. Method 2: [ ] 1 EY = E X > 1 0 = 1 x λe λ dx = λe λ 1 x f 1 X (x) dx = x λe λx dx > x dx = λe λ ln x 1 0 =, 0 1 x λe λx dx where the second inequality above comes from the fact that for x (0, 1), e λx > e λ. 10-3

34 ECS 315 HW Solution 10 Due: Nov /1 Problem 4. In wireless communications systems, fading is sometimes modeled by lognormal random variables. We say that a positive random variable Y is lognormal if ln Y is a normal random variable (say, with expected value m and variance σ 2 ). Find the pdf of Y. Hint: First, recall that the ln is the natural log function (log base e). Let X = ln Y. Then, because Y is lognormal, we know that X N (m, σ 2 ). Next, write Y as a function of X. Solution: We now show two methods that can be applied here to find f Y (y): Method 1 : Finding f Y (y) from F Y (y). Because X = ln(y ), we have Y = e X. We know that exponential function gives strictly positive number. So, Y is always strictly positive. In particular, F Y (y) = 0 for y 0. Next, for y > 0, by definition, F Y (y) = P [Y y]. Plugging in Y = e X, we have F Y (y) = P [ e X y ]. Because the exponential function is strictly increasing, the event [e X y] is the same as the event [X ln y]. Therefore, F Y (y) = P [X ln y] = F X (ln y). Combining the two cases above, we have { FX (ln y), y > 0, F Y (y) = 0, y 0. Finally, we apply f Y (y) = d dy F Y (y). For y < 0, we have f Y (y) = d 0 = 0. For y > 0, dy Therefore, f Y (y) = d dy F Y (y) = d dy F X (ln y) = f X (ln y) d dy ln y = 1 y f X (ln y). (10.1) f Y (y) = { 1 f y X (ln y), y > 0, 0, y < 0. At y = 0, if we know that Y is a continuous random variable, we can assign any value, e.g. 0, to f Y (0). Suppose this is the case. Then f Y (y) = { 1 f y X (ln y), y > 0, 0, otherwise. 10-4

35 ECS 315 HW Solution 10 Due: Nov /1 Method 2 : Finding f Y (y) directly from f X (x). Here we have Y = g(x) where g(x) = e x. Note that exponential function always gives positive number. So, for y 0, there is no (real-valued) x that satisfies g(x) = y. Therefore, f Y (y) = 0 for y 0. For y > 0, there is exactly one x that satisfies g(x) = y, namely x = ln y. At x = ln y, the slope g (x) is e x x=ln(y) = e ln y = y. So, f Y (y) = f X (x) g (x) = f X (ln (y)) y Combining the two cases above, we have = 1 y f X (ln (y)). f Y (y) = { 1 f y X (ln y), y > 0, 0, otherwise. Here, X N (m, σ 2 ). Therefore, f X (x) = 1 e 1 2( x m σ ) 2 2πσ and f Y (y) = { 1 2πσy e 1 2( ln(y) m σ ) 2, y > 0, 0, otherwise. Extra Question Here is an optional question for those who want extra practice. Problem 5. Consider a random variable X whose pdf is given by { cx f X (x) = 2, x (1, 2), 0, otherwise. Let Y = 4 X 1.5. (a) Find EY. (b) Find f Y (y). Solution: See handwritten solution 10-5

36 Q1 Tuesday, November 11, :15 PM ECS HW10 Page 1

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40 Q3b SISO (Method 2) Friday, November 21, :52 AM ECS HW10 Page 5

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42 Q5 Saturday, November 22, :54 PM ECS HW10 Page 7

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45 ECS 315: Probability and Random Processes 2014/1 HW 11 Due: Nov 14 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra question at the end is optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1. Consider a random variable X whose pdf is given by { cx, 0 < x < 1, f X (x) = 0, otherwise. (a) Find c. (b) Find f X (0.2) and f X (0.8). (c) Find P [0.19 < X < 0.21] and P [0.79 < X < 0.81]. (d) In class, we have seen that we may use f X (x) x to approximate the probability that the random variable X will be inside some small interval of length x near or around x. Use this approximation to calculate P [0.19 < X < 0.21] and P [0.79 < X < 0.81]. Compare your answers here with the exact answers in the previous part. Problem 2. Consider a random variable X whose pdf is given by { cx f X (x) = 2, 0 < x < 1, 0, otherwise. 11-1

46 ECS 315 HW 11 Due: Nov /1 (a) Find c. (b) Find f X (0.2) and f X (0.8). (c) Find P [0.19 < X < 0.21] and P [0.79 < X < 0.81]. (d) In class, we have seen that we may use f X (x) x to approximate the probability that the random variable X will be inside some small interval of length x near or around x. Use this approximation to calculate P [0.19 < X < 0.21] and P [0.79 < X < 0.81]. Compare your answers here with the exact answers in the previous part. Problem 3. Consider a random variable X whose pdf f X (x) is unknown. From an experiment, suppose we found that P [0.99 < X < 1.03] 0.1 and P [1.98 < X < 2.01] Use this information to approximate the ratio f X(2) f X (1). Problem 4. Suppose X U( 1, 4). Define a new random variable Y by Y = g(x) where the function g(x) is plotted in Figure g x x Figure 11.1: Function g(x) for Problem 4. (a) Find P [Y > 2]. (b) Find P [2 < Y < 3]. (c) Find P [1.9 < Y < 2.1]. (d) Find f Y (y) at y near y = 2. (e) Find f Y (y)

47 ECS 315 HW 11 Due: Nov /1 Problem 5. Let a continuous random variable X denote the current measured in a thin copper wire in milliamperes. Assume that the probability density function of X is { 5, 4.9 x 5.1, f X (x) = 0, otherwise. (a) Find the probability that a current measurement is less than 5 milliamperes. (b) Find and plot the cumulative distribution function of the random variable X. (c) Find the expected value of X. (d) Find the variance and the standard deviation of X. (e) Find the expected value of power when the resistance is 100 ohms? Extra Question Here are optional questions for those who want extra practice. Problem 6. The time until a chemical reaction is complete (in milliseconds) is approximated by the cumulative distribution function { 1 e F X (x) = 0.01x, x 0, 0, otherwise. (a) Determine the probability density function of X. (b) What proportion of reactions is complete within 200 milliseconds? Problem 7. Cholesterol is a fatty substance that is an important part of the outer lining (membrane) of cells in the body of animals. Its normal range for an adult is mg/dl. The Food and Nutrition Institute of the Philippines found that the total cholesterol level for Filipino adults has a mean of mg/dl and 84.1% of adults have a cholesterol level below 200 mg/dl. Suppose that the cholesterol level in the population is normally distributed. (a) Determine the standard deviation of this distribution. (b) What is the value of the cholesterol level that exceeds 90% of the population? (c) An adult is at moderate risk if cholesterol level is more than one but less than two standard deviations above the mean. What percentage of the population is at moderate risk according to this criterion? 11-3

48 ECS 315 HW 11 Due: Nov /1 (d) An adult is thought to be at high risk if his cholesterol level is more than two standard deviations above the mean. What percentage of the population is at high risk? Problem 8. Suppose that the time to failure (in hours) of fans in a personal computer can be modeled by an exponential distribution with λ = (a) What proportion of the fans will last at least 10,000 hours? (b) What proportion of the fans will last at most 7000 hours? [Montgomery and Runger, 2010, Q4-97] 11-4

49 Q1 Meaning of pdf Wednesday, November 12, :21 PM ECS HW11 Page 1

50 Q2 Meaning of pdf Wednesday, November 12, :21 PM ECS HW11 Page 2

51 Q3 Meaning of pdf Thursday, November 20, :27 PM ECS HW11 Page 3

52 Q4 SISO Thursday, November 20, :34 PM ECS HW11 Page 4

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57 Q5: pdf, cdf, expected value, variance - current and power Thursday, November 13, :01 AM ECS HW11 Page 9

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59 Q6: pdf and cdf - chemical reaction Thursday, November 13, :07 AM ECS HW11 Page 11

60 Q7: Gaussian RV - Cholesterol Thursday, November 13, :49 PM ECS HW11 Page 12

61 Q8: Exponential RV Friday, November 21, :02 AM ECS HW11 Page 13

62 ECS 315: Probability and Random Processes 2014/1 HW Solution 12 Due: Dec 4 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) ONE part of a question will be graded (5 pt). Of course, you do not know which part will be selected; so you should work on all of them. (b) It is important that you try to solve all problems. (5 pt) The extra questions at the end are optional. (c) Late submission will be heavily penalized. (d) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1. The input X and output Y of a system subject to random perturbations are described probabilistically by the following joint pmf matrix: x 1 3 y Evaluate the following quantities: (a) The marginal pmf p X (x) (b) The marginal pmf p Y (y) (c) EX (d) Var X (e) EY (f) Var Y 12-1

63 ECS 315 HW Solution 12 Due: Dec /1 (g) P [XY < 6] (h) P [X = Y ] Solution: The MATLAB codes are provided in the file P XY marginal.m. (a) The marginal pmf p X (x) is founded by the sums along the rows of the pmf matrix: p X (x) = 0.2, x = 1 0.8, x = 3 0, otherwise. (b) The marginal pmf p Y (y) is founded by the sums along the columns of the pmf matrix: p Y (y) = 0.1, y = , y = , y = 5 0, otherwise. (c) EX = x xp X (x) = = = 2.6. (d) E [X 2 ] = x x 2 p X (x) = = = 7.4. So, Var X = E [X 2 ] (EX) 2 = 7.4 (2.6) 2 = = (e) EY = y yp Y (y) = = = (f) E [Y 2 ] = y y 2 p Y (y) = = So, Var Y = E [Y 2 ] (EY ) 2 = = (g) Among the 6 possible pairs of (x, y) shown in the joint pmf matrix, only the pairs (1, 2), (1, 4), (1, 5) satisfy xy < 6. Therefore, [XY < 6] = [X = 1] which implies P [XY < 6] = P [X = 1] = 0.2. (h) Among the 6 possible pairs of (x, y) shown in the joint pmf matrix, there is no pair which has x = y. Therefore, P [X = Y ] =

64 ECS 315 HW Solution 12 Due: Dec /1 Problem 2. The input X and output Y of a system subject to random perturbations are described probabilistically by the following joint pmf matrix: x 1 3 y (a) Evaluate the following quantities: (i) E [XY ] (ii) E [(X 3)(Y 2)] (iii) E [X(Y 3 11Y Y )] (iv) Cov [X, Y ] (v) ρ X,Y Hint: Write down the formulas then use MATLAB or Excel to compute them. (b) Find ρ X,X (c) Calculate the following quantities using the values of Var X, Cov [X, Y ], and ρ X,Y that you got earlier. (i) Cov [3X + 4, 6Y 7] (ii) ρ 3X+4,6Y 7 (iii) Cov [X, 6X 7] (iv) ρ X,6X 7 Solution: The MATLAB codes are provided in the file P XY EVarCov.m. (a) (i) From MATLAB, E [XY ] = (ii) From MATLAB, E [(X 3)(Y 2)] = (iii) From MATLAB, E [X(Y 3 11Y Y )] = 104. (iv) From MATLAB, Cov [X, Y ] =

65 ECS 315 HW Solution 12 Due: Dec /1 (v) From MATLAB, ρ X,Y = (b) ρ X,X = Cov[X,X] σ X σ X (c) = Var[X] σ 2 X = 1. (i) Cov [3X + 4, 6Y 7] = 3 6 Cov [X, Y ] (ii) Note that ρ ax+b,cy +d = = Cov [ax + b, cy + d] σ ax+b σ cy +d accov [X, Y ] = ac a σ X c σ Y ac ρ X,Y = sign(ac) ρ X,Y. Hence, ρ 3X+4,6Y 7 = sign(3 4)ρ X,Y = ρ X,Y = (iii) Cov [X, 6X 7] = 1 6 Cov [X, X] = 6 Var[X] (iv) ρ X,6X 7 = sign(1 6) ρ X,X = 1. Problem 3. Suppose X binomial(5, 1/3), Y binomial(7, 4/5), and X the following quantities. (a) E [(X 3)(Y 2)] (b) Cov [X, Y ] (c) ρ X,Y = Y. Evaluate Solution: (a) First, because X and Y are independent, we have E [(X 3)(Y 2)] = E [X 3] E [Y 2]. Recall that E [ax + b] = ae [X]+b. Therefore, E [X 3] E [Y 2] = (E [X] 3) (E [Y ] 2) Now, for Binomial(n, p), the expected value is np. So, (E [X] 3) (E [Y ] 2) = (5 13 ) ( ) 2 = = 24 5 = 4.8. (b) Cov [X, Y ] = 0 because X = Y. (c) ρ X,Y = 0 because Cov [X, Y ] =

66 ECS 315 HW Solution 12 Due: Dec /1 Problem 4. Suppose we know that σ X = 21 10, σ Y = 4 6 5, ρ X,Y = (a) Find Var[X + Y ]. (b) Find E [(Y 3X + 5) 2 ]. Assume E [Y 3X + 5] = 1. Solution: (a) First, we know that Var X = σx 2 = 21, Var Y = 100 σ2 Y = 96, and Cov [X, Y ] = ρ 25 X,Y σ X σ Y = 2. Now, 25 Var [X + Y ] = E [ ((X + Y ) E [X + Y ]) 2] = E [ ((X EX) + (Y EY )) 2] = E [ (X EX) 2] + 2E [(X EX) (Y EY )] + E [ (Y EY ) 2] = Var X + 2Cov [X, Y ] + Var Y = = Remark: It is useful to remember that Var [X + Y ] = Var X + 2Cov [X, Y ] + Var Y. Note that when X and Y are uncorrelated, Var [X + Y ] = Var X +Var Y. This simpler formula also holds when X and Y are independence because independence is a stronger condition. (b) First, we write Y ax b = (Y EY ) a (X EX) (aex + b EY ). }{{} c Now, using the expansion we have (u + v + t) 2 = u 2 + v 2 + t 2 + 2uv + 2ut + 2vt, (Y ax b) 2 = (Y EY ) 2 + a 2 (X EX) 2 + c 2 2a (X EX) (Y EY ) 2c (Y EY ) + 2a (X EX) c. Recall that E [X EX] = E [Y EY ] = 0. Therefore, E [ (Y ax b) 2] = Var Y + a 2 Var X + c 2 2aCov [X, Y ] 12-5

67 ECS 315 HW Solution 12 Due: Dec /1 Plugging back the value of c, we have E [ (Y ax b) 2] = Var Y + a 2 Var X + (E [(Y ax b)]) 2 2aCov [X, Y ]. Here, a = 3 and b = 5. Plugging these values along with the given quantities into the formula gives E [ (Y ax b) 2] = = Extra Question Here are optional questions for those who want extra practice. Problem 5. The input X and output Y of a system subject to random perturbations are described probabilistically by the joint pmf p X,Y (x, y), where x = 1, 2, 3 and y = 1, 2, 3, 4, 5. Let P denote the joint pmf matrix whose i,j entry is p X,Y (i, j), and suppose that P = 1 71 (a) Find the marginal pmfs p X (x) and p Y (y). (b) Find EX (c) Find EY (d) Find Var X (e) Find Var Y Solution: All of the calculations in this question are simply plugging numbers into appropriate formula. The MATLAB codes are provided in the file P XY marginal 2.m. (a) The marginal pmf p X (x) is founded by the sums along the rows of the pmf matrix: 26/71, x = , x = 1 25/71, x = , x = 2 p X (x) = 20/71, x = , x = 3 0, otherwise 0, otherwise. 12-6

68 ECS 315 HW Solution 12 Due: Dec /1 The marginal pmf p Y (y) is founded by the sums along the columns of the pmf matrix: 13/71, y = , y = 1 8/71, y = , y = 2 21/71, y = , y = 3 p Y (y) = 15/71, y = , y = 4 14/71, y = , y = 5 0, otherwise 0, otherwise. (b) EX = (c) EY = (d) Var X = (e) Var Y = Problem 6. A webpage server can handle r requests per day. Find the probability that the server gets more than r requests at least once in n days. Assume that the number of requests on day i is X i P(α) and that X 1,..., X n are independent. Solution: [Gubner, 2006, Ex 2.10] [ n ] [ n ] P [X i > r] = 1 P [X i r] = 1 i=1 = 1 i=1 n P [X i r] i=1 ( n r ) ( r ) n α k e α α k e α = 1. k! k! i=1 k=0 k=0 Problem 7. Suppose X binomial(5, 1/3), Y binomial(7, 4/5), and X (a) A vector describing the pmf of X can be created by the MATLAB expression: x = 0:5; px = binopdf(x,5,1/3). What is the expression that would give py, a corresponding vector describing the pmf of Y? (b) Use px and py from part (a), how can you create the joint pmf matrix in MATLAB? Do not use for-loop, while-loop, if statement. Hint: Multiply them in an appropriate orientation = Y.