1 Basic continuous random variable problems

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1 Name M362K Final Here are problems concerning material from Chapters 5 and 6. To review the other chapters, look over previous practice sheets for the two exams, previous quizzes, previous homeworks and work out problems in the book that weren t on the homework. Contents 1 Basic continuous random variable problems 1 2 Expected value 2 3 Computational problems 2 4 Independent random variables 4 5 Marginal and conditional densities 5 6 Covariance/Variance/Expected Value 5 7 Chebyshev/Markov 9 8 Normal random variables 9 9 Normal approximations/ central limit theorem 11 1 Basic continuous random variable problems 1. The lifetime of certain battery is an exponential random variable X with mean 1 (in hours. Given that the battery has already lasted 1 hours, what is the probability that it will last another 1 hours? In other words, P (X 2 X 1 =? Hint: the density function of X is { 1 f X (x = 1 e x/1 if x otherwise Solution. By the memoryless property P (X 2 X 1 = P (X 1 = 1 = e x/1 ] 1 = e 1. e x/1 1 dx 2. Let X be a random variable with probability density function Find c. Then find P ( 2 < X < 1. f(x = ce x, < x <.

2 2 Expected value 3. A certain road is 1 miles long from point A to point B. There are bus repair stations at points A, B and C which is located 3 miles from point A. If the next accident occurs at a point that is uniformly distributed along the length of the road, what is the expected distance from the accident to the closest repair station? Solution. The question asks for E[g(X] where X is uniformly distributed on [, 1] and x x < 15 g(x = 3 x 15 x < 65 1 x 65 x. So the answer is E[g(X] = 1 = 1 ( 15 x dx + 1 = 1 1 g(x/1 dx x dx x 3 dx x dx ( 15 2 /2 + (3(15 ( /2 3(35 + ( /2 + (1(35 ( /2 4. You and your friend agree to meet at a coffee shop around noon. Suppose your arrivial time (in minutes after noon is exponential with mean 1 and your friends is also exponential with mean 1. Assuming the two arrival times are independent, how much time will you wait on average for your friend? It is OK to set up the integral without evaluating. 3 Computational problems 5. Let X be a continuous random variable with density { 1/x 1 < x < e f(x = otherwise (a Find the distribution function of X. (b P (X > 2 =? (c E[X] =? (d E[X 2 ] =? (e Find the distribution function of Y = X 2 (f Find the density function of Y. Solution. The distribution function of X is F X (t = P (X t = t 1 1/x dx = log(t for 1 t e (F X (t = if t < 1 and F X (t = 1 if t > e.

3 Let Y = X 2. So F Y, the distribution function of Y is F Y (y = P (Y y = P (X 2 y = P (X y = log( y = (1/2 log(y if 1 y e 2. The density function f Y is the derivative f Y (y = F Y (y = 1/(2y for 1 y e2 and f Y (y = otherwise. 6. Let X, Y be random variables with joint density function { cxy x y 2 f(x, y = otherwise where c > is a constant. In the questions below, it is OK to leave your answer in integral form without evaluating it. (a Find c. (b Find P (X + Y 2. (c Find E[X]. (d Find the density of Y. (e Find f X (x Y = 1, the density of X given that Y = 1. (f Are X and Y independent? Solution. Note 1 = So c = 1/2. 2 y For the second part, cxy dxdy = c X and Y are not independent. 2 P (X + Y 2 = y[x 2 /2] y dy = c 2 y 2 y 2 y 3 /2 dy = cy 4 /8] 2 = 2c. xy/2 dxdy. E[X] = xf X,Y (x, y dxdy f Y (y = f X,Y (x, y dx f X (x Y = 1 = f(x, 1 f(t, 1 dt. 7. Let X, Y be random variables with joint density function { c(x + y x 1, y 1 f(x, y = otherwise where c > is a constant. Compute the following. You can leave your answers in integral form. (a c =? (b P (X + Y 1

4 4 Independent random variables 8. Let X, Y be exponential random variables with parameters λ 1, λ 2. Suppose X and Y are independent. (a Compute the density function of X + Y. (b P (X + Y 1 =? 9. Let X and Y be uniformly distributed over the interval [, 1]. Compute the density function of X Y assuming X and Y are independent. Hint: find the distribution function of X Y first. 1. Let X, Y be independent exponential random variables such that E[X] = 1 and E[Y ] = 2. (a Find the joint density function of (X, Y. (b Find P (X 5, Y Suppose X, Y, Z are uniformly distributed over (, 1. Suppose they are also independent. (a Find P (X 2 + Y 2 Z. (b Compute P (X Y +Z. (It is OK to just set up the integral without computing. 12. You have two friends, Alice and Bob and they both promised to call you sometime after 7pm on Saturday. So you are sitting at home at 7pm on Saturday and you decide to go out with either Alice or Bob, whomever calls you first. Let A be the amount of time before Alice calls you, and B the amount of time before Bob calls you. Assume A and B are independent exponential random variables; the expectation of A is 6 minutes and the expectation of B is 3 minutes. What is the probability that you will go out with Alice? Solution. The question asks for P (B A. Note that the joint density function of (A, B is f(x, y = (1/6(1/3 exp( x/6 y/3 if x, y > are otherwise. So the answer is = = x y = (1/6(1/3 exp( x/6 y/3 dydx (1/6(1/3 exp( x/6 y/3 dxdy ( 1/3 exp( x/6 y/3 (1/3[exp( y/3 exp( y/6 y/3] dy = = 1 2/3 = 1/3. ] y x= dxdy (1/3[exp( y/3 exp( y/2] dy

5 5 Marginal and conditional densities 13. Suppose X, Y are random variables with joint density function { 4xy x, y 1 f X,Y (x, y = otherwise (a Find the density of X. (b Are X and Y independent? Solution. f X (x = f X,Y (x, y dy = 1 4xy dy = 2xy 2 ] 1 = 2x if x 1. It equals zero otherwise. Similarly, f Y (y = 2y for y 1 and equals zero otherwise. X and Y are independent because f X,Y (x, y = f X (xf Y (y. 14. Let (X, Y be uniformly distributed over the circle of radius 1 with center at (,. (a Compute the density of X. (b Compute f X (x Y = 6, the density of X conditioned on Y = 6. 6 Covariance/Variance/Expected Value 15. Suppose that every day, the price of a certain stock either increases by 3% or decreases by 1% and that the probability that it increases is 1/2. If it starts out at 1, after n days what is the expected value of the stock? You can assume the different days are independent. Solution. The answer is E[1 X 1 X n ] where each X i is a random variable satisfying P (X i = 1.3 = P (X i =.9 = 1/2. Because of independence this is 1 E[X 1 ] E[X n ]. Since E[X i ] = 1.1, the expected value of the stock is 1 (1.1 n. 16. Suppose X, Y, Z are pairwise uncorrelated (this means Cov(X, Y = Cov(Y, Z = Cov(X, Z = E[X] = E[Y ] = E[Z] = and Var(X = Var(Y = Var(Z = 1. Find Cov(X + Y, Z Y. Answer. Because they are uncorrelated Cov(X, Y = (for example. Because covariance is bilinear, Cov(X + Y, Z Y = Cov(X, Z Cov(X, Y + Cov(Y, Z Cov(Y, Y = Cov(Y, Y = Var(Y = 1.

6 17. Ten hunters are shooting at 2 different ducks. Each hunter aims at a duck at random and hits it with probability 1/2 independently of the other hunters. (a Let X be the number of hunters that hit a duck. What is E[X]? What is Var(X? (b What is the probability that the first hunter does not hit the first duck? (This means that either he does not aim at the first duck or he aims and misses. (c Let Y be the number of ducks hit. What is E[Y ]? What is Var(Y? Partial Solution. (a X = X X 1 where X i = 1 if the i-th hunter hits a duck and X i = otherwise. Since E[X i ] = P (X i = 1 = 1/2, we have E[X] = 5. (b This probability is 1 (1/2(1/2 = 39/4. (c Y = Y Y 2 where Y i = 1 if the i-th duck is hit. Since E[Y i ] = 1 P (Y i = and P (Y i = = (39/4 1 we have E[Y ] = 2(1 (39/ Suppose n people go to a dinner party. Each person checks in their hat. The host accidentally mixes up the hats and, in a vain attempt to avoid embarassment, hands them back to guests at random. Let X be the number of people that receive his/her hat back. Find the variance of X. (Hint: again you can write X = X X n for a suitable choice of X i s. Solution. Now Also if i j then Var(X = Var( i X i = n i=1 Var(X i + 2 i<j Cov(X i, X j. Var(X i = E[X 2 i ] E[X i ] 2 = E[X i ] E[X i ] 2 = (1/n (1/n 2. Cov(X i, X j = E[X i X j ] E[X i ]E[X j ] = (1/n(1/(n 1 (1/n 2. So ( n Var(X = n[(1/n (1/n 2 ] + 2 [(1/n(1/(n 1 (1/n 2 ] 2 = 1 (1/n + 1 (n 1/n = 1 1/n + 1/n = Let (X, Y be uniformly distributed over the triangle with vertices ( 1,, (1,, (, 1. Compute Cov(X, Y. Are X and Y positively correlated, negatively correlated or uncorrelated? Are X and Y independent? Solution. Because the triangle is symmetric about the y-axis, E[X] =. The joint density function is f(x, y = 1 for (x, y in the triangle and otherwise (because the triangle has area 1. So the covariance is Cov(X, Y = E[XY] E[X]E[Y] = 1 1 y y 1 xy dxdy.

7 Because (X, Y and ( X, Y have the same distribution we must have Cov(X, Y = Cov( X, Y. But this implies Cov(X, Y =. So we didn t actually have to compute it. X and Y are uncorrelated (because Cov(X, Y = but not independent. 2. Let (X, Y be uniformly distributed over the triangle with vertices (,, (1,, (, 1. Compute Cov(X, Y. Are X and Y positively correlated, negatively correlated or uncorrelated? Are X and Y independent? Solution. The area of the triangle is 1/2. So the density function = 2 on the triangle and outside of it. So 1 1 y ( 1 1 y ( 1 1 y Cov(X, Y = E[XY] E[X]E[Y] = 2xy dxdy 2x dxdy 2y dxdy. X and Y are negatively correlated (because Cov(X, Y < and so are dependent. 21. Let (X, Y be uniformly distributed over the triangle with vertices (,, (1,, (, 2. (a Compute Cov(X, Y. (You can leave your answer in terms of integrals without evaluating (b Are X and Y positively correlated, negatively correlated or uncorrelated? Solution. The area of the triangle is 1. So the density function = 1 on the triangle and outside of it. The triangle is bounded by the x-axis, the y-axis and the line y = 2 2x. Solving for x we have x = 1 y/2. So 2 1 y/2 ( 2 1 y/2 ( 2 Cov(X, Y = E[XY] E[X]E[Y] = xy dxdy x dxdy X and Y are negatively correlated (because Cov(X, Y < and so are dependent. The density of X is The density of Y is So you could also write Cov(X, Y = Note y/2 2 f X (x = f Y (y = 2 2x 1 y/2 ( 1 xy dxdy dy = 2 2x dx = 1 y/2. ( 2 x(2 2x dx y(1 y/2 dy. ] 1 x(2 2x dx = x 2 2x 3 /3 = 1 2/3 = 1/3. ] 2 y(1 y/2 dy = y 2 /2 y 3 /6 = 2 8/6 = 2/3. 1 y/2 y dxdy.

8 22. Roll two dice. Let X be the number on the first die and Y be the number of the second die. (a Compute Cov(X + Y, X Y. (b Are X + Y and X Y independent? Explain! Solution. Because of bilinearity, Cov(X + Y, X Y = Cov(X, X Cov(X, Y + Cov(Y, X Cov(Y, Y. Because X and Y are independent, Cov(X, Y = Cov(Y, X =. Because X and Y are identically distributed, Cov(X, X = Cov(Y, Y. So Cov(X + Y, X Y =. Are they independent? Well suppose X + Y = 12. Then we know X = Y = 6 and therefore X Y =. This is nontrivial information. So they cannot be independent. To be more rigorous, P (X + Y = 12, X Y = = 1/36. However, P (X + Y = 12P (X Y = = (1/36(1/6 = 1/216. So they are not independent. 23. Suppose an urn contains 3 red balls, and 7 blue balls. You pick 2 balls at random without replacement. Let X be the number of red balls chosen. Compute the variance of X. 24. Suppose an urn contains 3 red balls, 4 blue balls and 5 yellow balls (for a total of 12. You pick 12 balls at random. Let X be the number of red balls chosen. For the questions below, compute the answer exactly without any long summations. (a Assume that the balls are chosen with replacement. Find E[X]. (b Assume that the balls are chosen without replacement. Find E[X]. Solution. X = X 1 + +X 12 where X i = 1 if the i-th ball is red and X i = otherwise. Since E[X i ] = 3/12 = 1/4, E[X] = (12/4 = 3. This answer does not depend on whether or not the balls or chosen with replacement. Alternative Solution. X = X 1 + +X 3 where X i = 1 if the i-th red ball is selected at some time and X = otherwise. Then ( E[X i ] = ( 12 = = So E[X] = 3/1 = 3. Again, this does not depend on whether or not the balls or chosen with replacement. 25. (Balls and Boxes Suppose there are n balls and n boxes. Each ball is placed in a box at random. Each box can contain an unlimited number of balls. Moreover the function assigning to balls to boxes is equally likely to be any of the n n functions. Find the expected value and the variance of the number of empty boxes.

9 7 Chebyshev/Markov 26. If a post office handles, on average, 1, letters a day, then provide an upper bound on the probability that they will handle more than 2, letters tomorrow. Use Markov s inequality. Answer. P (X 2 E[X]/2 = 1/ If the number of letters a post office handles in any given day is a random variable with mean 1, and variance 2, then use Chebyshev s inequality to provide an upper bound on the probability that they will handle more than 2, letters tomorrow Answer. Because X always, P (X 2 = P ( X 1 1 = 2 1, = 1 5, =.2. 2, (1, 2 Notice how much smaller.2 is than 1/2. So this estimate is much better than the previous one. The moral of the story is that it helps to know the variance is of modest size! (We would not have achieved a better bound if the variance was 5,,. 28. Suppose X is a non-negative random variable and P (X 1 = 1/2. What is the smallest possible value of E[X]? Solution. Since 1/2 = P (X 1 E[X]/1 we have 5 E[X]. The smallest possible value is 5. This occurs when X = 1 with probability 1/2 and X = with probability 1/2. 8 Normal random variables 29. Suppose that X 1, X 2,... are i.i.d. normal random variables with mean, variance 1 and S n = n i=1 X i. Determine the smallest value of n so that P ( Sn n Hint: if χ is the standard normal then P ( χ 2 =.95. Solution. The standard deviation of S n is n. So S n / n is the standard normal. So P ( S n /n 1 = P ( S n / n n P ( χ n. So n = You and your friend agree to meet at noon. Your arrival time in minutes after noon is normally distributed with mean and standard deviation 3. Your friend s arrival time in minutes after noon is normally distributed with mean 1 and standard deviation 4. Assume that the two arrival times are independent. What is the probability that you will have to wait at least 4 minutes? Solution. Let X be your arrival time and Y your friend s arrival time. The question asks P (Y X 4. Note Y X is normally distributed with mean 1 and variance = 5 2. So P (Y X 4 = P ( Y X 1 5 3/5 = 1 Φ(3/5 = Φ( 3/5.

10 31. The distributions of the grades of the students of probability and calculus at a certain university are normally distributed with parameters µ = 65, σ 2 = 418 and µ = 72, σ 2 = 448 respectively. Dr. Olwell teaches a probability section with 22 students and a calculus section with 28 students. What is the probability that the difference between the averages of the final grades of these two classes is at least 2? You may assume independence. Write your answer in terms of the function Φ(t = 1 t /2 2π e x2 dx. Solution. Let X and Y be the averages of the final grades in the probability and calculus classes respectively. The question asks for P ( X Y 2. I m assuming that X = (X X 22 /22 where each X i is a normal random variable with mean 65 and variance 418 (and similarly for Y. Therefore X is a normal random variable with mean 65 and variance 418/22 (to see this observe that V ar(x 1 + +X 22 = because these random variables are independent; next use the fact that V ar(tz = t 2 V ar(z for any constant t and any random variable Z. Similarly, Y is a normal random variable with 72 and variance 448/28. So X Y is a normal random variable with µ = = 7 and variance σ 2 = 418 Now P ( X Y 2 = P (X Y 2 + P (2 X Y = P X Y P = Φ Φ X Y A weighted coin lands on heads with probability.8. Suppose it is flipped 1 times and let X denote the number of times it lands on heads. (a What is E[X]? (b What is Var(X? (c Use normal random variables to approximate P (X 9. Leave your answer in terms of the Φ-function. Answer. X is a binomial random variable, E[X] = np = 8, SD(X = np(1 p = (1(.8(.2 = 4. So P (X 9 = P ( X = 1 P ( X Φ( numbers are selected independently and uniformly from the interval [, 1]. Use the central limit theorem to approximate the probability that the average of these numbers is within.1 of 1/2. Hint: the standard deviation of the random variable X that is 1 uniformly distributed in [, 1] is 12. Leave your answer in terms of the Φ-function. 34. Toss a fair coin twice. You win $1 if at least one of the two tosses comes out heads. You neither win nor lose anything otherwise.

11 (a Assume that you play this game 3 times. What is, approximately, the probability that you win at least $25? (Use the normal approximation. Solution. Let S = X X 3 where X i = 1 is you win a dollar on the i-th game and X i = otherwise. Then E[X i ] = 3/4 and E[S] = 3 3/4 = 225. Note SD(X i = 3/4 1/4 = 3/4. So SD(S = 3 3/4 = 3/4 = 7.5. P (S 25 = P (S = P ( S P (χ = 1 Φ( (b Approximately how many times do you need to play so that you win at least $25 with probability at least.99? Solution. Let n be the number of times you play. So S = X X n and E[S] = 3n/4, SD(S = 3n/4. Then P (S 25 = P (S = P ( S 3n/4 3n/ n/4 3n/4 P (χ n/ n/4 = 1 Φ(. 3n/4 3n/4 So to solve this problem, find a number x so that 1 Φ(x =.99. Then solve for n in the equation n/4 3n/4 = x. 9 Normal approximations/ central limit theorem numbers are selected independently and uniformly from the interval [ 1/2, 1/2]. Use the normal approximation to approximate the probability that the average of these numbers lies in the interval (.1,.1. You can leave your answer in terms of the Φ- function. Solution. If S denotes the sum of the numbers then Var(S = 12/12 = 1. So the standard deviation of S is 1 = 1. ( P.1 S 12.1 = P ( 12 S1 12 P ( χ 12 = 2Φ( For a survey, we contact N people at random and ask them whether they plan to vote for candidate A. How large should N be so that there is at least a 99% chance that the sample mean determined by the survey is within 1% of the true mean? Use the normal approximation. Solution. If ˆp is percentage of people in our survey who plan to vote for candidate A and ˆσ is the standard deviation of ˆp then there is a 99% chance that p [ˆp 3ˆσ, ˆp + 3ˆσ].

12 Now So we want to choose N so that In other words, ˆσ = p(1 p N 1 2 N. N 3 2 N.1 ( 2 3 = Suppose we contact 1, people at random and ask them whether they plan to vote for candidate A. Suppose that 6% of them say yes. Give a 95% confidence interval for the true percentage of people that plan to vote for A. Solution. If ˆp is the sample mean then the standard deviation of ˆp is p(1 p p(1 p =. 1 1 We don t know what the true percentage p is. However, we can either substitute.6 for p above or use the fact that p(1 p 1/4 for any p. In the first case we end up with [ [.6 2 σ, σ] = , ] With the second method we end up with [.59,.61] for our 95% confidence interval. 38. A fair die is rolled 1 times. Use the central limit theorem to approximate the probability that the sum of the outcomes is more than 4. Leave your answer in terms of the Φ-function. You do not have to simplify. Hint: if X is the value of a single die roll then E[X] = 3.5 = 7 35 and Var(X = Solution. Let X i be the result of the i-th roll and S = X X 1. So E[S] = 1(3.5 = 35 and Var(S = 1 35 = 35. So S 35 P (S 4 = P (S 35 5 = P ( (35/ Φ(. (35/12 5 (35/ An astronomer is trying to estimate the distance from her observatory to a certain star. She makes 1 different measurements of this distance and takes the average of these 1 measurements to be her estimate. Suppose that the error in each measurement is a continuous random variable with mean and variance σ 2 = 4 light years. Suppose also that the measurements are independent. Use the Central Limit Theorem to estimate the probability that her estimate of the distance is less than the true distance by at least 1 light-year. Leave your answer in terms of the Φ-function. Hint: if X i is the error in the i-th measurement, then difference between her estimate and the true error is the mean Y = (X X 1 /1. The question asks P (Y 1 (which is the same thing as P (Y 1.

13 Solution. Let X i be the error in the i-th measurement. This is the measured distance minus the true distance. So E[X i ] = and V ar(x i = 4. Let S = X X 1. So E[S] =, V ar(s = 4 and SD(S = 2. This random variable is continuous, so there is no need to make a histogram correction. Her estimate of the distance is S/1 plus the true distance. So we are interested in P (S/1 1 (this is the probability that her estimate of the distance is less than the true distance by at least 1 light-year: P (S/1 1 = P (S/2 5 P (χ 5 = Φ( 5 = 1 Φ(5. 4. Suppose that 4% of all voters in this county are Republicans. If 1 are selected at random what is the probability that over 5 Republicans are selected? Use normal random variables to obtain an approximation. Leave your answers in terms of the Φ-function. Solution. The question asks P (S 1 > 5 where S 1 is a Binomial random variable with n = 1, p =.4. Using the histogram correction, this is the same as ( S P (S = P. 1(.4(.6 1(.4(.6 By the Central Limit Theorem, this is approximately equal to ( P χ 1.5 ( 1.5 = 1 Φ To get a rough estimate, note that 24 5, so this is about 1 Φ( A fair die is rolled 2 times. Use the central limit theorem to approximate the probability that the sum of the outcomes is between 65 and 75. Leave your answer in terms of the Φ-function. Hint: if X denote the value of a single die roll then E[X] = 7 and 2 V ar(x = Solution. Let X i be the result of the i-th roll and S = X X 2. So E[S] = = 7 and V ar(s = = 5 35 = 175. So ( P (65 S 75 = P (64.5 S 75.5 = P ( S 7 = P SD(S n 5.5 SD(S n SD(S n S 7 SD(S n P ( χ 5.5 SD(S n = 2P (χ 5.5/ 175/3 1 = 2Φ(5.5/ 175/ SD(S n To get a rough estimate, 175/3 58 and and 5.5/7.5 = 11/15 = 66/9.63.

14 Some helpful formulae (this will be on the exam; no need to memorize it Type of Random Variable Density or Mass Function Expected Value Variance Binomial(n, p P (X = k = ( n k p k (1 p n k ( k n np np(1 p λ λk Poisson(λ P (X = k = e (k N λ λ k! Geometric(p P (X = k = (1 p k 1 1 p p, (k N 1/p p 2 Exponential(λ f(x = λe λx 1 1, (x λ λ 2 Normal(µ, σ f(x = 1 /2σ 2 2πσ 2 e (x µ2 µ σ 2 Uniform over an interval [a, b] Var(X = E[(X µ 2 ] = E[X 2 ] E[X] 2. f(x = b a a+b (x [a, b] 2 2 Cov(X, Y = E[(X E[X](Y E[Y ]] = E[XY ] E[X]E[Y ]. (Markov s inequality P (X P (X t E[X]/t. (Chebyshev s inequality P ( X µ t Var(X t 2. (b a 2 12