(4) Suppose X is a random variable such that E(X) = 5 and Var(X) = 16. Find

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1 Chapter 5 Exercises A random variable X has the distribution: X 6 PX Find the expected value, variance, and standard deviation of X. A number Y is chosen at random from the set S = {,, }. Find the expected value, variance, and standard deviation of Y. A hotel has two conference rooms and on any given day, the number of conference rooms that will be rented out for events is a random variable X. Let X have the following distribution X PX 7 8 a Find EX and VarX. b Suppose each conference room rents for $ a day. Let Y stands for the rents collected on any given day. Write down the probability distribution of Y. c Find EY and VarY. Suppose X is a random variable such that EX = 5 and VarX = 6. Find a EX b EX c E X d Var X 5 Min works for a computer company that sells very expensive computers. Let X, the number of computers she sells in a month have the following distribution: X PX a Find EX and VarX. b Every month, Min earns a commission that is equal to $X for X computers sold. Let Y be Min s monthly commission. Write down the probability distribution of Y. c Find EY and VarY. d In addition to her commission, Min receives a fixed salary of $ per month. Let Z denote Min s month income fixed salary + commission. Compare VarZ to VarY and comment. e For this month, Min has already sold one computer. Let W be the total number of computers sold this month. Find the distribution of W and use this distribution to find the expected commission this month. The expected commission is called a conditional expectation. 6 Let Y be the sum of the numbers on the roll of two fair dice. a Find the probability function of Y.

2 b Find EY and VarY. 7 Continued from Chapter, Question. Recall the joint probability function of X and Y : Y P X, Y X Find a EX and VarX. b EY and VarY. c Based on your answer in Chapter, Question d, calculate EX+Y and VarX+Y. Compare VarX + Y to VarX + VarY and comment. d If the daily rental rate of a passenger limousine is $ and a 5 passenger limousine is $6, find the expected value of the revenue, U and its variance. e Find the conditional expectation EY X = and use the answer to find the new expected revenue. 8 Suppose X is a random variable. For each of the following probability density function on which X is defined, find the expected value and variance of X. {, a fx =, < x < otherwise. { x, b fx =, < x < otherwise. { x, c fx =, < x < otherwise. { d fx = x, < x <, otherwise. e fx = { e x., < x <, otherwise. 9 Fiji is an archipelago of about islands that lies in the South Pacific. The islands are renowned for their natural beauty and every year, tourists from all over the world visit the islands for its many coral reefs and atolls. Unfortunately, being in the tropics, Fiji is subject to the mercy of tropical storms. An insurance company insures against losses from two types of storms: tropical cyclones and severe tropical cyclones. Let X represents the deviation of the medium losses from a tropical cyclone and let Y be similarly defined for a severe tropical cyclone. Suppose it is known that VarX =.7 and VarY =.5 both in million Fijian dollars. The probability density function of X, Y are given in the attached plot. Which of the two densities belong to X Y? justify your answer.

3 PDF An economist on vacation on the islands wants to study the impact of severe cyclones on the wealth of the population. From his training, he is aware of the famous Pareto distribution for modelling wealth distribution. Using the distribution, wealth X in a population has PDF: { x α α m fx = x, x > x α+ m, otherwise where x m is the minimum wealth in the population and α > is a constant. He knows x m = in Fijian dollars and α = would be appropriate and for that value of α EX = αx m α, VarX = αx m α α. In any given year, the probabilities of, and severe cyclones are.5,. and., respectively. He estimates that the impact of Y cyclones is. expy of the wealth of each individual, assuming for simplicity that everyone in the population is affected equally. Find the mean reduction in wealth of the population due to severe cyclones. Hint: assume Y and any transformation of Y and X are independent. A major industry in Fiji is sugar production. In a particular sugarcane plantation, each worker is required to fill two tractor loads of sugarcane every morning. The times in hours X and Y, to fill up the first and second tractor, respectively, have a bivariate Weibull distribution with joint density function: fx, y = κ κ α x κ α y κ α s α α + αs α e sα, s = x κ κ α + y α where κ =, κ =, α =.5 The corresponding joint CDF F x, y of X, Y is F x, y = e xκ e yκ + e sα. To encourage the workers, a bonus of Fijian dollars is given for each tractor that is filled within half an hour. a Find PX /, Y /, PX /, Y > /, PX > /, Y / and hence PX > /, Y > /. Please round off to the rd decimal place and make sure probabilities add to. After Wilfried Fritz Pareto, 88-9

4 b Let W and W be the bonus received on the first and second tractor, respectively. What are the possible values of W and W? c Using your results in a and b, create a table of the joint distribution of W and W. d Are W and W independent? e Find EW, EW, VarW and VarW. f Find CovW, W using both formulae we learned in class. Which formula is easier to use? g For a particular worker, let W be the total bonus received in a morning, find EW and VarW.

5 Answers: EX = + 6 VarX =.975 = SDX = VarX = = Based on the question, the distribution of X is X - PX EX = VarX = = 9 + SDX = VarX = =. + a. EX = VarX = =. 7 = = 5 56 = b. X Y = X = = = PY = PX 7 8

6 c. EY = [ = = EX =. VarY = = [ = VarX = ] ] 8 8 a.varx = EX EX EX = VarX + EX = 6 + =. b. EX = EX = =. c. E X = EX = 5. d. Var X = E X E X = E X [ EX] = EX EX = VarX = 6 5 a. EX = =.5. VarX = =.75. b. X Y = X = = = = 9 PY = PX c. EY = =. VarY = = 7.5. d. There are two ways to solve this problem. First way, we can use the identity that VarX + a = VarX for any constant a, therefore, VarZ = VarY + = VarY. Second way, we can work from the distribution of Z directly. distribution of Z: We can easily write down the

7 Y = 9 Z = Y PZ = PY EZ = = EY +. VarZ = {[ + EY + ]}. + {[ + EY + ]}.75 +{[ + EY + ]}.75 + {[9 + EY + ]}. = [ EY ]. + [ EY ].75 + [ EY ].75 + [9 EY ]. = VarY. Since is a constant amount, therefore, it is a part of Z that has no variability and hence, the only variable component in Z comes from Y, so VarZ = VarY. e. Notice the distribution of W is actually the conditional distribution of X given X, hence the possible values of W are,,, and its distribution can be obtained as follows: PW = = PX = X = = 7 Similarly, we can obtain PW =, PW =, and the results can be summarized as follows: W PW The expected value of W can then be obtained easily: EW = The expected commission is EW = EW = 7 = This problem can be solved by first forming a table of outcomes of rolling two dice. The numbers within the table are the possible values of Y, the sum of the numbers on the two dice. Die Die a. Based on this table, we can deduce the probability distribution of Y by finding the proportions of,,...,,, in the table, which gives: Y PY 5 Note that Y = is not possible. 6 5

8 b. EY = + VarY = = = = We first need to work out the marginal probability distributions of X and Y. They can be easily derived from the table of joint probabilities: a. Y P X, Y PX X PY EX = =.. VarX = =.9. b. EY = =.65. VarY = =.775. c. EX + Y = EX + EY = =.75. Since X and Y are not independent from Question d in Chapter, We need to work out the last term: VarX + Y = VarX + VarY + CovX, Y. CovX, Y = {a EX}{b EY }PX = a, Y = b X=a Y =b = abpx = a, Y = b EXEY. X=a Y =b We will use the second expression because it is easier to evaluate. CovX, Y = abpx = a, Y = b EXEY X=a Y =b = =.9.85 =.85. Therefore, VarX + Y = =.875. In this problem, VarX + Y > VarX + VarY because CovX, Y >. Since CovX, Y > is an indication that X and Y are positively related, or in other words, when the value of X is

9 high, the value of Y will likely be high; the consequence is that the sum of X + Y having a higher variation that the sum of VarX and VarY. This result generalizes so that whenever X and Y are positively related, the induced variance of X + Y is higher than the sum of the variances of X and Y. On the other hand, if X and Y are negatively related, the induced variance of X + Y would be lower than the sum of the variances of X and Y. If X and Y are independent, then VarX + Y = VarX + VarY. d. EU = EX + 6Y = EX + 6EY = =. VarU = VarX + 6Y = VarX + 6 VarY + 6CovX, Y = = 8. e. The conditional distribution of Y given X = can be obtained by calculating the conditional probability PY X = : etc., giving the following: Y PY X = 6 PY = X = = PY = X = = =, =.5, Notice that we do not need to list Y = since it has probability zero given X =. EY X = = + + = 6 6 Expected revenue is: 8 a. EX + 6Y X = = EX X = + E6Y X = = + 6 EX = EX = [ ] x xdx = = [ ] x x dx = = VarX = EX EX = = = 8. 6 b. EX = x dx = ] [ x = [ ] x EX = x dx = = VarX = EX EX = 9 = 8

10 c. EX = EX = x xdx = x xdx = x x dx = x x dx = VarX = EX EX = 6 9 = 8. [ { x [ }] x = { }] x x = 6 d. EX = EX = [ ] x dx = 8 x x dx = [ x5 ] = = 5 VarX = EX EX = 5 = 5 e. EX = EX = xe x dx = [ xe x] + e x dx ] [ e x = = x e x dx = [ x e x] + xe x dx = xe x dx = [ xe x] + e x dx = ] [ e x VarX = EX EX = 9 9 = 9 9 = 9 9 Since Y has a much larger variance than X, therefore, the spread of the data of Y is much bigger than that of X. In other words, there are more chance of extreme values for Y than for X. A probability density function PDF gives the probability distribution of a random variable; a PDF that is wide and flat characterizes a situation where there is significant probability for extreme values. Hence, the dashed blue curve should correspond to the PDF of Y.

11 PDF X Y Let Y denote the number of severe cyclones; furthermore consider the impact of the cyclones on wealth be gy =.e Y. Then the PDF of Y and gy is given in the following table. Then Y gy.e.e.e PY.5.. E[gY ] =.5. e +.. e +.. e =.68. Furthermore, assuming X and Y are independent, then X and gy are independent, so that E[gY X] = E[gY ]EX =.68.5 thousand dollars. Since κ =, κ =, α =.5, and F x, y = e xκ e yκ + e sα = e x e y + e s.5 We first evaluate s when x = /, y = /, Similarly, when x = /, y =, we have s = x κ α + y κ α = x.5 + y.5 = x + y 8. s = / + / 8 = 7/6, e s.5 = s = / + 8 =, e s.5 = and using the idea, for x =, y = / or x =, y = e s.5 =. a. We can draw a figure to aid us in determining the regions in which we would like to evaluate the probabilities.

12 In this problem, we need to use the partition rule repeatedly. PX /, Y / = F /, / = e / e / + e s.5 = PX /, Y > / = PX / PX /, Y / = PX /, Y PX /, Y / = F /, F /, / = e / e = PX > /, Y / = PY / PX /, Y / = PX, Y / PX /, Y / = F, / F /, / = e e / = PX > /, Y > / = [PX /, Y / + PX /, Y > / + PX > /, Y /] =.77 b. The possible values of W and W are and, corresponding to no bonus or dollars of bonus. c. The possible combinations of W, W are,,,,, and, and they correspond to, respectively, the situations that X /, Y /, X /, Y > /, X > /, Y / and X > /, Y > /. Hence the joint distribution of W and W is

13 W P W, W W d. Since PW =, W = =.55 PW = PW = =..6 =.5, W and W are not independent. e. EW = PW = + PW = = =. EW = PW = + PW = = =.6 VarW = EW [EW ] = [ PW = + PW = ]. = 7.76 VarW = EW [EW ] = [ PW = + PW = ].6 = f. First way: CovW, W = E{[W EW ][W EW ]} = =.58 Second way: CovW, W = EW W EW EW = =.58 The second way is easier to use.

14 g. Since W is the total bonus received in a morning, hence, W = W + W. EW = EW + W = EW + EW =. +.6 =.8 VarW = VarW + W = VarW + VarW + CovW, W = =.9