Questions and Answers #7 Random Variables

Questions and Answers Sheet 7 Random Variables Question #1 Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability p, then he or she will receive a score of 1 − (1 − 𝑝)2 if it does rain 1 − 𝑝2 if it does not rain We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability p*, what value of p should he or she assert so as to maximize the expected score? Answer: Define random variable X that marks the score of a certain meteorologist at some day. If the says that it will rain (he says it with probability p* ) and it really rains (with probability p), he gets score 1 − (1 − 𝑝)2 . Also, if he says that it will not rain and it really does not rain (with probability 1-p* ) and it really does not rain (with probability 1-p), he get score 1 − 𝑝2 . Hence 𝐸(𝑋) = 𝑝 ∗⋅ (1 − (1 − 𝑝)2 ) + (1 − 𝑝 ∗) ⋅ (1 − 𝑝2 ) = 2𝑝𝑝 ∗ −𝑝 ∗ −𝑝2 + 1 We now have to analyse the function 𝑝 ↦ 𝐸(𝑋) and find its maximum on (0,1). We have that 𝑑𝐸(𝑋) 𝑑𝑝 = 2𝑝 ∗ −2𝑝 = 0 𝑟𝑖𝑔ℎ𝑡 ⇒ 𝑝 = 𝑝 ∗ So we have that the meteorologist maximizes it chances when 𝑝 = 𝑝 ∗ , which is according to our intuition. Question #2 Give correct answer when have two random variables, X and Y. X is the value of a fair die, Y is the result of a coin flip, with heads being 1 and tails being 0. 𝑘 7 1 7 𝐸[𝑋] = ∑6𝑘=1 = , and 𝐸[𝑌] = . Thus 𝐸[𝑋]𝐸[𝑌] = . 6 2 2 4 That the expectation of XY is not multiplicative, i.e.: E[X]E[Y] is not necessarily equal to E[XY]. But I'm confused about what E[XY] means in the first place. That is, is E[XY] each possible value of the two events combined, multiplied by the probability that the two events occur? 1 𝑘 1 𝑘 7 That is, is 𝐸[𝑋𝑌] = (1) ∑6𝑘=1 + (0) ∑6𝑘=1 = ? If not, what is it? 2 6 2 6 4 1 𝑘 1 𝑘 2 6 2 6 Due to a typo the last equation [𝑋𝑌] = (1) ∑6𝑘=1 + (0) ∑6𝑘=1 7 7 was evaluated as , when I believe it 2 should be . I'd appreciate it if responders told me whether this is a correct value for E[XY]. 4 That concerned with how to calculate E[XY] in the quickest way possible, but how to interpret what E[XY] means. E[X]E[Y]=E[XY] for independent events doesn't concern me as much as why that is the case, and how to manually evaluate E[XY] in order to prove that indeed E[XY]=E[X]E[Y]. Answer: We should separate the meaning of E(XY) from whatever devices we use to compute the expectation. What does XY mean? It is a random variable. In our particular example, imagine the following game. We roll a fair die and a fair coin. If the die shows k and the coin shows Head, we get (k)(1) dollars. If the die shows k and the coin shows Tail, we get (k)(0) dollars, that is, 0. The random variable XY is the amount of money we get. It takes on values 0,1,2,3,…,6, with various probabilities, it has a certain distribution. Then E(XY) is the expected value (mean) of XY, in the usual sense. 1 1 2 12 In this case, 𝑃𝑟(𝑋𝑌 = 0) = , 𝑎𝑛𝑑 𝑃𝑟(𝑋𝑌 = 𝑛) = 1 1 2 12 have 𝐸(𝑋𝑌) = (0) + (1) + 1 12 (2) + ⋯ + 1 12 𝑓𝑜𝑟 𝑛 = 1,2, … ,6. Now we can compute E(XY). We (6). But in this case, there is a shortcut. Since X and Y are independent, we have 𝐸(𝑋𝑌) = 𝐸(𝑋)𝐸(𝑌), and therefore there is no need to find the distribution of XY. Question #3 The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter 𝜆 = 1 . What is the probability that a repair time exceeds 2 hours? Answer: Let X be exponential random variable that represents time required to repair a machine, 𝑋 ∼ 𝑒𝑥𝑝(1). We use what we know about exponential distribution. 𝑃(𝑋 > 𝑥) = 1 − 𝐹(𝑥) = 1 − (1 − 𝑒 −𝜆𝑥 ) = 𝑒 −𝜆𝑥 and for conditional probability: 𝑃(𝑋 > 𝑡 + 𝑠|𝑋 > 𝑡) = 𝑃(𝑥 > 𝑠) 𝑃(𝑋 > 2) = 𝑒 −2 = 0.135 Question #4 The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter 𝜆 = 1 . What is the conditional probability that a repair takes at least 3 hours, given that its duration exceeds 2 hours? Answer: Let X be exponential random variable that represents time required to repair a machine, 𝑋 ∼ 𝑒𝑥𝑝(1). We use what we know about exponential distribution. 𝑃(𝑋 > 𝑥) = 1 − 𝐹(𝑥) = 1 − (1 − 𝑒 −𝜆𝑥 ) = 𝑒 −𝜆𝑥 and for conditional probability: 𝑃(𝑋 > 𝑡 + 𝑠|𝑋 > 𝑡) = 𝑃(𝑥 > 𝑠) 𝑃(𝑋 > 3|𝑋 > 2) = 𝑃(𝑋 > 1 + 2|𝑋 > 2) = 𝑃(𝑋 > 1) = 𝑒 −1∗1 = 𝑒 −1 = 0.368 Question #5 Let X and Y be continuous random variables with joint density function x  + cy 0  x  1,1  y  5 f ( x, y ) =  5 0 otherwise where c is a constant. What is the value of c? Answer: Since it has to be ∫ ∫𝑅𝟚 𝑓 = 1, we have that 1 5 1 𝑥 4𝑥 2 1 = ∫ ∫ ( + 𝑐𝑦𝑑𝑦) 𝑑𝑥 = ∫ ( + 12𝑐) 𝑑𝑥 = + 12𝑐 5 5 0 1 5 0 which yield that 𝑐 = Result: 1 𝑐= 20 1 20 . Question #6 Let X and Y be continuous random variables with joint density function x  + cy 0  x  1,1  y  5 f ( x, y ) =  5 0 otherwise where c is a constant. Are X and Y independent? Answer: Since the joint PDF of X and Y can not be written as the product of some two PDFs, we can conclude that X and Y are not independent. Result: No. Question #7 Let X and Y be continuous random variables with joint density function x  + cy 0  x  1,1  y  5 f ( x, y ) =  5 0 otherwise where c is a constant. Find P(X+Y>3). Answer: We have that 𝑃(𝑋 + 𝑌 > 3) = ∫ ∫𝑥+𝑦>3 𝑓 𝑑𝑥 𝑑𝑦 1 5 𝑥 𝑦 = ∫0 ∫3−𝑥 ( + 𝑑𝑦) 𝑑𝑥 5 20 1 (𝑥+2)(7𝑥+8) = ∫0 = 11 40 𝑑𝑥 15 Result: 11 15 Question #8 Find the following probabilities for a standard normal random variable z : 𝑃(−.42 < 𝑧 < 1.96) Answer: Definitions Complement rule: 𝑃(𝐴𝐶 ) = 𝑃(𝑛𝑜𝑡 𝐴) = 1 − 𝑃(𝐴) Addition rule for disjoint or mutually exclusive events: 𝑃(𝐴 ∪ 𝐵) = 𝑃(A or B) = 𝑃(𝐴) + 𝑃(𝐵) Solution The probability 𝑃(0 ≤ 𝑧 ≤ 1.96) is given in the row with z=1.9 and in the column with .06 of the normal probability table in the appendix: 𝑃(0 ≤ 𝑧 ≤ 1.96) = 0.4750 The probability 𝑃(0 ≤ 𝑧 ≤ 0.42) is given in the row with z=0.4 and in the column with .02 of the normal probability table in the appendix. The standard normal distribution is symmetric about 0: 𝑃(−0.42 ≤ 𝑧 ≤ 0) = 𝑃(0 ≤ 𝑧 ≤ 0.42) = 0.1628 Add the corresponding probabilities: 𝑃(−0.42 ≤ 𝑧 ≤ 1.96) = 𝑃(0 ≤ 𝑧 ≤ 1.96) + 𝑃(−0.42 ≤ 𝑧 ≤ 0) = 0.4750 + 0.1628 = 0.6378 Question #9 The random variable x is normally distributed with mean 𝜇 = 174 and standard deviation 𝜎 = 20 Find the indicated probability. 𝑃(𝑥 > 182) Answer: Given: 𝜇 = 174 𝜎 = 20 We need to determine P(x>182). x=182 The z-score is the value x decreased by the mean and then divided by the standard deviation. 𝑥−𝜇 182−174 8 2 𝑧= = = = = 0.40 𝜎 20 20 5 The probability to the left of z=0.40 is then given in the row with 0.4 and in the column with 0.00 of the standard normal table in the appendix. 𝑃(𝑥 < 182) = 𝑃(𝑧 < 0.40) = 0.6554 Use the Complement rule: 𝑃(𝐴𝑐 ) = 𝑃(𝑛𝑜𝑡 𝐴) = 1 − 𝑃(𝐴) 𝑃(𝑥 > 182) = 1 − 𝑃(𝑥 < 182) = 1 − 0.6554 = 0.3446 Result: 0.3446 Question #10 A population of values has a normal distribution with 𝜇 = 67.9 and 𝜎 = 44.8 . You intend to draw a random sample of size n=17. Find the probability that a single randomly selected value is greater than 94 Answer: Step 1 Introduction: Define X as the random variable of interest here. It is given that X has a normal distribution with population mean 𝜇 = 67.9 , and population standard deviation 𝜎 = 44.8 . A sample of size n = 17 is selected from the population. Step 2 Calculation: The probability that a single randomly selected value is greater than 94 is calculated below: 𝑃(𝑋 > 94) = 1 − (𝑋 ≤ 94) = 1−𝑃( 𝑋−𝜇 𝜎 ≤ 94−𝜇 𝜎 ) [By standardizing] = 1 − 𝑃 (𝑍 ≤ [𝑍= 𝑋−𝜇 𝜎 94−67.9 44.8 ) is the standard normal variable.] ≈ 1 − 𝑃(𝑍 ≤ 0.58) [The z-score is rounded to two decimal places.] = 1 − 0.7190 [Using the standard normal distribution table] = 0.2810 Thus, the probability that a single randomly selected value is greater than 94 is 0.2810.