ESE 425, Spring 2018 Homework Set #1 (10 problems) Due Tuesday, Jan. 23 1. Short proofs. (a) Prove Theorem 4. That is, if both A and B are subsets of S, and A ⊂ B, then P B( )≥ P A( ). Hint: axioms 1 and 3 and set algebra. (b) Prove Theorem 5. That is, for any A ⊂ S, 0 ≤ P A( )≤1. Hint: axiom 2 and theorems 2 and 4. 2. The prevalence of a certain disease in the general population is one case in a population of 1,000. A test for the disease gives either a positive or negative result. • The test has a “success rate” of 99%, meaning of every 100 persons with the disease who are tested, the test will give a “positive” result 99 times on average. • The test has a “false positive rate” of 2%, meaning of every 100 persons without the disease who are tested, the test will give a “positive” result 2 times on average. (a) What is the probability that you have the disease, given you are tested and the test gives a “positive” result? (b) Repeat part (a), but for a “success rate” of 95% and a “false positive rate” of 0.1% (1 in 1,000). 3. The discrete random variables (rvs) X and Y may each take on integer values 1, 3 and 5. The joint probabilities are given below. Y X 1 3 5 1 1/18 1/18 1/18 3 1/18 1/18 1/6 5 1/18 1/6 1/3 (a) Find marginal probabilities P(X) and P(Y) for all possible values of X and Y. (Hint: complete the table.) (b) Are X and Y independent? (c) Find pY |XY = 5 | X = 3( ).ESE 425, Spring 2018 4. The discrete rvs X and Y may each take on only the integer values 0 and 1. The total probability P(X=0) = 0.75. The conditional probabilities of Y are given here: P Y = 0 | X = 0( )= 0.9P Y =1 | X = 0( )= 0.1P Y = 0 | X = 1( )= 0.2P Y =1 | X = 1( )= 0.8 (a) Fill in the numerical values of the joint and marginal probabilities of this table. Y X 0 1 P(X) 0 pXY(X=0,Y=0) pXY(X=0,Y=1) pX(X=0)=? 1 pXY(X=1,Y=0) pXY(X=1,Y=1) pX(X=1)=? P(Y) pY(Y=0)=? pY(Y=1)=? (b) Find the conditional probabilities P(X=0|Y=1) and P(X=0|Y=0). 5. A continuous random variable, X, is uniformly distributed from a to b (b > a). (a) Find its pdf. (b) Find its cdf. (c) Find the probability that X ≤2a + b3. (d) Let a = -1 and b = 2. Find the probabilities that |X| < c, for c = ½ and then for c = 1½. 6. A continuous random variable, x, has the following pdf., where θ is some given parameter (constant) such that 0 < θ < 1. fXx( )=θk0!"#$#0 ≤ x <θθ≤ x ≤1otherwise Find (as functions of θ): (a) k, (b) the cdf. 7. A moving particle’s velocity is uniformly distributed over [-1,1]. Let V denote the random variable that characterizes the particle’s velocity. Write and sketch both the cdf and pdf of V. 8. The kinetic energy, k, of a particle is related to its velocity, v, by the relationship k =12mv2. Suppose a particle has mass m = 2 and its velocity is characterized by the random variable V with cdf and pdf as found in problem 7. Let K denote the random variable characterizing the particle’s kinetic energy. Find and sketch the cdf and then the pdf of K.ESE 425, Spring 2018 9. Consider this joint pdf. fXYx, y()=e− x+ y( )0⎧⎨⎪⎩⎪x ≥ 0 and y ≥ 0otherwise (a) Find P(X < ½). (b) Find P[(X+Y) < 1)]. (c) Find P[(X or Y) > 1)]. (d) Find P[(X and Y) > 1)]. Hint: Do double integrals of the joint pdf over the appropriate region in x-y; sketch the region of integration for each. 10. Consider this joint pdf. ( )( )00000,1<<≥≥⎩⎨⎧=+−yorxyandxxeyxfyxXY (a) Find fX(x). (b) Find fY(y). (Hint: Laplace trick.) (c) Are X and Y independent? Why? (d) Find fX|Y(x|y). (e) Find fY|X(y|x). (f) What is the probability that x < 1 given y = 1? (Hint: Table of integrals.) (g) What is the probability that y < 1 given x = 1? (h) What is the probability that both x and y are < 1?ESE 425, Spring 2018 Solutions 1. Short proofs. (a) Prove Theorem 4. That is, if both A and B are subsets of S, and A ⊂ B, then P B( )≥ P A( ). Hint: axioms 1 and 3 and set algebra. (b) Prove Theorem 5. That is, for any A ⊂ S, 0 ≤ P A( )≤1. Hint: axiom 2 and theorems 2 and 4. (a) Express B as B = A ∪ A ∩ B()where A ∩ A ∩ B()=/0. Then by axiom 3, P B( )= P A( )+ P A ∩ B(). But by axiom 1, P A ∩ B()≥ 0. Therefore, P B( )≥ P A( ). (b) Since /0 ⊂ A ⊂ S, by Theorem 4, P/0( )≤ P A( )≤ P S( ). Then by Theorem 2 and axiom 2, 0 ≤ P A( )≤1. 2. The prevalence of a certain disease in the general population is one case in a population of 1,000. A test for the disease gives either a positive or negative result. • The test has a “success rate” of 99%, meaning of every 100 persons with the disease who are tested, the test will give a “positive” result 99 times on average. • The test has a “false positive rate” of 2%, meaning of every 100 persons without the disease who are tested, the test will give a “positive” result 2 times on average. (a) What is the probability that you have the disease, given you are tested and the test gives a “positive” result? (b) Repeat part (a), but for a “success rate” of 95% and a “false positive rate” of 0.1% (1 in 1,000). Use Bayes Rule. Let T denote the event of a positive test result and D having the disease. a( )P T | D( )=99100P T | D( )=2100P D( )=11000P D( )=1− P D( )=9991000P D | T( )=P T | D( )P D( )P T | D( )P D( )+ P T | D( )P D( )=9910011000( )9910011000( )+21009991000( )=9999 + 2 999( )≅ 0.0472 = 4.72% b( )P T | D( )=95100P T | D( )=11000P D | T( )=9510011000( )9510011000( )+110009991000( )=950950 + 999≅ 0.487 = 48.7%ESE 425, Spring 2018 3. The discrete random variables (rvs) X and Y may each take on integer values 1, 3 and 5. The joint probabilities are given below. Y X 1 3 5 1 1/18 1/18 1/18 3 1/18 1/18 1/6 5 1/18 1/6 1/3 (a) Find marginal probabilities P(X) and P(Y) for all possible values of X and Y. (Hint: complete the table.) (b) Are X and Y independent? (c) Find pY |XY = 5 | X = 3( ). Complete the table. Y X 1 3 5 P(X) 1 1/18 1/18 1/18 1/6 3 1/18 1/18 1/6 5/18 …