Page 1 of 9Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) Solutions to Quiz 1: Spring 2006 Problem 1: Each of the following statements is either True or False. There will be no partial credit given for the True False questions, thus any explanations will not be graded. Please clearly indicate True or False in your quiz booklet, ambiguous marks will receive zero credit. Consider a probabilistic model with a sample space Ω, a collection of events that are subsets of Ω, and a probability law P() defined on the collection of events—all exactly as usual. Let A, B and C be events. (a) If P(A) ≤ P(B), then A ⊆ B. True False False. As a counterexample, consider the sample space associated with a biased coin which comes up heads with probability 1 3 . Then, P(heads) ≤ P(tails), but the event heads is clearly not a subset of the event tails. Note that the converse statement is true, i.e. if A ⊆ B, then P(A) ≤ P(B). (b) Assuming P(B) > 0, P(A | B) is at leas t as large as P(A). True False False. As a counterexample, consider the case where A = BC and 0 < P(B) < 1. Clearly P(A B) = 0, but P(A) = 1 − P(B) > 0. So, in this case P(A B) < P(A).| | Now let X and Y be random variables defined on the same probability space Ω as above. (c) If E[X ] > E[Y ], then E[X2] ≥ E[Y 2]. True False False. Let X take on the values 0 and 1 with equal probability. Let Y take on the values −1000 and 1000 with equal probability. Then, E[X] = .5, and E[Y ] = 0, so E[X] > E[Y ]. However, E[X 2] = .5, while E[Y 2] = 1000000. Thus, E[X2] < E[Y 2] in this case. An additional c ounterexample using degenerate random variables is the trivial case of X = −1, Y = −2. (d) Suppose P(A) > 0. Then E[X] = E[X | A] + E[X AC ]. True False |False. This resembles the total expectation theorem, but the P(A) and P(AC ) terms are missing. As an explicit counterexample, say X is independent of A, and E[X] = 1. Then, the left-hand side is 1, while the right-hand side of the equation is 1 + 1 = 2. (e) If X and Y are independent and P(C) > 0, then pX,Y C (x, y) = pX C (x) pY C (y). True False | | |False. If X and Y are independent, we can conclude pX,Y (x, y) = pX (x) pY (y). The statement asks whether X and Y are conditionally independent given C. As we have seen in class, independence does NOT imply conditional independence. For example, let X take on the values 0 and 1 with equal probability. Let Y also take on the values 0 and 1 with equal probability, independent of X. Let C be the event X + Y = 1. Then, clearly X and Y are not independent conditioned on C. To see this, note that the joint PMF of X and Y conditioned 1on C puts probability 4 on each of the outcomes (0, 1) and (1, 0). Thus, conditioned on C, telling you the value of X determines the value of Y exactly. On the other hand, conditioned on C, if you don’t know the value of X, Y is still equally likely to be 0 or 1.Page 2 of 9Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) (f) If for some constant c we have P({X > c}) = 1 2 , then E[X] >c . True False 2 False. Let X take on the values −1 and 1 with equal probability. Then, P({X > 0}) = .5, and E[X] = 0 >�1 . Incidentally, if we res trict X to be nonnegative, the statement is true. 4 We will study this more carefully later in the term. The interested reader can look up the Markov inequality in section 7.1 of the textbook. In a simple game involving flips of a fair coin, you win a dollar every time you get a head. Suppose that the maximum number of flips is 10, however, the game terminates as soon as you get a tail. (g) The expected gain from this game is 1. True False False. Consider an alternative game where you continue flipping until you see a tail, i.e. the game does not terminate at a max of 10 flips. We’ll refer to this as the unlimited game. Let G be your gain, and X be the total number of flips (head and tails). Realize that X is a geometric random variable, and G = X− 1 is a shifted geometric random variable. Also E[G] = E[X] − 1 = 2 − 1 = $1, which is the value of the unlimited game. The truncated game effectively scales down all pay-offs of the unlimited game which are > $10. Thus the truncated game must have a lower expected value than than the unlimited game. The expected gain of the truncated game is 1023 1024 ≈ .99902 < 1. Let X be a uniformly distributed continuous random variable over some interval [a, b]. (h) We can uniquely describe fX (x) from its mean and variance. True False True. A uniformly distributed continuous random variable is completely specified by it’s (b−a)2 range, i.e. by a and b. We have E[X] = (a+b) , and var(X) = , thus given E[X] and2 12 var(X) one can solve for a and b. Let X be an exponentially distributed random variable with a probability density function fX (x) = e−x . (i) Then P ({0 ≤ X ≤ 3} ∪ {2 ≤ X ≤ 4}) = 1 − e−4 True False True. Note {0 ≤ X ≤ 3} ∪ {2 ≤ X ≤ 4} = {0 ≤ X ≤ 4}, so we have P ({0 ≤ X ≤ 4}) = FX (4) = 1 − e−4 Let X be a normal random variable with mean 1 and variance 4. Let Y be a normal random variable with mean 1 and variance 1. (j) P(X < 0) < P(Y < 0). True False False. Since X ∼ N(1, 4), X−1 ∼ N(0, 1). Similarly, Y − 1 ∼ N (0, 1). So, P(X < 0) = 2 Φ( 0−1 ), and P(Y < 0) = Φ(0 − 1). We know that any CDF is monotonically nondecreasing, 2 1so Φ(−2 ) ≥ Φ(−1). This shows that the statement in false. An alternative solution:Page 3 of 91 Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) False. Since X ∼ N(1, 4), X −1 ∼ N (0, 1). Similarly, Y − 1 ∼ N (0, 1). Let Z be a standard 2 normal random variable. Then, P(X < 0) = P(Z < 0−1 ), and P(Y < 0) = P(Z < 0 − 1).2 1Now, the event {Z < −1} is …
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