Probabilistic Methods of Signal and System Analysis Solutions
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The document discusses two probability problems involving apple picking by Bob, who picks apples from an orchard with gala and honey crisp varieties. It calculates probabilities and distributions related to the number of apples picked, their ripeness, and the independence of random variables representing apples from different trees. The document includes mathematical formulations for probability mass functions, expectations, variances, and joint distributions of related variables.
Probabilistic Methods of Signal and System Analysis Solutions
- 1. PROBABILISTIC METHODS OF SIGNAL AND SYSTEM ANALYSIS SOLUTIONS https://www.liveexamhelper.com/ support@liveexamhelper.com +1 (315) 557-6473 4112 Thrash Trail, Purdon, Texas- 76679
- 2. Problem 1: Bob goes apple picking. The orchard he goes to grows two varieties of apples: gala and honey crisp. The proportion of the gala apples in the orchard is p (0 <p< 1), the proportion of the honey crisp apples is 1 - p. The number of apples in the orchard is so large that you can assume that picking a few apples does not change the proportion of the two varieties. Independent of all other apples, the probability that a randomly picked gala apple is ripe is g and the probability that a randomly picked honey crisp apple is ripe is h. (a) Let n = 20, and suppose that Bob picked exactly 10 gala apples. What is the probability that the first 10 apples that Bob picked were all gala? Solution: Let A be the event that the first 10 apples picked were all gala and let B be the event that exactly 10 gala apples were picked out of the 20 apples. https://www.liveexamhelper.com/
- 3. The probability of the event B can be computed by using the result in (a), where n =20 and k = 10. P(B)= 20 p10(1− p)10. P(A ∩ B) can be computed by seperating the 10 event of A and B into disjoint events. The event {A ∩B}= {1st 10 apples picked are gala and the last 10 apples are honey crisp}. P(A ∩B)= p10(1−p)10. Therefore, https://www.liveexamhelper.com/
- 4. Problem 2. Next, Bob tries a different strategy. He starts with a tree of the gala variety and picks apples at random from that tree. Once Bob picks an apple off the tree, he carefully examines it to make sure it is ripe. Once he comes across an apple that is not ripe, he moves to another gala tree. He does this until he encounters an unripe apple on that second tree. Assume that each tree has a very large, essentially infinite, number of apples. (a) Let Xi be the number of apples Bob picks off the ith tree, (i =1, 2). Write down the PMF, expectation, and variance of Xi. Solution: Bob picks apples from each tree until he finds one that is not ripe, and so if we think of the event “picking an unripe apple” as a “success”, then Xi is the number of apples picked(i.e. trials) until we get our first success. Therefore Xi is a geometrically distributed random variable with probability of success 1 −g. Since each tree has a random collection of apples, the apples Bob picks from the first tree are independent of the apples he picks from the second tree, and therefore X1 and X2 are independent and identically distributed random variables. The PMF of Xi is https://www.liveexamhelper.com/
- 5. the expectation of Xi is and the variance of Xi is (b) For i =1, 2, let Yi be the total number of ripe apples Bob picked from the first i trees. Find the expectation and the variance of Y2. (Note that Y1 = X1 − 1 and Y2 =(X1 − 1) + (X2 − 1).) Solution: Given that Y2 =(X1 −1)+(X2 −1), we can use the linearity property of expectations to find the expectation of Y2. https://www.liveexamhelper.com/
- 6. Since adding or subtracting a constant from a random variable has no effect on its variance, and since X1 and X2 are independent, we have (c) Find the joint PMF of Y1 and Y2. Solution: Let k ≥ 0 and ℓ ≥ k. The event “Y1 = k and Y2 = ℓ” is identical to the event “X1 = k + 1 and X2 = ℓ − k + 1”. Since X1 and X2 are independent, the desired probability is (1 − g)gk+1−1(1− g)gℓ−k+1−1 = (1− g)2gℓ, when 0 ≤ k ≤ ℓ, and zero otherwise. https://www.liveexamhelper.com/
- 7. Note that we can interpret this result as the probability of ℓ failures (i.e. picking ℓ ripe apples) and two successes (i.e. the two unripe apples that cause Bob to stop at each tree). Note also that although the expression for the joint PMF doesn’t depend on k, the sample space does depend on k because ℓ ≥ k: Bob cannot pick more ripe apples in the first tree than he does in the first two trees combined. (d) In the following, answer just “yes” or “no.” (Explanations will not be taken into account in grading.) (i) Are X1 and Y2 independent? (ii) Are X2 and Y1 independent? Solution: (i) No. X1 and Y2 are not independent because Y2 ≥ X1−1 and so knowing X1 gives us information about Y2. Specifically, if we know that Bob picked 10 total apples from the first tree (i.e. X1 = 10), then we also know that Bob must have picked atleast9 ripe apples from the first two trees combined (i.e. Y2 ≥ 9). (ii) Yes. X2 relates only to the second tree and Y1 relates only to the first tree, and since the picking of apples is independent across trees, the two random variables are independent. https://www.liveexamhelper.com/