Hw3 solutions - hw 3 solu PDF

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Fall 2018 S&DS 241: Probability Theory with Applications Homework 3 Due: Sep 30, 2018 in class Prof. Yihong WuSolutions by Ganlin. We haveP(X= 1) = 1−p 1 ,P(X=j) =p 1 p 2 ..− 1 (1−pj) for 2≤j≤6, andP(X= 7) = p 1 p 2 .. 6. LetXbe the value of your most valuable prize. The support is 5,6,...,1...

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Fall 2018 S&DS 241: Probability Theory with Applications Homework 3 Due: Sep 30, 2018 in class Prof. Yihong Wu Solutions by Ganlin. 1. We have P (X = 1) = 1 − p1 , P (X = j) = p1 p2 ...pj−1 (1 − pj ) for 2 ≤ j ≤ 6, and P (X = 7) = p1 p2 ...p6 . 2. Let X be the value of your most valuable prize. The support is 5,6,...,100 since the worst case is getting the $1, . . . , $5 prizes. The event X = k says that you got the $k prize and 4 less valuable prizes. So the PMF of X is k−1  4  P (X = k) = 100 5

for k = 5, 6, ..., 100.

3. (a) Since each pair of people are friends with probability p independently, by linearity of expectation,   n(n − 1) n p= p. E(X) = 2 2 (b) Since each pair of people are friends with probability p independently, probability of each triple who are all friends is p3 . Then by linearity of expectation,   n 3 n(n − 1)(n − 2) 3 p = p . E(X) = 3 6 4. (a) If p = 1, then α = 0; If p = 0, then α = 1. (b) α = P {eventually fall down the cliff} By law of total probability, P {eventually fall down the cliff} = P {eventually fall down the cliff, walk to the left} + P {eventually fall down the cliff, walk to the right} = P {eventually fall | walk to the left}P {walk to the left} + P {eventually fall | walk to the right}P {walk to the right}. Since this process starts at 1, P {eventually fall | walk to the left} = P {eventually return to 0 | starting at 0} = 1. And, P {eventually fall | walk to the right} = P {eventually return to 0 | starting at 2} = P {eventually return to 1 | starting at 2} × P {eventually return to 0 | starting at 1}, Since the drunkard can only move to the neighboring states within one operation. Also since the state space is infinite, P {eventually return to 1 | starting at 2} = P {eventually return to 0 | starting at 1} = α, 1

which is equivalent to ‘falling down the cliff twice’. Then since P {walk to the left} = 1 − p and P {walk to the right} = p, we can obtain the following equation α = pα2 + 1 − p. (c) By solving this equation, we can get α = 1 or

1−p . p

Since α ∈ [0, 1], then we let α = 1 when p ∈ [0, 1/2), and α = (1 − p)/p when p ∈ [1/2, 1]. The surprising point here is that when p = 1/2, the drunkard still has a probability of 1 falling down the cliff. 5. P {X = 0} = P {of players 1,2: 2 has the higher number} =

P {X = 1} = P {of players 1,2,3: 3 has largest, 1 the next largest} = P {X = 2} = P {of 1,2,3,4: 4 has largest, 1 the next largest} =

1 1 1 · = 3 2 6

1 1 1 · = 12 4 3

P {X = 3} = P {of 1,2,3,4,5: 5 has largest, 1 the next largest} = P {X = 4} = P {of 1,2,3,4,5: 1 has largest} =

1 2

1 1 1 · = 5 4 20

1 5

Hence, PMF of X is as follows: x pX (x)

0 1 2

1 1 6

2 1 12

3 1 20

4 1 5

6. (a) The random variable X can assume the values +1, 0, and −1. P (X = +1) = P ((X, Y ) = (+1, 0)) = 1/4. P (X = −1) = P ((X, Y ) = (−1, 0)) = 1/4. P (X = 0) = P ((X, Y ) = (0, −1)) + P ((X, Y ) = (0, +1)) = 1/4 + 1/4 = 1/2. (b) By definition, E(X) = (+1) · P (X = +1) + 0 · P (X = 0) + (−1) · P (X = −1) = 0. By linearity of expectation and symmetry, E (X + Y ) = E(X) + E (Y ) =0+0 = 0. 2

Note that (X + Y )2 = 1 and hence 1 = E(X + Y )2 = E(X 2 ) + E(Y 2 ) + 2E(XY ) = 1/2 + 1/2 + 2E(XY ) = 1 + 2E(XY ). Thus E(XY ) = 0. (c) No. P (X = 0) = P (Y = 0) = 1/2 but P (X = 0, Y = 0) = 0. Remark: This is an example where independence is a sufficient for E(XY ) = E(X)E(Y ) but need not be necessary. (d) Yes. Since P (U = 1, V = 1) = 1/4 and P (U = 1) = P (V = 1) = 1/2, we have P (U = 1, V = 1) = P (U = 1)P (V = 1). Similar for other three points. So (U, V ) are independent.

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