math 302 homework 4 solutions

Math 302, assignment 4 solutions 1. a. In a game, a player draws three cards from a deck. He wins $1 for each heart card chosen. What is the expected amount won? b. If additionally the player gets $2 for each queen (so the queen of hearts is worth $3), what is the expected amount won? solution. a. The expected amount won from each card is $1/4. By additivity of expectation, the total from three cards is $3/4. b. Each card is a queen with probability 1/13, so this adds $2/13 to the expectation from each card, which is now 1/4+2/13 = 21/52. The total from three cards is now $63/52. 2. For a geometric random variable X, show that P(X>n + m|X>n)= P(X>m). (We say that the geometric random variable has no memory.) 3. For a geometric random variable X, show that P(X>n + m|X>n)= P(X>m). (We say that the geometric random variable has no memory.) solution. We have P(X>n)= ∞ X k=n+1 q(1 - q) k-1 = (1 - q) n . This can also be seen since this is the probability that an experiment fails n times. Since the intersection of {X>n + m} and {X>n} is {X>n + m}, we get P(X>n + m|X>n)= P(X>n + m and X>n) P(X>n) = (1 - q) n+m (1 - q) n = (1 - q) m = P(X>m). 4. a. Prove that if EX = 0 then f (t)= E[(X - t) 2 ] is minimized at t = 0. b. Prove that if EX = μ then f (t)= E[(X - t) 2 ] is minimized at t = μ. solution. Part a is a special case of b when μ = 0. b. By linearity of expectation, f (t)= E[(X - t) 2 ]= E(X 2 - 2tX + t 2 )= t 2 - 2tEX + E(X 2 ). Taking derivatives we get f 0 (t)=2t - 2EX, which is 0 when t = E(X)= μ. Since the quadratic tends to +∞ as t → ±∞, this is a minimum.

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UBC, MATH 302, 2014/2015

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Math 302, assignment 4 solutions

1. a. In a game, a player draws three cards from a deck. He wins $1 for each heart card chosen. What isthe expected amount won?b. If additionally the player gets $2 for each queen (so the queen of hearts is worth $3), what is theexpected amount won?

solution. a. The expected amount won from each card is $1/4. By additivity of expectation, thetotal from three cards is $3/4.b. Each card is a queen with probability 1/13, so this adds $2/13 to the expectation from each card,which is now 1/4 + 2/13 = 21/52. The total from three cards is now $63/52.

2. For a geometric random variable X, show that P(X > n+m|X > n) = P(X > m). (We say that thegeometric random variable has no memory.)

3. For a geometric random variable X, show that P(X > n+m|X > n) = P(X > m). (We say that thegeometric random variable has no memory.)

solution. We have

P(X > n) =

k=n+1

q(1 q)k1 = (1 q)n.

This can also be seen since this is the probability that an experiment fails n times.

Since the intersection of {X > n+m} and {X > n} is {X > n+m}, we get

P(X > n+m|X > n) = P(X > n+m and X > n)P(X > n)

=(1 q)n+m(1 q)n = (1 q)

m = P(X > m).

4. a. Prove that if EX = 0 then f(t) = E[(X t)2] is minimized at t = 0.b. Prove that if EX = then f(t) = E[(X t)2] is minimized at t = .

solution. Part a is a special case of b when = 0.b. By linearity of expectation, f(t) = E[(X t)2] = E(X2 2tX + t2) = t2 2tEX +E(X2). Takingderivatives we get f (t) = 2t 2EX, which is 0 when t = E(X) = . Since the quadratic tends to +as t , this is a minimum.