University Of Chicago

• STAT 24400

• Stat 244 Autumn 2020 — HW2 Due on Canvas, Tuesday 10.13.20 by 5pm 1. Let X be the number of red cards, when two cards are drawn from a standard deck of 52 cards (as usual, cards are drawn at random and without replacement). Let Y be the number of Heads, when a fair coin is flipped twice. (a) Do X and Y have the same expected value? Justify your answer. Solution: The probabilities of having RR, RB, BR, BB when drawing two cards are 26 52 25 51 , 26 52 26 51 , 26 52 26 51 , 26 52 25 51 respectively. Therefore, E(X) = 2 · 26 52 25 51 + 1 · 26 52 26 51 + 1 · 26 52 26 51 + 0 = 1. (1) The probabilities of having HH, HT, T H, HH when flipping a coin twice are all 1 4 . Therefore E(Y ) = 2 · 1 4 + 1 · 1 4 + 1 · 1 4 + 0 = 1. (2) Note that a simpler way to calculate E(X) and E(Y ) is to write X = X1 + X2 as in part (c) below, then E(X) = E(X1)+E(X2) = 0.5+0.5 = 1, and we can do the same type of calculation for Y . (b) Do X and Y have the same distribution? Justify your answer. Solution: X and Y do not have the same distribution, since for example P(Y = 2) = 1 2 · 1 2 = 1 4 while P(X = 2) = 26 52 · 25 51 6= 1 4 . (c) We can write X = X1 + X2 and Y = Y1 +Y2 where X1 is the indicator variable for the event that the first card is red, X2 is the same for the second card, and similarly Y1 is the indicator variable for the event that the first coin is Heads, and Y2 is the same for the second coin. Do X1 and Y1 have the same distribution? Do X2 and Y2 have the same distribution? Solution: It is clear to see that Y1 and Y2 each have a Bernoulli(0.5) distribution—their only values are 0 and 1 and each has a 50% probability. The same is true for X1 and X2. Both are Bernoulli—0 and 1 are the only values. The first card clearly has a 50% chance of being red, so X1 is Bernoulli(0.5). While it’s a bit less intuitive, this is also true for the second card. The easiest argument is by symmetry—it’s not possible for a red second card to be more likely than a black second card, and vice versa. (We will often be tempted to think that the probability of X2 = 1 is not 1/2, since for example of the first card is red then X2 has probability 25 51 of equalling 1. However, when we think this way, we are actually computing P(X2 = 1|X1 = 1), a conditional probability, while the question asks for the distribution of X2, not for a conditional distribution—i.e., we are interested in computing P(X2 = 1), not P(X2 = 1|X1 = 1).) 2. A cell divides into two parts. Assume that the total volume of the cell is 1, and the volume of each part is distributed uniformly on (0, 1).