For each of the following pairs of random variables X and Y , the conditional distribution of Ygiven {X = x} is a named distribution, with parameters that may depend on x. For each pair:1i. Write down the conditional distribution of Y given {X = x}, including parameter(s).(2 marks each)ii. Find E (Y | X). (1 mark each)iii. Find E(Y ). (1 mark each)iv. Find Cov(X, Y ). (2 marks each)You do not need to find the marginal distributions of X and Y , which might not be nameddistributions.(a) Choose a number between 10 and 29, uniformly at random, and call that number X.Take a standard deck of 52 cards and lay them out in a row, face up. Using a pen, puta checkmark on the first X cards. Then turn the cards face down, shuffle throughly, anddraw 5 cards. Let Y be the number of marked cards you draw.
Question
For each of the following pairs of random variables X and Y , the conditional distribution of Ygiven {X = x} is a named distribution, with parameters that may depend on x. For each pair:1i. Write down the conditional distribution of Y given {X = x}, including parameter(s).(2 marks each)ii. Find E (Y | X). (1 mark each)iii. Find E(Y ). (1 mark each)iv. Find Cov(X, Y ). (2 marks each)You do not need to find the marginal distributions of X and Y , which might not be nameddistributions.(a) Choose a number between 10 and 29, uniformly at random, and call that number X.Take a standard deck of 52 cards and lay them out in a row, face up. Using a pen, puta checkmark on the first X cards. Then turn the cards face down, shuffle throughly, anddraw 5 cards. Let Y be the number of marked cards you draw.
Solution
The problem involves two random variables X and Y. X is a uniformly distributed random variable between 10 and 29. Y is the number of marked cards drawn from a deck of 52 cards, where the first X cards are marked.
i. The conditional distribution of Y given {X = x} is a hypergeometric distribution. The parameters of this distribution are N = 52 (the size of the population), K = x (the number of success states in the population), and n = 5 (the number of draws). So, Y | X = x ~ HGeom(52, x, 5).
ii. The expected value E(Y | X) of a hypergeometric distribution is n(K/N), where n is the number of draws, K is the number of success states in the population, and N is the size of the population. So, E(Y | X) = 5 * (X/52).
iii. To find E(Y), we need to take the expectation of E(Y | X). Since X is uniformly distributed between 10 and 29, we can write E(Y) = ∑[x=10 to 29] P(X = x) * E(Y | X = x) = ∑[x=10 to 29] (1/20) * 5 * (x/52) = 5/52 * ∑[x=10 to 29] (x/20) = 5/52 * (19.5) = 1.875.
iv. To find Cov(X, Y), we use the formula Cov(X, Y) = E(XY) - E(X)E(Y). We already found E(Y). E(X) is the expected value of a uniform distribution, which is (a+b)/2 = (10+29)/2 = 19.5. E(XY) = E(X)E(Y) in this case because X and Y are independent. So, Cov(X, Y) = E(XY) - E(X)E(Y) = 19.5 * 1.875 - 19.5 * 1.875 = 0.
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