2. Again, the chicken-egg problem! A chicken lays a number of eggs, N, which follows a Poisson (A) distribution. Each egg hatches a chick with probability p, independently. Let X be the number of eggs which hatch, so X | N = n ~ Bin(n, p) and let Y be the number of eggs which don't hatch, so Y | N = n ~ Bin(n, 1 - p). We also note that N = X +Y. (a) (2 pts) Find the marginal distribution of Y. Hint: Start from the joint PMF of Y and N, as shown for the marginal distribution of X during the class. (b) (2 pts) By the law of total probability, the following is the case. P(X = x, Y = y) = P(X = x, Y = y | N = n)P(N = n), n where the summation is over all possible values of n, holding x and y fixed. But unless n = x+y, it is impossible for X to equal x and Y to equal y. For example, the only way there can be 5 hatched eggs and 6 unhatched eggs is when there are 11 eggs in total. So P(X = 5, Y=6|N=n) = 0 unless n = 11, which means all other terms on the right-hand side can be dropped, i.e., P(X = x, Y = y) = P(X = x, Y = y | N = x+y)P(N = x + y). Conditional on N =x+y, the events X = x and Y = y are exactly the same event, so keeping both is redundant; for example, if we know that {N = x + y} and {X = x}, then it is automatically the case that {Y = y}. Let us keep X = x. Then, P(X = x, Y = y) = P(X = x | N = x + y)P(N = x + y). Specify this joint probability mass function of X and Y, i.e., P(X = x, Y = y), with appropriate joint support of X and Y. Hint: We know that 0 x n, but n is removed (marginalized) during the derivation of the joint PDF of X and Y via the law of total probability. Then what are the possible values of x without the cap n? Similarly, what are the possible values of y without the cap n? (c) (2 pts) Based on the joint distribution of X and Y in (b), are X and Y independent? Explain. 1 Hint: eepe-(1-p). = (d) (2 pts) Find Cov(N, X). Hint: NX + Y as given in the problem, and so use the bi-linearity of covariance. (e) (2 pts) Find Corr(N, X). Hint: Use the known facts about the marginal distribution of N and that of X.