Consider the following:

(a) If X and Y are i.i.d. continuous r.v.s with CDF F(x) and PDF f(x), then M = max(X, Y) has PDF 2F(x)f(x). Now let X and Y be discrete and i.i.d., with CDF F(x) and PMF f(x). Explain in words why the PMF of M is not 2F(x)f(x).

(b) Let X and Y be independent Bernoulli(1/2) r.v.s, and let M = max(X, Y), L = min(X, Y ). Find the joint PMF of M and L, i.e., P(M = a, L = b), and the marginal PMFs of M and L.

(a) If X and Y are i.i.d. continuous r.v.s with CDF F(x) and PDF f(x), then M = max(X, Y) has PDF 2F(x)f(x). Now let X and Y be discrete and i.i.d., with CDF F(x) and PMF f(x). Explain in words why the PMF of M is not 2F(x)f(x).

(b) Let X and Y be independent Bernoulli(1/2) r.v.s, and let M = max(X, Y), L = min(X, Y ). Find the joint PMF of M and L, i.e., P(M = a, L = b), and the marginal PMFs of M and L.

Solution:
(a) The PMF is not 2F(x)f(x) in the discrete case due to the problem of ties: there is a nonzero chance that X = Y. We can write the PMF as P(M = a) = P(X = a, Y < a) + P(Y = a,X < a) + P(X = Y = a) since M = a means that at least one of X, Y equals a, with neither greater than a. The first two terms together become 2f(a)P(Y < a), but the third term may be nonzero and also P(Y < a) may not equal F(a) = P(Y =< a).
(b) In order statistics notation, L = X(1), M = X(2). Marginally, we have X(1) ` Bern(1/4), X(2) ~ Bern(3/4). The joint PMF is
P(X(1) = 0, X(2) = 0) = 1/4;
P(X(1) = 0, X(2) = 1) = 1/2;
P(X(1) = 1, X(2) = 0) = 0;
P(X(1) = 1, X(2) = 1) = 1/4
Note that these are nonnegative and sum to 1, and that X(1) and X(2) are dependent; how does this relate to the problem of the probability of both of two children being girls, given that at least one is a girl?

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