problem 40

(a): Suppose, toward a contradiction, that X and Y do not have the same PMF. Then there is at least one k in the support of X such that $P (X = k)$ and $P (Y = k)$ are not equal.
Note that if $P (X = Y) = 1$ , then $P (X = Y | X = k) = P (X = Y | Y = k) = P (X = k | Y = k) = P (Y = k | X = k) = 1$ , as an event with probability 1 will still have probability 1 conditioned on any non-zero event.
Using the above and examining Bayes’ theorem, we have $P (X = k | Y = k) = P (Y = k | X = k) * P (X = k) ∕ P (Y = k)$ , which simplifies to $1 = P (X = k) ∕ P (Y = k)$ as the conditional probabilities equal 1 as previously shown. However, this equality is impossible if $P (X = k) = ∕ = P (Y = k)$ . This contradicts the assumption that $P (X = Y) = 1$ - therefore, X and Y must have the same PMF if they are always equal.
(b): Let X, Y be r.v.s with probability 1 of equalling 1, and probability 0 of equalling any other value.
Then for $x = y = 1$ $P (X = x \land Y = y) = 1 = P (X = x) P (Y = y)$ , and for all other possible pairs of values $x, y$ , $P (X = x \land Y = y) = 0 = P (X = x) P (Y = y)$ . Therefore, X, Y can be independent in this extreme case.