2. Let X Exp() for > 0. Let Y = [X] be the result of rounding up X to the nearest integer, so that e.g. [3.1] = [3.8] = 4

(a) Note that Y takes values in N, i.e. P(Y N) = 1. Compute pY , the probability mass function of Y . Hence show that Y follows a geometric distribution with parameter p = 1 exp( )

(It's OK to jump this question since I am confident about it)

(b) Compute GX|Y (a|k) = P(X a|Y = k) for any k N and any a > 0

(G is the generating function)

(c) Differentiate GX|Y (a|k) with respect to a and hence compute the expected value E[X|Y = k] of X conditional on Y = k where k N

(d) Show that EY [E[X|Y ]] = E[X] holds by considering the left and the right-hand side of this expression independently. Do not rely on the iterated conditional expectation formula for this calculation. Produce of word-processed (not handwritten!) explanation of why the area under the curve g(x) = xfX(x) can be written as EY [E[X|Y ]]. Your explanation should connect the notion of integral, area under the curve g and how this area is split up by EY .

Please answer the above question as this is regarded as one whole question. Really appreciate your help and support.