((3) Suppose V;, i = 1,. . . ,n, are independent exponential random variables with rate 1. Denote X=max{n:ZVgA}, 5:1 50 X can be thought of as being the maximum number of exponentials having rate 1 that can be summed and still be less than or equal to A. (i) (ii) (iii) Using properties of a Poisson process with rate 1, explain Why X has a Poisson distibution with parameter A. Let V} = ~ log [15, U,- ~ Uniform(0, 1), 3' = 1, . . . ,n. Show that X=max{n:UsZe_'\\}, (1) {:1 where H0 U; = 1. i=1 It can be shown that (1) is equivalent to f?- erin{n:HU.- {9)} 1. 5:1 This result may be used to simulate a Poisson random variable with parameter A. If we continue generating Uniform (0, 1) random variables U.- until their product falls below (3\