You observe particles emitted from a radioactive sample. Your task is to determine whether the sample is composed of an element a or of an element

b. It turns out that the emission patterns of these two elements are well approximated by independent Poisson processes with parameters (or rates as we say) a and b , respectively. From past data you also know that element a is roughly twice as common as element b. As a result, the number of emitted particles at time t can be modeled by

N(t)={Na(t)if=1Nb(t)if=0

where is a Bernouilli random variable with parameter 2/3, Na (t) is a Poisson random process with parameter a and Nb (t) is a Poisson random process with parameter b . In addition, we assume that Na(t) and Nb(t) are independent.

a) What is the expected number of particle emissions between t = 0 and t = 5 as a function of a and b ?

b) For i = 0, 1, 2, ... we denote by T (i) the ith interarrival time of N(t), that is the time between the i-th emission and the (i + 1)-th emission. What is the first-order pdf of the discrete-time continuous-state process T?

c) You observe that y particles are emitted up to time 10, i.e. N(10) = y and we are given

that b > a . We want to check whether it is more probable that = 1 as oppose to = 0, given this information. Find a real-valued quantity g (a , b ) depending only on a , b for which

P(=1N(10)=y)>P(=0N(10)=y)

is equivalent with simply checking

g(a , b ) > y (equation 1)

(Hint: Applying logarithms should be useful.) Note that when y is "extremely large" such a condition (equation 1) cannot be satisfied and therefore = 0 is more probable than = 1 given N(10) = y. Can you give an intuitive explanation of it?