Let {mi(t), t > =0}, i=1,2,3 be independent poisson processes with respective rates λi, i=1,2, and set: n1(t)=m1(t)+m2(t), n2(t)= m2(t)+m3(t)the stochastic process {(n1(t), n2( t> =0} is called a bivariate poisson processa) find p{n1(t)=n, n2(t)=m}b) find cov(n1(t), n2(t)}

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560