I did not understand the explanation given for question 3.9(i) {ie Why do the functions f and g have to be increasing functions?} for model 5.5 of Markov chains. Also, I did not get the proof of how cumulative number of accidents is a markov chain ie how P[Y(n+1)=1 | Y1=x1, Y2=x2-x1,...,Yn=xn-x(n-1)]=f(xn)/g(n).
I'm not sure there's that much I can add to the solution to Q3.9(i). In essence, f and g should be increasing functions in order for the probabilities obtained by dividing f by g to be sensible in the context of the model described. It might help you to look at an example with numbers in it - I'd suggest April 2012, Q11. Starting from: P(Y(n+1)=1 | Y1=x1, Y2=x2-x1,...,Yn=xn-x(n-1)) This is equal to (using the definition of Xn in terms of Yn): P(Y(n+1)=1 | Y1=y1, Y2=y2,....,Yn=yn) But we're told at the start that this equals: f(y1+y2+...+yn)/g(n) which equals: f(xn)/g(n) as stated.
Hi Am abit confused, what do the functions f and g stand for in this section of the core reading i. Chapter 3 part 5.5.
f(x_n)=f(y_1+y_2+......+y_n) is a function of (y_1,y_2,.........,y_n) random variables with having an accident=1 or not=0 in a period upto n periods. g(n) is a function of period n.
I, too, am a little confused about the core reading here. If you solely look at the first part of core reading on page 24 (forget the bits before hand), what follows seems to show that absolutely any discrete process can be made Markov by considering the partial sum of that process. (along with ignoring the very final line =f(x_n)/g(n)) ??
