1.CMP CS2-CH1-question1.4-(iii)-c&d How to prove compound poisson process is weakly stationary? X_i=sum^i=N_i=1(Y_i),Y~iid,N~Exp(lambda) I know definition of compound poisson process. But N is a random variable. As N increase,X increase. The expection and variance are always chaging. Weakly stationary ask that the mean are always constant over time and covariance are depend on lag. As X increase,the mean increase. Is the mean constant? 2.CMP-CS2-CH1-question1.10-(ii) Is continuous S compound Poisson process=any special compound Poisson? Doesn't it a discrete space and discrete time set process? Isn't discrete space S=seperate value? Isn't discrete time set J=seperate value? 3.CMP-CS2-CH1-question1.10-(ii) Isn't white noise process all discrete state space and discrete time set? 4.CS2-Assignment X1-queation1.6-(i) Why can't I use Poisson(lambda*t) directly? As long as time~Exp(lambda*t),doesn't it mean probability of event~Poi(lambda*t)? P(x>1)=1-p(x=0)-p(x=1)=1-e^-0.25*5-(0.25*5)*e^-0.25*5=0.3553642071 5.CS2-Assignment X1-queation1.10-(iv)-a I don't understand the logic of why we used 1 and 1+m to represend proportion of changed and unchanged after a change. Doesn't m="the expected number of quarters until the rating changes"not="the expected number of quarters before the rating changes"? Shouldn't it be m=(original porportion of B)*P(X_n not=B|X_n-1=B)+(original proportion of B+the proportion of changes from A to C to B at the previous point in time)*P(X_n=B|X_n-1=B)?=>m=1*0.21+(1+0.21)*0.79? Is the expected number of quarters until the rating changes=the proportion of changes from A to C to B at the previous point in time? 6.Does Time-homogeneous poisson have to be Markov jump chain? 7.Does the jump chain of a Markov jump process have to be time-homogeneous Markov jump process?
8.CMP-CS2-CH4-question4.9-(iv) If E(I_s) not the expected amount of time spent offline over the period [0,t], what is it mean?
Hello See some comments below: 1. It isn't weakly stationary. The question asks us to state whether it is or isn't, not prove that it is. 2. I'm not sure exactly what you're asking here. The state space of a compound Poisson process can be discrete or continuous. Recall that this is looking at the aggregate total of values associated with arrivals (eg claim amounts). The values associated with each arrival could be discrete or continuous for any particular situation. 3. You can have a white noise process in continuous or discrete time with a continuous or discrete state space. 4. You're thinking of a time-homogeneous Poisson process. The structure is similar in that you start from 0 and can only go up to the next integer. However, here the transition rates are not the same at each point (and the state space only goes to 10, rather than to infinity). This approach would actually work if the transition from state 1 to state 2 was also 1/4. However, because it is 1/2, the probability of the process not having jumped to 2 in the given time, for example, is not quite that given by your formula for a Poisson probability. 5. Splitting out the expectation into two parts based on the first transition: E[quarters before rating changes] = E[quarters before rating changes | rating changes after 1 quarter] P(rating changes after one quarter) + E[quarters before rating changes | rating changes after more than 1 quarter] P(rating changes after more than one quarter) Let m = E[quarters before rating changes] then E[quarters before rating changes | rating changes after more than 1 quarter] = m + 1. This is because if we are still rated B in quarter 2, then the expected number of quarters before the rating changes from that point on is just m. So the total expected number is m + 1. This gives: m = 1 0.21 + (1 + m) * 0.79 6 / 7 I'm not entirely sure what you're asking here. It may help to go back through my other response regarding what the jump chain represents. A Poisson process is a Markov jump process. There is an underlying jump chain but the Poisson process itself is not a jump chain. The jump chain of a Markov jump process is a Markov chain not a jump process. 8 E[Is] is equivalent to the probability that we are in the offline state at time s. Hope this helps! Andy
Thank you for your response. I still have some questions need your response. 2.I can't distinguish continuous/discrete state space in diagram. When I saw question 1.1,I though continuous state space may be cumulated state and it won't decrease. I will ask this question because I don’t know much about the characteristics of these processes on the diagram. Can you point out the key to question 1.11 to distinguish between continuous/discrete time space and continuous/discrete state space? This chapter give no diagram in text to explain 4 kind of combination of S&J. It's hard for me to imagine it. 5.I still can't understand that formula totally. I try to seperate it to find my blind points. According to "This equation is constructed by considering what happens after the first quarter.", after the first quarter occurs, part of B goes to other quarters, and part stays in B. In my understanding, 0.79 is the probability of staying in B for all run. 0.21 is the probability of changing from B to something else for all run. Since 0.21 is multiplied by 1, then 1 represents all of B before the change.Is it right? Since m=E[quarters before rating changes],does this rating changes represent "before each change" or "only before the first change"?
Hello 2. For the time set, remember that the stochastic process consists of a collection of random variables, one for each time in the time set. So, B and C relate to discrete time processes (as we have a discrete set of observations) and A and D relate to continuous time processes (as we have a continuous set of observations). For the state space, if it is discrete, this means that the set of possible values is a discrete set. If it is continuous then it can take any values in an interval(s) of the number line. So, A and C look like a discrete set of possible values where as B and D look less so. 5. I'm not quite sure what you mean here. What do you mean for all run? The one multiplied by the 0.21 represents the 1 quarter before the change. We are only considering the first change. Let's perhaps take it back to first principles. Let X be the discrete random variable with possible values 1, 2, 3, 4, ... etc to infinity representing the number of quarters until we move from B. The expectation of a discrete random variable X is sum(x * P(X = x)). Also, P(X = x) is given by 0.79^(x-1) * 0.21 (ie stay in B for x-1 quarters then move away). So m = E[X] = sum(x * 0.79^(x-1) * 0.21) We can recognise the RHS as the derivative of the sum of a geometric sequence. Ie let: S = a + a * r + a * r^2 + .... = a / (1 - r) (for mod(r) < 1) Then dS / d r = a + 2 * a * r + 3 * a * r^2 + ... but d (a / (1-r)) / dr = a / (1-r)^2. So m = a / (1 - r)^2 = 0.21 / (1 - 0.79)^2 = 0.21 / (0.21^2) = 1/0.21. All the best Andy
