We are told that the aggregate claims are normally distributed with mean 0.7P and standard deviation 2P. P = 300 * 5000 = 1 500 000 = 1.5m According to the solutions, the standard deviation is 0.173m. How did we end up with standard deviation = 0.173m and not standard deviation = 2 * 1.5 = 3m
Here, aggregate claims for a particular type of policy is ~N[0.7P, (2P)²] So for n policies, S~ N(n×mu, n×sigma²) P is the annual premium here and there are 300 policies, given P=£5000. Then mu= 0.7×5000 and sigma=2×5000=10000 So..S~N(300×0.7×5000, 300×10000²) i.e. S~N[1.05m, (0.173m)²]
I was doing this question and also have a question but in regards to the premiums my reading of the problem made me think that the premiums were annual therefore for the policies sold in the first year we get 2x premiums because we are evaluating probability of ruin at the end of the 2nd year for the expenses they were specified that you only pay expenses on writing of the policy. So U(2) = 0.1m + 100x500x2 + 200x5000 - 300x0.2x5000, am I interpreting the question wrong?
You seem to have assumed that the 100 policies that were sold in the first year have a premium in the second year. they are one year policies. So they expire.
Not necessarily. They're just saying the premium income they get every year. If one year policies then this will be just different policies.
But sir, how do we know it? They have just asked the prob. of ruin at the end of 2nd year and considered both years. And in q9.6 also, they have asked to find the prob. of ruin at the end of 2nd year and considered only 2nd year?
No in Q9.6 they used the premium gained over 2 years. But it was expressed as 100 policies in the first year and 200 policies in the second year. Whereas the next example the just use "premium income per year" - so we aren't looking at the number of policies and which year but just the income that comes in using the standard formula.
