In Que 10.9, we are told that 46 deaths occur over 5 years and the popn remains constant at 7500. I calculated the rate of mortality as annual - ie. D=46/5 and then divided by Ex0=7500. In the answer they kept D=46 and Ex0 as 5x7500. Same answer for force of mortality but different once we come to calculating the variance. What have I missed in my approach? Thanks.
The observed value of D is d=46 and the central exposed to risk is 5*7,500. This still leads to an ANNUAL rate of mortality. We take the number of deaths (46) and divide by the number of observed people years (5*7500) In your approach, you've changed these figures so that your D is the average number of deaths per year. This is an unusual move but will still lead to the same varaince for mu if we do it from first principles: Let's define X = D/5. Now D~Poi(5*7,500mu), so the variance of X is var[D]/25 = 7500*5mu/25 = 7500mu/5 So, var[mu] = var[X/7,500] = (7,500mu/5)/7,500^2 = mu/7,500*5 (same answer) John
Thanks John. So keeping deaths at 46 and multiplying the population makes more sense to me now. So does it make sense to call it an annual rate of mortality? I think that's where I was confused originally but as I understand it now, its an instantaneous rate - is that right?
No, you were right that mu is annualised but, as you say, it only applies to the next very small period of time. mu * amount of time (IN YEARS) = Probability that person dies in that same amount of time The smaller the amount of time, the more accurate this calculation becomes. eg I am just gone 36, so I look up mu36 in AM92, 0.000706. I take a bit off as I don't drink, smoke, I eat healthily and I'm sporty. Let's say my mu is 0.0004. (This is a 4th method of graduation not mentioned in the notes!) This means the probability that I die in the next 3 hours is roughly... 0.0004 * amount of time (IN YEARS) = 0.0004 * (1/365)*(3/24) = 0.00000013698 Thankfully, this is a very small number. Though note how it's still much more likely than winning the national lottery, which I still stupidly play ;-) John
