Devon
You are correct. My method is equivalent to using the identify matrix. This is okay to do as independence implies linear correlation = 0 as well.
I guess if you want to do it using aggregate mean and variance, you are effectively assuming that the aggregate distribution is gamma (the question does not tell you that explicitly but you would need to assume that).
I did some number crunching at home:
If we do it my way (as described in an earlier post): DivCap = 110,560
The UndivCap would be just sum of each component capital i.e. 181,442 ;
=> Div = 70,882
If we fit a gamma to the aggregate distribution: DivCap = 120,211 (note this number has been computed in Excel)
You have to assume UndivCap as sum of capital computed earlier i.e. 181,442
=> Div = 61,231
[But: You would invariably be making an 'incorrect' assumption that sum of gamma is gamma where the scale parameters of individual components are actually different. Note you have computed the individual capital numbers assuming gamma distribution]
If you want to avoid this assumption you would have to compute the un-diversified capital in some other way which I am not sure what it can be. Also, do we really need to re-do part (ii)!
PS: If we fit an approximate 'Normal' distribution to the aggregate distribution: DivCap = 104,031. This is much lower than any of the other numbers as Normal is a thin tailed distribution
Last edited: Oct 11, 2012