Hello Asingh
This was a devil of a question. I'm sure if you'd (correctly) used a continuity correction for Investor 2, you would not have been penalised. However, I don't quite agree your figures
Here's how I would do it:
Investor 2
On each bond, Investor 2 loses 0 with probability 0.96 and loses 1 with probability 0.04.
Let Li be the loss on the ith bond.
Li has a Bin(1, 0.04) distribution.
Let X be the loss on the total investment.
X = L1 + L2 + ... + L1000
X has an approx Normal(40, 38.4) distribution.
For the 95% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.05
Applying continuity corrections:
P[X > 49] = P[Z > (49.5 - 40)/sqrt(38.4)] = 1 - phi(1.5331) = 0.0626
P[X > 50] = P[Z > (50.5 - 40)/sqrt(38.4)] = 1 - phi(1.6944) = 0.0451
so x = 50.
For the 90% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.1
P[X > 47] = P[Z > (47.5 - 40)/sqrt(38.4)] = 1 - phi(1.2103) = 0.1131
P[X > 48] = P[Z > (48.5 - 40)/sqrt(38.4)] = 1 - phi(1.3717) = 0.0851
so x = 48.
Just to complete the picture:
Investor 1
Let X be the loss on the total investment for Investor 1.
X = 0 with probability 0.96
X = 1000 with probability 0.04
For the 95% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.05
P[X > 0] = P[X = 1000] = 0.04
x = 0
For the 90% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.10
P[X > 0] = P[X = 1000] = 0.04
x = 0
Does this help?
Anna