Question

Suppose X is a normal with zero mean and standard deviation of $10 million.

a) Find the value at risk for X for the risk tolerances h=0.01, 0.02, 0.05, 0.10, 0.50, 0.60, and 0.95.

b) Is there a relation between VaR for values of h <= 0.50 and values for h>= 0.50?

Answer 1

a) We are given the distribution here as:

From standard normal tables, we have here:
P(Z < -2.326) = 0.01,
P(Z < -2.054) = 0.02,
P(Z < -1.645) = 0.05,
P(Z < -1.282) = 0.1,
P(Z < 0) = 0.5,
P(Z < 0.253) = 0.6,
P(Z < 1.645) = 0.95

Therefore, now the Value at risk values for various risk tolerances are computed here as:
Var(0.01) = -2.326*10 = -23.26 million.
Var(0.02) = -2.054*10 = -20.54 million.
Var(0.05) = -1.645*10 = -16.45 million.
Var(0.1) = -1.282*10 = -12.82 million.
Var(0.5) = 0*10 = 0 million.
Var(0.6) = 0.253*10 = -2.53 million.
Var(0.95) = 1.645*10 = 16.45 million.

b) Yes, the Var values for h <= 0.5 are negative of the Var values for h >= 0.5 for same distance from 0.

For example we have from the above part:
Var(0.05) = -1.645*10 = -16.45 million.
Var(0.95) = 1.645*10 = 16.45 million.