Question

Value at Risk (VAR) has become a key concept in financial calculations. The VAR of an investment is defined as that value v such that there is only a 1 percent chance that the loss from the investment will be greater than v.

(a) If the gain from an investment is a normal random variable with mean 10 and variance 49 determine the VAR. (IfX is the gain, then ?X is the loss.)
(b) Among a set of investments all of whose gains are normally distributed, show that the one having the smallest VAR is the one having the largest value of mean minus standard deviation*2.33?

Answer 1

Var = 6.31

Explanation:

The Value at Risk (VAR)

`P(X < x_o) = 0.01`

By using normal distribution

Mean
`\mu` = 10

Variance = 49

Standard deviation
`\sigma = √(49 ) = 7`

This implies that:

`P\Big ( (X - \mu)/(\sigma ) < (x_o - \mu )/(\sigma)\Big) = 0.01 \\ \\ P\Big ( Z < (x_o - \mu )/(\sigma)\Big) = 0.01 \\ \\ (x_o - \mu )/(\sigma) = invNorm(0.01) \\ \\ x_o = \mu + \sigma * invNorm (0.01)`

Using the z-table;

`x_o = 10 + 7 * (-2.33) \\ \\ x_o = -6.3100`

Hence, there exist 1% chance that X < -6.31 or the loss from investment is > 6.31

From the calculated value above;

`V = \mu -\sigma * 2.33`; Since the result is negative, then it shows that the greater the value(i.e the positive or less negative it is ) the lower is the value of VAR. Thus, the least value of VAR is accepted by the largest value of

`min( \mu -\sigma *2.33,0)`