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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?

Sagot :

ajeigbeibraheem

Answer:

Var = 6.31

Step-by-step explanation:

The Value at Risk (VAR)

[tex]P(X < x_o) = 0.01[/tex]

By using normal distribution

Mean [tex]\mu[/tex] = 10

Variance = 49

Standard deviation [tex]\sigma = \sqrt{49 } = 7[/tex]

This implies that:

[tex]P\Big ( \dfrac{X - \mu}{\sigma } < \dfrac{x_o - \mu }{\sigma}\Big) = 0.01 \\ \\ P\Big ( Z < \dfrac{x_o - \mu }{\sigma}\Big) = 0.01 \\ \\ \dfrac{x_o - \mu }{\sigma} = invNorm(0.01) \\ \\ x_o = \mu + \sigma \times invNorm (0.01)[/tex]

Using the z-table;

[tex]x_o = 10 + 7 \times (-2.33) \\ \\ x_o = -6.3100[/tex]

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

From the calculated value above;

[tex]V = \mu -\sigma \times 2.33[/tex]; 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

[tex]min( \mu -\sigma \times2.33,0)[/tex]

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