Financial Risk Manager Part 1

Financial Risk Manager Part 1

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According to the Jarque-Bera (JB) test, the test statistic is given by:

JB=(Tβˆ’1)(S^26+(k^βˆ’3)224)\text{JB} = (T - 1) \left( \frac{\hat{S}^2}{6} + \frac{(\hat{k} - 3)^2}{24} \right)

So,

JB=(100βˆ’1)(0.3526+(3.04βˆ’3)224)=2.028\text{JB} = (100 - 1) \left( \frac{0.35^2}{6} + \frac{(3.04 - 3)^2}{24} \right) = 2.028

Recall that JB ~ Ο‡22\chi^2_2 so that critical value at 5% with 2 degrees of freedom (df) is 5.991 (which can be seen from a chi-square table).

Since the test statistic is less than the critical value, we fail to reject the null hypothesis that the returns are normal, and hence the analyst can use the normal distributions in analyzing the returns.

What is the calculated Jarque-Bera test statistic value?_

TTanishq





Explanation:

The Jarque-Bera test statistic is calculated as:

JB=(Tβˆ’1)(S^26+(k^βˆ’3)224)\text{JB} = (T - 1) \left( \frac{\hat{S}^2}{6} + \frac{(\hat{k} - 3)^2}{24} \right)

Given:

  • T = 100 (sample size)
  • S^\hat{S} = 0.35 (skewness)
  • k^\hat{k} = 3.04 (kurtosis)

Substituting the values:

JB=(100βˆ’1)(0.3526+(3.04βˆ’3)224)\text{JB} = (100 - 1) \left( \frac{0.35^2}{6} + \frac{(3.04 - 3)^2}{24} \right) JB=99(0.12256+0.001624)\text{JB} = 99 \left( \frac{0.1225}{6} + \frac{0.0016}{24} \right) JB=99(0.0204167+0.0000667)\text{JB} = 99 \left( 0.0204167 + 0.0000667 \right) JB=99Γ—0.0204834=2.028\text{JB} = 99 \times 0.0204834 = 2.028

Therefore, the correct Jarque-Bera test statistic is 2.028. Since this value (2.028) is less than the critical value of 5.991 at the 5% significance level with 2 degrees of freedom, we fail to reject the null hypothesis that the returns are normally distributed.

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