Standard and Log-Normal Distributions
The function obtained is called standard normal distribution with probability density function, the graph was obtained with,
The second normal curve was constructed with as follows,
The data from part (a) was used to draw the following graph with anti-log on, the nature of the graph changed from normal to positive skewness. The graph looked like a Ch-square plot.
Log normal distribution with mean 8.55 and standard deviation 0.975 was drawn as below.
Pearson’s correlation for MKT and RATE as levels was calculated to be 0.065, and correlation for MKT and RATE as returns was evaluated as 0.222. Both of the correlation values are positive but low for any practically significant relation. Correlation for MKT and RATE as levels was reflected almost no correlation between the variables.
Joint probability distribution was evaluated by dividing all the observed frequencies in each cell by the total number of observation = 253.
The marginal probabilities were calculated as sum of the joint probabilities for each case of RATE and MKT, individually as below.
Overall analysis yielded no significant correlation or dependency between the variables. But, from joint distribution table it was observed that MKT and RATE have significant probability of increasing or rising together. A substantial probability was also noted for explanation of both of the rates decreasing together. Again, no explanation or relation was observed when MKT and RATE were moving in the opposite direction.
The descriptive summary of MKT for daily returns as follows.
The value of Z-statistics at 95% confidence level is 1.96. From the summary, the 95% confidence interval was calculated as
- Null Hypothesis: H0:
Right tail Alternate Hypothesis: HA:
Level of significance was considered at 5%,
Test statistic was chosen as Student’s
Calculated value of the statistics was
At 5% level, critical one tail t-value = 1.6509, and the p-value was calculated as .
Hence, at 5% level of significance the null hypothesis cannot be rejected following the evidences of the hypothesis testing. Therefore, it was concluded that average returns from MKT was not significantly higher than the zero percentage mark.
The Excel output has been included below.
Jarque-Bera test of normality for the MKT rates was conducted. The skewness and excess kurtosis were collected from the descriptive summary. The formula of was used to find he test statistics. Skewness measure has been denoted by S and Kurtosis measure has been denoted by K in the formula.
Here, n = 253, S =1.066, K = 4.505, and
The p-value was obtained from Chi-Square distribution in Excel as =CHIDIST (J9, 2) = 0.000
Hence, the null hypothesis of jointly skewness and kurtosis being zero was strongly rejected. Consequently, it was evident that MKT rates were not normally distributed.
The excel output was as below.
Considering the fact that the MKT rates were not normal, the confidence interval and hypothesis testing with t-statistic were stands out to be invalid. But, at the same time from central limit theorem it can be inferred that for a large set of observation, a dataset tends to a normal distribution. Hence, for a much longer time period the results hold good.
Joint Probability Distribution
Null Hypothesis: The order of the change and no change in daily return rate is normal.
Alternate Hypothesis: The order of the change and no change in daily return rate is not normal.
Level of significance Alpha = 5%.
R = 138 continuous runs were noted with 129 times of no change in return, and 124 times where returns changed from the previous day. The Z-statistics was calculated as where the p-value for two tail alternate hypothesis was evaluated as p = 0.933.
Implication: Hence, there was not enough evidence for rejecting the null hypothesis that order of the change and no change in daily return rate is normal. Therefore, market return fluctuations are in accordance with normal distribution pattern.
- Three approaches for regression were adopted with sample data, population data, and forecasting. The three outputs have been provided below.
Approach 1
Approach 2
Approach 3
- The regression equation was assessed by normal equations and regression coefficients. The initial values were set as, and the solver tool was used to minimize (the error / residuals). The optimal values were obtained as. The Excel output has been incorporated below.
(ii) The normal equations were and
It was observed from the excel output that both the equations were satisfied.
(iii) The data analysis tool of Excel was used to evaluate the regression model, and the regression equation was found as RATE = 0.00005 MKT + 2.425. The values of were in line with the previously evaluated values.
- The regression model was reconstructed with continuous MKT and RATE. The estimated equation was found to be RATE = 1.19 MKT + 2.702
- The regression model was re-estimated with continuous RATE and MKT as below. The regression equation was estimated as RATE = 0.519 MKT +0.00017.
- Among the three models, the last model with continuous market return was found to be statistically significant (F = 12.99, P < 0.05) at 5% level of significance. The MKT (S&P200) was found to be statistically significant in the final model (t = 3.6, p < 0.05) only. Hence, the regression model with continuous return was the only valid model with MKT as the predictor.
ANS: The objective function
Constraint is where price are and
Now, and
and
The Lagrangian function was formed as
Hence, the optimization would be done from solving the following equations, which are,
Here is the Lagrange multiplier.
Solving above three equations we got,
The Excel output is provided as below, where the initial solution for the two decision variables and. The Lagrange Multiplier = indicated that a unit increases in the given value of the Expected utility function of 2000 will increase the value of the optimal level of risk by 0.0924 units.
- Dimension of A , dimension of , and dimension of
- B and C were not conformable for multiplication.
- Excel outputs are as below.
- The C-inverse matrix was found using Excel MINVERSE function as below, and the inverse matrix was also obtained.
ANS: The requisite work was carried out in MS Excel and the following outputs are obtained. Outputs have been explained as follows.
The average daily and yearly returns for the oil stocks were obtained as follows.
The variance-covariance matrix was evaluated as follows. The diagonal values are the variances of the individual stocks.
The model was optimized (minimized) for risk of the portfolio and different values of expected return levels. Initially, the solver tool was used to find the risk for 1% expected return and gradually it was increased by a margin of 1% till 10% expected return. The answer and sensitivity report at expected return of 1% and 10% have been included in the report.
For 1% expected return, the final value of risk was calculated to be 10%. The risk improved from the initial value of 21% risk associated with initial fund allocation. The final fund distribution for 10% risk was found at final value of the adjustable cells. It was observed that for 10% risk, 55.8% fund was advised to be allocated to AGL energy. All the constraints were satisfied by the model.
The sensitivity report reflected that the Lagrange Multiplier was calculated as 0.08427 for expected return of the portfolio. In the precedent the estimation of 0.08427 shows that a unit increments in the given estimation of the Expected Portfolio Returns of 1% will increase the estimation of the ideal level of hazard by 0.08427 units.
For 10% expected return, the final value of risk was calculated to be 15%. The risk improved from the initial value of 21% risk associated with initial fund allocation. The final fund distribution for 15% risk was found at final value of the adjustable cells. It was observed that for 15% risk, 50.65% fund was advised to be allocated in Mineral Resources and 31.31% to at Woodside petroleum. All the constraints were satisfied by the model.
The sensitivity report reflected that the Lagrange Multiplier was calculated as 1.71547 for expected return of the portfolio. In the precedent the estimation of 1.71547 shows that a unit increments in the given estimation of the Expected Portfolio Returns of 10% will increase the estimation of the ideal level of hazard by 1.71547 units.
The ten levels of expected return from portfolio, and corresponding risk obtained from the solver was used to plot the efficient frontier as below. The Lagrange Multiplier in t he a current problem did not came to be negative, even for Pret = 0.0001, that is for expected return of 0.01%.
The optimum risk with expected return was observed at 1% value of expected return with 10% risk (Guerard Jr, Markowitz, & Xu, 2015).
ANS: The two pass or Fama-MacBeth approach was used to test the CAPM model (Fernandez, 2015).
- For the first Pass the systematic risks were calculated for betas for all 23 finance firms with respect to the value weighted index. The market model was estimated in Excel as follows. Using the LINEST function the betas were found and STEYC command was used for evaluating the variances of the all 23 finance firms. S&P 200 or MKT was considered as the dependent factor.
- For the second pass, the estimated model was found to be where beta was found to be a statistically non significant predictor of expected return (t = -0.21, p = 0.83).
- The intercept of -0.0000065 indicated almost zero expected return for no systematic risk. The slope of beta denoted that for one percent increase in systematic risk, the expected return will decrease by 0.01%. now, this result was not at all in line with the assumptions of CAPM model.
- The Multiple Regression Model yielded estimated regression equation as
The nature of the equation reflected that returns will increase for systematic risk increases.
- The regression model was not significant (F = 1.444, P = 0.261). From the p-values of the regression model, it was evident that none of the predictors were statistically significant in predicting the return of the portfolio of finance companies in considered time period.
- t-test for linearity of Beta
Null Hypothesis:
Alternate Hypothesis:
At 5% level of significance, the t-statistic value was calculated as t = 0.9163 ( p = 0.3710)
Hence, the null hypothesis failed to get rejected , and it was possible to conclude that there was no significant linear relation of systematic risk with expected return.
Similarly, no linear relation between the expected return and any other variable was founded.
- The results contradict with the fact that there was no significant linear relationship between the expected return and systematic risk. Also, there was no significant linear relation of variance of error terms with expected return of the finance shares.
References
Fernandez, P. (2015). CAPM: an absurd model. Business Valuation Review, 34(1), 4-23.
Guerard Jr, J. B., Markowitz, H., & Xu, G. (2015). Earnings forecasting in a global stock selection model and efficient portfolio construction and management. International Journal of Forecasting, 31(2), 550-560.
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