FIN 6307 HW 4

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6307

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Finance

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Jan 9, 2024

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FIN 6307 HW 4 4. Suppose you invest in a product whose returns follow a uniform distribution between −40% and +60%. What is the expected return? What is the 95% VaR? The expected shortfall? The expected return: [(−40%)+(+60%)]/2= +10% Since the density is 1, then 95% VaR is just 5% from the worst outcome of -40%, which is −40% +5%(60%− −40%)= −35%, meaning that the density between -35% and -40% is 5.0% of the total 100% density between 60% and -40%. The expected shortfall: [(−35%)+(−40%)]/2= −37.5% 5. You are the risk manager for a portfolio with a mean daily return of 0.40% and a daily standard deviation of 2.3%. Assume the returns are normally distributed (not a good assumption to make, in general). What is the 95% VaR? For a normal distribution, 5% of the weight is corresponding to value −1.64 (Based on the Exhibit 7.1 from textbook). The 95% VaR is 0.40% − 1.64*2.30%= − 3.38% (−0.00337), Or use Excel: NORM.INV (5%, 0.40%, 2.3%) which gives the same value. 7. In the previous question, you were told that the actual standard deviation was 40%. If, instead of 40%, the measured standard deviation turns out to be 50%, how confident can you be in the initial assumption? State a null hypothesis and calculate the corresponding probability. Null hypothesis is H0: σ =40% Test statistic: (33−1) 0.50 2 /0.40 2 = 50 By calculating P(X>50), with 32 degree of freedom, only 2.33% (0.02229) of the distribution is great than 50, or use Excel: 1−CHISQ.DIST (50, 32, TRUE) which gives the same value. At a 95% of confidence level, we would reject the null hypothesis. 8. A hedge fund targets a mean annual return of 15% with a 10% standard deviation. Last year, the fund returned –5%. What is the probability of a result this bad or worse happening, given the target mean and standard deviation? Assume the distribution is symmetrical. According to Chebyshev’s inequality, the probability of being greater than two standard deviations from the mean is less than or equal to 25%. P[|X−15%|>= 2*10%] <=1/2 2 =25% Because the distribution is symmetrical, half of these extreme returns are greater than +2 standard deviations, and half are less than –2 standard deviations. So the probability of a result this bad or worse happening is 12.5%.
9. A fund of funds has investments in 36 hedge funds. At the end of the year, the mean return of the constituent hedge funds was 18%. The standard deviation of the funds’ returns was 12%. The benchmark return for the fund of funds was 14%. Is the difference between the average return and the benchmark return statistically significant at the 95% confidence level? Standard deviation of the mean: σ µ = σ/ n, σ µ = 12%/ 36= 2% (0.02) The difference between the average fund return and the benchmark return is 4% (=18%-14%). For a T-distribution with 35 degree of freedom, use Excel: 1-T.DIST (2, 35, TRUE)= 2.67% (0.0266) Since 2.67 % < 5%, so we can reject the null hypothesis, H0: μ =14%, at the 95% confidence level. Therefore, the difference between the average return and the benchmark return is statistically significant at the 95% confidence level.
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