Answers _ Mathematics and Statistics for Financial Risk Management

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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 1/40 Answers CHAPTER 1 1. a. y = 5 a. y = ln(1) – ln( e ) = 0 – 1 = –1 b. y = ln(10) + ln( e ) = ln(10) + 1 = 3.3026 2. Annual rate = 5.12%; semiannual rate = 5.05%; continuous rate = 4.99%. 3. 4. 5. 6. 7. 8. ln(ln(10)) = 0.8340.
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 2/40 9. 10. 11. The bond will pay 10 coupons of $2, starting in a year's time. In addi- tion, the notional value of the bond will be returned with the final coupon payment in 10 years. The present value, V, is then: We start by evaluating the summation, using a discount factor of δ = 1/1.05 0.95: Inserting this result into the initial equation we obtain our final result: Note that the present value of the bond, $78.83, is less than the notional value of the bond, $100. This is what we would expect, given that there is no risk of default, and the coupon rate is less than the discount rate. CHAPTER 2
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 3/40 1. Probability that both generate positive returns = 60% × 70% = 42%. Probability that both funds lose money = (1 – 60%) × (1 – 70%) = 40% × 30% = 12%. 2. 88%. The sum of all three events—upgrade, downgrade, and no change —must sum to one. There is no other possible outcome. 88% + 8% + 4% = 100%. 3. 50%. The outcomes are mutually exclusive; therefore, 20% + 30% = 50%. 4. 5. 6. 32.14%. By applying Bayes’ theorem, we can calculate the result: Even though the model is 90% accurate, 95% of the bonds don't default and it predicts that 10% of them will. Within the bond portfolio, the mod- el identifies 9.5% of the bonds as likely to default, even though they won’t. Of the 5% of bonds that actually default, the model correctly identifies 90%, or 4.5% of the portfolio. This 4.5% correctly identified is over- whelmed by the 9.5% incorrectly identified.
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 4/40 7. Given the density function, we can find c by noting that the sum of probabilities must be equal to one: 8. First we check that this is a valid CDF, by calculating the value of the CDF for the minimum and maximum values of x : Next we calculate the PDF by taking the first derivative of the CDF: 9. We first calculate the CDF by integrating the PDF:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 5/40 We first try to find c using the fact that the CDF is zero at the minimum value of x , x = 0. As it turns out, any value of c will satisfy this constraint, and we cannot use this to determine c . If we use the fact that the CDF is 1 for the maximum value of x , x = e , we find that c = 1: The CDF can then be expressed simply as: 10. P (both bonds default) = 9%. P (one defaults) = 42%. P (neither defaults) = 49%. 11. We can start by summing across the first row to get W : In a similar fashion, we can find X by summing across the second row: To calculate Y , we can sum down the first column, using our previously calculated value for W : Using this result, we can sum across the third row to get Z :
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 6/40 The completed probability matrix is: The last part of the question asks us to find the conditional probability, which we can express as: We can solve this by taking values from the completed probability matrix. The equity underperforms in 40% of scenarios. The equity underper- forms and the bonds are downgraded in 15% of scenarios. Dividing, we get our final answer, 37.5%. 12. The probability that a B-rated bond defaults over one year is 2%. This can be read directly from the last column of the second row of the ratings transition matrix. The probability of default over two years is 4.8%. During the first year, a B-rated bond can either be upgraded to an A rating, stay at B, be down- graded to C, or default. From the transition matrix, we know that the probability of these events is 10%, 80%, 8%, and 2%, respectively. If the bond is upgraded to A, then there is zero probability of default in the sec- ond year (the last column of the first row of the matrix is 0%). If it re-
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 7/40 mains at B there is a 2% probability of default in the second year, the same as in the first year. If it is downgraded to C, there is a 15% probabili- ty of default in the second year. Finally if a bond defaulted in the first year it stays defaulted (the last column of the last row is 100%). Putting this all together we have: CHAPTER 3 1. Mean = 6.43%; median = 5%. 2. Mean = 3.00%; standard deviation = 6.84%. 3. 4. Using the results of Question 3, we first calculate the variance of the es- timator of the mean: where σ is the standard deviation of r . For the last step we rely on the fact that, because the data points are i.i.d., the covariance between different data points is zero. We obtain the final answer by taking the square root of the variance of the estimator:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 8/40 5. Covariance = 4.87%; correlation = 82.40%. 6. Series #1: Mean = 0, standard deviation = 39, skewness = 0. Series #2: Mean = 0, standard deviation = 39, skewness = –0.626. 7. Series #1: Mean = 0, standard deviation = 17, kurtosis = 1.690. Series #2: Mean = 0, standard deviation = 17, kurtosis = 1. 8. The variance is approximately 5.56. From a previous example we know the mean to be 20/3; thus the variance can be found as: 9. We start by expanding the mean: By carefully rearranging terms, we are left with: Assuming that all the different values of X are uncorrelated with each other, we can use the following two relationships: Then:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 9/40 10. First we note that the expected value of X A plus X B is just the sum of the means: Substituting into our equation for variance, and rearranging, we get: Expanding the squared term and solving: Using our definition of covariance we arrive at our final answer: 11. If the bond does not default, you will receive $100. If the bond does default, you will receive 40% × $100 = $40. The future value, the expected value of the bond at the end of the year, is then $94: The present value of the bond is approximately $89.42: CHAPTER 4
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 10/40 1. The number of times XYZ Corporation exceeds consensus estimates fol- lows a binomial distribution; therefore: 2. The cumulative number of exceedance events follows a binomial distri- bution; therefore: 3. Because the annual returns of both funds are normally distributed, the difference in their returns is also normally distributed: The mean of this distribution is 10%, and the standard deviation is 50%. At the end of the year, the difference in the expected returns is 92%. This is 82% above the mean, or 1.64 standard deviations. Using Excel or con- sulting the table of confidence levels in the chapter, we see that this is a rare event. The probability of more than a 1.64 standard deviation event is only 5%. 4. The average number of defaults over five months is 10; therefore: 5. If the returns of the fund are normally distributed with a mean of 10% and a standard deviation of 15%, then the returns of $200 million invest- ed in the fund are also normally distributed, but with an expected return of $20 and a standard deviation of $30. A loss of $18.4 million represents a –1.28 standard deviation move:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 11/40 This is a one-tailed problem. By consulting the table of confidence inter- vals or using a spreadsheet, we determine that just 10% of the normal dis- tribution lies below –1.28 standard deviations. 6. The return of –30% is approximately a –1.64 standard deviation event: According to the table of confidence intervals, 5% of the normal distribu- tion lies below –1.64 standard deviations. The probability of a return less than –30% is then 5%. 7. For the mean: From a previous example, we know that c = 1/( x 2 x 1 ); therefore: For the variance: Substituting in for c and μ from above: For the final step, we need to know that:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 12/40 Substituting in and solving, we have: 8. Using integration by substitution, define a new variable y and solve: 9. Using the same substitution as in the previous question: 10. Using the same substitution as before: For the final step we need to know that:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 13/40 Using this result, we achieve the desired result: 11. First we note that the mean of X A is zero: Similarly, the mean of X B is zero. Next, we want to calculate the variance. In order to do that, it will be use- ful to know two relationships. First we rearrange the equation for vari- ance, Equation 3.20 , to get: Similarly, we can rearrange our equation for covariance, Equation 3.26 , to get: With these results in hand, we now show that the variance of X A is one: Next we calculate the covariance of X A and X B :
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 14/40 Putting the last two results together completes the proof: 12. For the portfolio consisting of 50% A and 50% B , we can proceed two ways. The PDF of the portfolio is a triangle, from –0.5 to +0.5, with height of 2.0 at 0. We can argue that the mean is zero based on geometric argu- ments. Also, because both distributions are just standard uniform vari- ables shifted by a constant, they must have variance of 1/12; 50% of each asset would have a variance of ¼ this amount, and—only because the variables are independent—we can add the variance of the variable, giv- ing us: Alternatively, we could calculate the mean and variance by integration:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 15/40 This confirms our earlier answer. For the 50/50 mixture distribution, the PDF is bimodal and symmetrical around zero, giving a mean of zero: For the variance we have: Notice that, while the mean is the same, the variance for the mixture dis- tribution is significantly higher. CHAPTER 5 1. Mean = 45.0; standard deviation = 29.3; standard deviation of mean = 9.3. For the hypothesis that the mean is greater than 40, the appropriate t - statistic has a value of 0.54. For a one-sided t -test with 9 degrees of free- dom, the associated probability is 70%. There is a 30% chance that the true mean is found below 40, and a 70% chance that it is greater than 40. 2. The mean is 6.9%, and the standard deviation of the returns is 23.5%, giving a standard deviation of the mean of 7.4%. The t -statistic is 0.93. With 9 degrees of freedom, a one-sided t -test produces a probability of 81%. In other words, even though the sample mean is positive, there is a 19% chance that the true mean is negative. 3. A negative return would be greater than 2 standard deviations. For a normal distribution, the probability (one-tailed) is approximately 2.28%. If we do not know the distribution, then, by Chebyshev's inequality, the
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 16/40 probability of a negative return could be as high as 12.5% = ½ × 1/(2 2 ). There could be a 25% probability of a +/–2 standard deviation event, but we're interested only in the negative tail, so we multiply by ½. We can only perform this last step because we were told the distribution was symmetrical. 4. The expected return is +10%. The 95% VaR is 35% (i.e., 5% of the re- turns are expected to be worse than –35%). The expected shortfall is 37.5% (again the negative is implied). 5. For a normal distribution, 5% of the weight is less than –1.64 standard deviations from the mean. The 95% VaR can be found as: 0.40% – 1.64 · 2.30% = –3.38%. Because our quoting convention for VaR, the final answer is VaR = 3.38%. 6. We can use Equation 5.4 to calculate the expected variance of the sam- ple variances. Because we are told the underlying distribution is normal, the excess kurtosis can be assumed to equal zero and n = 33; therefore: The standard deviation of the sample variances is then 4.0%. 7. An appropriate null hypothesis would be: H 0 : σ = 40%. The appropriate test statistic is: Using a spreadsheet, or other program, we calculate the corresponding probability for a chi-squared distribution with 32 degrees of freedom. Only 2.23% of the distribution is greater than 50. At a 95% confidence lev- el, we would reject the null hypothesis. 8. Answer: 12.5%. This is a –2 standard deviation event. According to Chebyshev's inequality, the probability of being more than 2 standard de-
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 17/40 viations from the mean is less than or equal to 25%. Because the distribution of returns is symmetrical, we assume that half of these extreme events are greater than +2 standard deviations, and half are less than –2 standard deviations. This leads to the final result, of 12.5%. 9. The standard deviation of the mean is 2%: This makes the difference between the average fund return and the benchmark, 18% – 14% = 4%, a +2 standard deviation event. For a t distri- bution with 35 degrees of freedom, the probability of being more than +2 standard deviations is just 2.67%. We can reject the null hypothesis, H 0 : μ = 14% at the 95% confidence level. The difference between the average return and the benchmark return is statistically significant. 10. To find the 95% VaR, we need to find v , such that: Solving, we have: The VaR is a loss of 90. Alternatively, we could have used geometric argu- ments to arrive at the same conclusion. In this problem, the PDF de-
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 18/40 scribes a rectangle whose base is 200 units and whose height is 1/200. As required, the total area under the PDF, base multiplied by height, is equal to one. The leftmost fraction of the rectangle, from –100 to –90, is also a rectangle, with a base of 10 units and the same height, giving an area of 1/20, or 5% of the total area. The edge of this area is our VaR, as found by integration before. 11. In the previous question we found that the VaR, v , was equal to –90. To find the expected shortfall, we need to solve the following equation: Solving, we find: The final answer, a loss of 95 for the expected shortfall, makes sense. The PDF in this problem is a uniform distribution, with a minimum at –100. Because it is a uniform distribution, all losses between the (negative) VaR, –90, and the minimum, –100, are equally likely; therefore, the average loss, given a VaR exceedance, is halfway between –90 and –100. 12. To find the 95% VaR, we need to find v , such that: By inspection, half the distribution is below 5, so we need only bother with the first half of the function:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 19/40 We can use the solution to the quadratic equation: Because the distribution is not defined for π < –15, we can ignore the neg- ative, giving us the final answer: The one-day 95% VaR for Pyramid Asset Management is approximately 8.68. CHAPTER 6 1. 2. For the first part of the question, because matrix addition is commuta- tive and associative, the order in which we perform the operations does not matter:
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 20/40 3. 4. 5. A matrix multiplied by an appropriately sized identity matrix is itself. This is true when the identity matrix is multiplied by itself, too. For U 2 : For AU :
10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 21/40 6. To prove that J is the inverse of K , we need to show that the two matri- ces multiplied together produce an identity matrix. 7. To solve this problem we could multiply M by itself five times. Alternatively, we can reexpress M as the product of a constant and an identity matrix: 8. At the end of the year, it is expected that 61% of the bonds will have an A rating, 36.4% B, 2.2% C, and 0.4% D. To get the answer, we can proceed one rating at a time. Of the 60% of bonds that are rated A at the start of the year, we expect 95% will still be rated A at the end of the year. Of the 40% of bonds that are rated B at the start of the year, we expect 10% to have been upgraded to A by the end of the year. Putting the two together, we have: We can calculate the other three ratings similarly: We can check our answer by noting that the sum of the answers is 100%. At the end of the year each bond must be either A, B, C, or D; therefore, the sum of the expected values must be 100%.
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 22/40 9. To calculate the two-year transition matrix, we simply square the one- year matrix. Using T 1 and T 2 to denote our one-year and two-year matri- ces, respectively, we have: Though not necessary, we can reformat this to match the original one- year matrix: 10. We can use our Cholesky algorithm to find the elements of the matrix: We can express the full lower triangular matrix as:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 23/40 We can verify this answer by noting that LL ′ is indeed equal to our origi- nal covariance matrix, Σ . CHAPTER 7 1. Vectors a and b are not orthogonal, but b and c are orthogonal. We know this from their inner products, which we can calculate as follows: 2. In order for A to be an orthonormal basis, we require that the column vectors are orthogonal and have a magnitude of one. For the two column vectors to be orthogonal, we require that their inner product is zero: We next check that the column vectors have a magnitude of one:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 24/40 Both vectors are normal; therefore, the solution: makes A an orthonormal basis. 3. In order for B to be an orthonormal basis, we require that the column vectors are orthogonal and have a magnitude of one. For the two column vectors to be orthogonal, we require that their inner product is zero: Using the fact that the magnitude of the first column vector must be one: Substituting in our previous result: Both the positive and negative root are legitimate solutions. There are ac- tually two possible final answers.
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 25/40 4. Because B is an orthonormal basis, we can find the coordinate vector for x : 5. A coordinate vector, c , for x should satisfy the following equation: Working through produces three simultaneous equations: By solving and substituting in, we arrive at the final answer: c 1 = 3, c 2 = 1, c 3 = 4. CHAPTER 8 1. The expected return of XYZ is 6.01%: 2. The expected value of r XYZ is 0.07%: The variance of r XYZ is:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 26/40 To get to the last line, we use the fact that the covariance between the re- gressor and the disturbance term is zero in a linear regression, which implies: Taking the square root of the variance, we get a standard deviation of 2.06%. 3. We start by calculating the covariance: The correlation is then: 4. The R 2 is 20%: 5. The corresponding F -statistic is 12:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 27/40 Using a spreadsheet or other program, we see that the probability associ- ated with this F -statistic is 0.11%; that is, there is only a 0.11% chance that an F -statistic of this magnitude (or greater) would have happened by chance. The F -statistic is significant at the 95% confidence level. 6. We compute the adjusted R 2 for each model. The univariate model has two regressors, including the constant. The second has four: On this basis, the original univariate model is slightly better. 7. 30.5%: 8. One possible solution is to drop X 3 from the model: Another possibility, if the spread between X 2 and X 3 is of interest, is: where β 5 is taken from the regression of X 2 on X 3 . Based on the assump- tion of the OLS model, the term in parentheses will be uncorrelated with X 2 . 9. We start by writing the equation for the covariance of X and Y :
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 28/40 Using our linear regression equation and making use of the OLS assump- tions, we see that the second term can be expressed in terms of X , β , and : Substituting this into our covariance equation: All that remains is to divide both sides by the variance of X , and to ex- pand the correlation term: 10. First, we find the optimal value of α , α *: Next, we solve for β :
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 29/40 At this point we substitute in our optimal value of α , α *: CHAPTER 9 1. The models are: a. AR(2) b. AR(1) c. ARMA(2,1) d. Drift-diffusion 2. 3. The expected value of r t is 0.80% the next period, and 0.84% the follow- ing period: To get the two-period-ahead forecast, we can use the previous result:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 30/40 Alternatively, we can substitute the original equation into itself to get r t in terms of r t –2 and r t –3 : 4. The expected log return over one year is 0.0%. The standard deviation of the annual log return is 24%. We can get this result by recognizing the annual return as a collection of i.i.d. variables, and using our square root rule to calculate the standard deviation. More formally, we can construct the annual return series (re- member, log returns are additive): where r 256, t is our 256-day annual return. We can find the expected value as follows: We can then calculate the variance as follows:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 31/40 For each term in the final summation, we can determine the value by not- ing the following: Now we have: The variance of the annual returns is 256 times as great as the variance of the daily returns. To get the standard deviation, we just take the square root of both sides: 5. The expected log return over one year is 25.6%. The standard deviation of the annual log return is 24%. As before, we can get this result by recognizing the annual return as a col- lection of i.i.d. variables, and using our square root rule to calculate the standard deviation. More formally, we can construct the annual return series (remember, log returns are additive):
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 32/40 where r 256, t is our 256-day annual return. We can find the expected value as follows: Using this result, we can then calculate the variance as follows: This is exactly the same as what we had in the previous question. The ad- dition of the drift term does not impact the variance or standard devia- tion. The final result is the same as before: 6. The expected log return over two days is 0.40%. The standard deviation of the two returns is 3.0%. For the case where λ equals –0.50, the mean of the two-day return would be approximately 0.13%, and the standard de- viation would be approximately 1.73%. We start by expressing the original AR(1) equation as an infinite sum of lags of the disturbance term:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 33/40 Constructing the two-period return is fairly straightforward. Paying care- ful attention to the time subscripts, we can group the disturbance terms into one summation: where r 2, t is our two-day return. We can find the expected value as follows: We then proceed to find the variance: The standard deviation of the two-day return is then: 7. The unconditional mean of the model is equal to θ , 4%. If interest rates start out at 6%, then we would expect interest rates to be 5.00%, then 4.50%, and then 4.25% over the next three periods. This result is obtained
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 34/40 by noting that the conditional expectation for the next period's interest rate is, in this case, simply the average of the previous period's rate and the long-term mean of 4%: 8. By iteratively substituting the equation into itself, we see that this process can be written as an infinite moving average: The unconditional mean is 0: Similarly, we can find the unconditional variance. First we note that, be- cause the covariance between different disturbance terms is zero and the expected value of any individual disturbance terms is zero, we have the following: Using this and the fact that the unconditional mean is also zero: 9. Using the results from the previous question, we first derive an expres- sion for the covariance:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 35/40 For the last line we were able to eliminate the first term in the expecta- tions by noting that all the products contained different disturbance terms. From the preceding problem we know that the expected value of these cross products is zero. Because the unconditional variance is the same for both r t and r t n , finding the correlation is just a matter of divid- ing Var[ r t ]: 10. We start by expressing both r t , and r t –1 as infinite series: Next we find the mean of both series:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 36/40 Because the error terms are uncorrelated, we know that: Using this and the previous results, we calculate the variances and covariance: Finally, the correlation is: 11. To annualize the log return, we simply multiply by the number of months in a year, 12. To get the annualized standard deviation, we multi- ply by the square root of the number of periods. For skewness and kurto- sis, we divide by the square root of 12 and 12, respectively. This gives: mean = 24%; standard deviation = 5.20%; skewness = –0.29; kurtosis = 0.20. CHAPTER 10 1. We need to find h , such that: Solving, we find:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 37/40 Alternatively, the formula for the half-life can be expressed as: 2. We start by computing decay factors and values for x 2 : For the mean, using Equation 10.2 , we have: For the variance, using Equation 10.12 , we have: Finally, we can take the square root of our answer for the variance, to get the standard deviation, 22.48. 3. We start by calculating the following values:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 38/40 We then use Equation 10.2 to calculate our estimates of the mean: mean (no decay) = 0.5236; mean (decay = 0.99) = 0.5263; mean (decay = 0.90) = 0.5486. 4. We start by expanding the table from our answer to question #3: In the last line, we have used our estimate of the mean (no decay) from the previous problem. For the first estimator with no decay factor we can use Equation 3.19 to calculate the variance: For the second and third estimators, we use Equation 10.12 , and our esti- mates of the mean from the previous question:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 39/40 Taking the square root of the variances, we arrive at our final answers: standard deviation (no decay) = 0.2688; standard deviation (decay = 0.99) = 0.2676; standard deviation (decay = 0.90) = 0.2610. 5. The new estimates are 10.10%, 9.82%, and finally 9.78%. These can be found as follows: 6. The half-lives are: 7. The half-life of the EWMA estimator is approximately 11.11 days. A rec- tangular window with 22 days would have the most similar half-life, 11 days. 8. 9. Approximately 19.72%. We first update our estimate of the variance, and then take the square root:
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10/23/23, 1:31 PM Answers | Mathematics and Statistics for Financial Risk Management https://learning.oreilly.com/library/view/mathematics-and-statistics/9781118170625/OEBPS/9781118170625_epub_bm_06.htm#ans-exs-0009 40/40 10. Approximately 10.25%. We can use our updating rule: to calculate successive estimates of the variance. The estimate of the stan- dard deviation is just the square root of the variance estimator:
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