P1.T2. Quantitative Analysis

Bionic Turtle FRM Practice Questions

Global Topic Drill

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GLOBAL TOPIC DRILL .................................................................................................................. 1 P1.T2.316. DISCRETE DISTRIBUTIONS (TOPIC REVIEW) .............................................................................. 3 P1.T2.317. CONTINUOUS DISTRIBUTIONS (TOPIC REVIEW) ......................................................................... 6 P1.T2.318. DISTRIBUTIONAL MOMENTS (TOPIC REVIEW) ............................................................................ 9 P1.T2.319. PROBABILITIES (TOPIC REVIEW) ........................................................................................... 12 P1.T2.320. STATISTICAL INFERENCE: HYPOTHESIS TESTING AND CONFIDENCE INTERVALS (TOPIC REVIEW)...... 15 P1.T2.321. UNIVARIATE LINEAR REGRESSION ......................................................................................... 17 P1.T2.322. MULTIVARIATE LINEAR REGRESSION ..................................................................................... 20 P1.T2.323. MONTE CARLO SIMULATION AND GBM (TOPIC REVIEW) ......................................................... 24 P1.T2.324. ESTIMATING VOLATILITY (TOPIC REVIEW) .............................................................................. 28

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P1.T2.316. Discrete distributions (Topic review) 316.1. Assume a 99.0% daily value at risk (VaR) model is perfectly accurate. Specifically, among 100 days, we expect a loss that exceeds VaR on exactly one day. The probability mass function of a binomial distribution is given by:

Over a series of 20 trading days, which is nearest to the probability that the daily loss will exceed VaR on exactly two (2) days?

a) 0.44% b) 0.93% c) 1.59% d) 2.36%

316.2. The frequency of "Damage to Physical Assets," an operational risk event type, is characterized by a Poisson distribution. Over an average year, a company expects 36 of these particular loss events. During the next month, which is nearest to the probability the company will experience exactly zero (none) of these events?

a) 3.4% b) 5.0% c) 7.5% d) 9.1%

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316.3. The current price of an asset is S(0) and its future evolution is modeled with a binomial tree. At each node, there is a 62% probability of an (jump-up) increase. If the price increases, the next price is given by S(T+1)=S(T)*1.050; if the price decreases, the next price is given by S(T+1)=S(T)/1.050. Over the next year, there are twelve time steps, one for each month. Which is nearest to the probability that the final price of the asset is exactly the same as its current current price, S(0)?

a) 15.80% b) 19.33% c) 22.25% d) 33.67%

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Answers:

316.1. C. 1.59% if k=2, p=1% and n=20, then Pr(X=2) = C(20,2)*1%^2*99%^18 = 190*0.00010*0.83451 = 1.5856% 316.2. B. 5.0% Poisson Prob[X=0 | mean = 36/12 = 3] = exp(-3)*3^0/0! = exp(-3) = 4.9787% 316.3. A. 15.80% To recombine to the current price, the asset must increase six times and decrease six times (the sequence does not matter). The up/down magnitudes are not relevant here. This probability is given by the binomial: Pr [X = 6 | p = 62% & n = 12] = C(12,6)*62%^6*38%^6 = 15.8024% Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-316-discrete-distributions-topic-review.7220/

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P1.T2.317. Continuous distributions (Topic review) 317.1. For a sample of ten (n = 10) market-neutral hedge funds, analyst Peter finds their sample average monthly excess return to be +0.890% with a sample standard deviation of 1.00%. He wants to conduct a one-tailed test of significance, specifically: his null hypothesis is that the true excess return is equal to, or less than, zero. Here is a snippet of the student's t lookup table:

Which is nearest to his p-value? (bonus: use the p-value in a sentence)

a) 0.01% b) 0.50% c) 1.00% d) 2.00%

317.2. Let X and Y represent the rates of return (in percent) on two stocks. Their distributions are approximately normal: X ~ N(6, 3) and Y ~ N(10, 8) assuming conventional notation of N(mean, variance). The variables are positively correlated with a correlation coefficient of 0.20. Each of the following is true except which is false?

a) (X+Y) ~ N(16, 13) b) (X-Y) ~ N(-4, 13) c) 5X ~ N(30, 75) d) (4X + 5Y) ~ N(74, 287)

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317.3. The chart below graphs a mixture distribution, X, of two normal components: N(0, 2) which receives a weight of 30% and N(4,1) which receives the remainder weight of 70%:

Which is nearest to probability that the mixture distribution outcome is less than zero; i.e., Prob[X < 0]?

a) 15.0% b) 30.0% c) 47.5% d) 70.0%

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Answers: 317.1. C. 1.00% The standard error = 1%/SQRT(10) = 0.003162; Tip: use STO 1, then RCL 1 to store values in your calculator. The computed t = (0.89% - 0)/0.003162 = 2.8144. At 10 - 1 = 9 degrees of freedom, this is very close to 2.821 and therefore corresponds to a one-sided p-value of 1.0% ... to be exact, T.DIST.RT(9 df, 2.8144) = 1.0115% In a sentence, we could say: "We are approximately 99.0% confident [i.e., 98.988% confident] that we should reject the null and decide the true average hedge fund excess return is greater than zero." 317.2. B. (X-Y) ~N(-4, 9) not N(-4, 13). In regard to (A), (C), and (D), each is TRUE. 317.3. A. 15.0% = 30%*50%. Prob(mixture)[X < 0] = 0.30*Prob(1st component)[X < 0] * 0.70*Prob(2nd component)[X < 0]. In regard to the 1st component, Z = (0-0)/sqrt(2) = 0 In regard to the 2nd component, Z = (0-4)/sqrt(1) = -4 = -4.0. Such that: Prob(mixture)[X

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P1.T2.318. Distributional moments (Topic review) 318.1. Consider a discrete distribution where Pr(X = -1) = 10%, Pr(X = 0) = 50%, and Pr(X = 1) = 40%, as below:

If the distribution's variance is 0.410, which is nearest to this distribution's skew?

a) -0.37 b) 0.00 c) +0.25 d) +1.08

318.2. Over the previous five days an asset produced the following daily returns, given in percentages (%): -2.0, +5.0. -4.0, +2.0 and +6.0. Assume we calculate the daily volatility with an assumption of zero mean (i.e., excluding the mean) and also assume we will rely on a maximum likelihood variance estimate (MLE); that is, we will divide the sum of squared returns by the sample size (n) rather than (n - 1). Which is nearest to the daily standard deviation, in percentage (%) terms?

a) 2.65 b) 3.88 c) 4.12 d) 5.90

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318.3. The exhibit below shows moment calculations for a Poisson distribution truncated at x = 30: f(x) is computed for x = {0, 1, ... 30} where, in this distribution, P(x = 30) is virtually zero.

Which are nearest to the distribution's skew (S) and kurtosis (K)?

a) S = -0.3 and K = 2.75 b) S = +0.5 and K = 3.25 c) S = +0.1 and K = 5.90 d) S = -1.6 and K = 3.00

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Answers:

318.1. A. -0.37 The mean = 0.30 and the third (3rd) central moment about the mean = (-1.0 - 0.30)^3*10% + (0 - 0.30)^3*50% + (1.0 - 0.30)^3*40% = -0.0960. Skew = -0.0960/[0.41^(3/2)] = -0.3657 318.2. C. 4.12 MLE variance, assuming zero mean = [(-2)^2 + 5^2 + (-4)^2 + 2^2 + 6^2]/5 = 17, and standard deviation = SQRT(17) = 4.1231 318.3. B. S = +0.5 and K = 3.25 Skew = 3rd central moment / StdDev^3 = 4.0/[4^(3/2)] = 0.5. Kurtosis = 4th central moment / StdDev^4 = 52.0/4^2 = 52/16 = 3.25. Note, the mean and variance of this distribution are 4.0 such that lamba is 4.0. The skew of a Poisson = 1/sqrt(lambda) and excess kurtosis = 1/lambda; in this case, skew = 1/sqrt(4) = 0.5 and excess kurtosis = 1/4 = 0.25. Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-318-distributional-moments-topic-review.7232/

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P1.T2.319. Probabilities (Topic Review) 319.1. The current yield is 4.0% (with annual compounding) such that the price of a 12-year zero-coupon bond is $62.46, because 62.46 = 100/1.04^12. Assume the yield is normally distributed with basis point volatility of 85 basis points. For reference, here is a snippet of the standard normal CDF lookup table:

Which is nearest to the probability the bond price will fall below $60.00? (note: variation on FRM handbook discussion 2.3.5; single period assumed)

a) 29.5% b) 34.1% c) 48.0% d) 55.3%

319.2. The following is a probability matrix for X = {1, 2, 3} and Y = {1, 2, 3}; i.e., the Joint Prob (X = 3, Y = 3) = 18.0%:

Each of the following is true EXCEPT which is false?

a) X and Y are independent b) The covariance(X,Y) is non-zero c) The probability Y = 3 conditional on X = 1 is 10.0%; i.e., Prob (Y = 3, X = 1) = 10.0% d) The unconditional probability that X = 2 is 50.0%; i.e., Prob (X = 2) = 50.0%

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319.3. A young credit agency has been rating bonds for the last five years. Peter analyzed their history. Among their coverage universe, 10% were rated "investment grade," 30% were rated "junk" and 60% were "Unrated." As such these are unconditional probabilities; i.e., Prob(I) = 10%, Prob(J) = 30%, and Prob (U) = 60%. He also computed conditional probabilities, as shown below. For example, the probability of an upgrade (within five years) conditional on an Investment rating is 8.0%; i.e., Prob (U | I) = 8%.

If a bond was downgraded, which is nearest to the probability it was rated Junk; i.e., what is Prob (J | D)?

a) 14.0% b) 28.0% c) 52.5% d) 63.6%

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Answers:

319.1. B. 34.1% As P = 100/(1+r)^T, r = (100/P)^(1/T) - 1. To breach 60, the rate must rise above 4.3488% = (100/62.46)^(1/12)-1. As Z = (4.3488% - 4.0%)/0.85% = 0.41034, the probability the normally distributed rate will go above 4.3488% is given by 1 - Pr(Z 4.3488%) = 1 - Prob(Z < 0.41034) = 1 - 0.6591 = 34.08% 319.2. B. Because X and Y are independent, their covariance is zero (it is not true that we can infer independence from zero covariance, however!). In regard to (A), X and Y are independent because for all cells in the matrix: the Joint Prob (X, Y) = Prob(X)*Prob(Y). In regard to (C) and (D), these are TRUE. 319.3. C. 52.50%, see below:

Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-319-probabilities-topic-review.7239/

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P1.T2.320. Statistical inference: hypothesis testing and confidence intervals (topic review) 320.1. Recently 25 banks were surveyed. Their sample average total capital is 8.40% (i.e., Tier 1 plus Tier 2 as a percentage of risk-weighted assets, RWA) with a sample standard deviation of 1.0%. Our one-sided null hypothesis is that the population's "true" average total capital is less than or equal to 8.0%. With 95.0% confidence, can we accept the one-sided alternative hypothesis that the population's average total capital is greater than 8.0%?

a) No, we do not accept the alternative because our computed test statistic of 2.797 is greater than the critical (lookup) t value of 2.000

b) No, we do not accept the alternative because our computed test statistic of 2.000 is less than the critical (lookup) t value of 2.064

c) Yes, we accept the alternative because our computed test statistic of 2.000 is greater than the critical (lookup) t value of 1.711

d) Yes, we accept the alternative because our computed test statistic of 1.318 is less than the critical (lookup) t value of 2.492

320.2. Sixteen (16) financial institutions are surveyed. Their average overnight rate is 40.0 basis points with a sample standard deviation of 9.0 basis points. Which is nearest to the width of the two-sided 99.0% confidence interval (C.I.) for the "true" population mean overnight rate, where width of C.I. is upper limit minus lower limit?

a) 4.75 bps b) 9.59 bps c) 13.26 bps d) 24.07 bps

320.3. A random sample of 18 daily returns generated by a low-volatility exchange-traded fund (ETFs) produces a sample daily volatility of 10 basis points. What is the two-sided 95.0% confidence interval for the fund's true (population) daily return volatility?

a) 4.0 to 16.0 b) 7.0 to 13.0 c) 7.5 to 15.0 d) 8.5 to 16.3

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Answers:

320.1. C. Yes, he accepts alternative because his computed test statistic of 2.000 is greater than the critical (lookup) t value of 1.711 The standard error = 1.0%/SQRT(25) = 0.20%. The test statistic (computed t) = (8.40% - 8.0%)/0.20% = 2.00. The one-sided 95.0% confident critical (lookup) t value with 24 degrees of freedom is 1.711 = T.INV(0.95,24) As 2.0 > 1.711, Peter rejects the null in favor of the alternative; the one-tailed p-value is 2.85%, such that he does would not reject at 99.0% however. 320.2. B. 9.59 bps As the upper/lower limits = sample mean +/- (critical t)*(standard error), the two-sided width of the C.I. = 2*(critical t)*(standard error). In this case, width of C.I. = 2 * 2.947 * 9.0/SQRT(16) = 13.2602, as 2.947 = T.INV.2T(1% sig,15 df). Two-sided CI = (33.37, 46.63) 320.3. C. 7.5 to 15.0 The critical chi-square values are = CHISQ.INV.RT(97.5%, 17 df) = 7.56 and CHISQ.INV97.5%, 17 df) = 30.19, such that: Lower limit, variance = 10^2*17/30.19 = 56.31; and Upper limit, variance = 10^2*17/7.56 = 224.74. As variance CI = (56.31, 224.74), the volatility CI = (SQRT[56.31], SQRT[224.74]) = (7.50, 14.99). Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-320-statistical-inference-hypothesis-testing-and-confidence-intervals-topic-review.7251/

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P1.T2.321. Univariate linear regression 321.1. Using Stock & Watson's CPS08 dataset, we regressed wage and age data for 7,711 workers. The following is the resulting regression model of average hourly earnings (AHE) against worker's age:

Are the slope and intercept coefficients significant with 95% confidence?

a) Neither is significant with 95% confidence b) The intercept (1.0823) is significant, but the slope is insignificant c) The slope (0.6050) is significant, but the intercept is insignificant d) Both are significant with 95% confidence

321.2. Samantha Xiao is trying to get some insight into the relationship between the stock returns of Facebook (FB) and the NASDAQ-100 (QQQ). Using historical returns, she estimates the following:

The mean return for FB is 9.0% with volatility of 30.0% The mean return for QQQ is 6.0% with volatility of 20.0% The correlation coefficient between FB and QQQ is 0.40

If she uses the same data to regress FB returns (dependent variable) against QQQ returns (independent variable), which of the following models will she obtain? (note: variation on GARP's 2013 Practice Exam Part 1, Question 8)

a) return(FB) = 1.00% - 0.10*return(QQQ) + e b) return(FB) = 3.20% + 0.30*return(QQQ) + e c) return(FB) = 5.40% + 0.60*return(QQQ) + e d) return(FB) = 7.80% + 0.90*return(QQQ) + e

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321.3. Using Stock & Watson's dataset (Growth), we performed a univariate regression of average annual percentage GDP growth (Growth, the dependent regressand) against average number of years of schooling of the country's adult residents (YearsSchool, the independent regressor) for 65 countries. The regression model is given by the following:

Each of the following is true about this regression model EXCEPT which is false?

a) The coefficient of determination, R^2, is approximately 0.1094 b) The 95.0% confidence interval for the "true" slope coefficient is approximately 0.158

to 0.336 c) SER = 1.80 = SQRT(SSR/df) = SQRT(205.1/63) d) The correlation coefficient, r, is approximately 0.331

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Answers:

321.1. C. The slope (0.6050) is significant, but the intercept is insignificant The t (test) statistic for the slope coefficient = (0.6050 - 0)/0.0399 ~= 15.2 with p value ~ 0. The t (test) statistic for the intercept coefficient = (1.0823 - 0)/1.1843 ~= 0.914 with p value ~ 36%. 321.2. C. return(FB) = 5.40% + 0.60*return(QQQ) Slope = cov(FB,QQQ)/var(QQQ) = 20%*30%*0.40/20%^2 = 0.60. Intercept = E(FB) - slope*E(QQQ) = 9.0% - 0.60*6.0% = 5.40% 321.3. B. 95% CI for slope = 0.2470 +/- 0.0887*1.96 = 0.0731 to 0.4209 In regard to (A), (C) and (D), each is TRUE. Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-321-univariate-linear-regression.7256/

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P1.T2.322. Multivariate linear regression 322.1. Using the dataset from Stock & Watson (S&W), we used Excel's LINEST() function to regress student test scores (TestScore, the dependent regressand) against three independent variables: student teacher ratio (STR); percent of of English learners (EL_PCT); and percent qualifying for reduced-price lunch (MEAL_PCT). The multivariate regression results, which are also given by S&W 7.19, are displayed below:

The dataset includes 419 observations. Each of the following is true about the regression model EXCEPT which is false?

a) Each of the single regression coefficients are individually significant with 95.0% confidence

b) The standard error of regression (SER) = SQRT[33,910/(419 - 3 - 1)] c) The adjusted R^2 is less than the unadjusted R^2 of 0.777= 117,908/(117,908 +

33,910) d) The atypical relationship between the F ratio and the R^2 suggests a hypothesis of

(i.e., provide evidence for) omitted variable bias

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322.2. The following regression model is from Stock & Watson's (S&W) "Growth" dataset. The model regresses annual growth in real Gross Domestic Product (GROWTH) against five independent regressors: real per capital GDP (RGDP); average share of trade in the economy (TRADE); average number of years of schooling of adult residents (SCHOOL); average annual number of revolutions or insurrections (REVOL); and average annual number of political assassinations (ASSASSIN):

The dataset contains 65 countries (i.e., SSR df = 65 - 5 partial slope estimates - 1 intercept estimate) and the displayed R^2 of 0.359 is an unadjusted R^2 as suggested by ESS and SSR. Each of the following is true about this regression model EXCEPT which is false?

a) The partial effect of SCHOOL on GROWTH (the dependent variable) is positive and significant

b) A 95.0% confidence interval for the true value of SCHOOL is about 0.30 to 0.85 c) The correlation coefficient between GROWTH and the independent variables is about

0.60 = SQRT(0.359) d) The adjusted R^2 = 0.305 = 1 - [(148/231)*(65-1)/(65-5-1)]

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322.3. Each of the following statements is true about the classical linear regression model (CLRM) which uses the method of ordinary least squares (OLS) to estimate the coefficients in a multivariate regression EXCEPT which is false?

a) Omitted variable bias occurs if both (i) an omitted variable is a determinant of the dependent variable and (ii) at least one of the included independent variables (regressors) is correlated with the omitted variable

b) If we add an additional regressor and the adjusted R^2 increases, then this implies (i) the regressor is statistically significant, (ii) the regressor is a true cause of the dependent variable and (iii) there is no omitted variable bias

c) If the regression error is heteroskedastic, the OLS estimators are no longer "efficient" (they are linear and unbiased, but not BLUE); however, heteroskedasticity-robust standard errors are available and advisable

d) The CLRM OLS assumes no perfect multicollinearity (i.e., no exact collinearity between regressors), however imperfect does not necessarily prevent estimation of of the regression nor does it imply a logical problem with the choice of regressors

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Answers:

322.1. D. False. The high F ratio suggests we can reject a joint hypothesis that the coefficients are (jointly) zero; the high R^2 (which is high even when adjusted) suggests "merely" that the included regressors are good at predicting the dependent variable in the sample of data. In regard to (A), (B) and (C), each is TRUE. In regard to (A), the t-stats and p-value are: MEAL_PCT: t-stat = 25.35, p-value ~= 0.00% EL_PCT: t-stat = 3.88, p-value = 0.01204% STR: t-stat = 4.27, p-value = 0.00243%, i-cept: t-stat = 150.08, p-value ~= 0.00000% In regard to (C), the adjusted R^2 = 1 - [(n-1)/(n-k-1)]*SSR/TSS = 1 - [(419-1)/(419-3-1)]*33,910/(33,910+117,908) = 0.775 < 0.777; i.e., the adjusted R^2 is always less than the unadjusted R^2 322.2. C. False. The correlation coefficient (r) is between two variables; we can retrieve r = SQRT(R^2) only in a univariate regression. In regard to (A), (B) and (D), each is TRUE. 322.3. B. None of the three implications are true! In regard to (A), (C), and (D), each is TRUE. Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-322-multivariate-linear-regression.7264/

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P1.T2.323. Monte Carlo Simulation and GBM (Topic Review) Note: For the questions below, we provide the following snippet from the standard normal CDF lookup table:

323.1. Assume a stock price following a geometric Brownian motion (GBM) process such that the natural log of the future stock price, S(t), is approximately normal:

The current stock price, S(0), is $48.00 with volatility of 36.0% per annum. The expected return (mu) is 18.0% per annum. Assuming this model, which is nearest to the probability that the stock price will end the year (T = 1.0) above $60.00? [Sidebar bonus question: Why is the answer different than N(d2) in BSM, or is it the same?]

a) 17.03% b) 24.55% c) 38.21% d) 44.82%

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323.2. Analyst Peter needs to price an American call option with one year to maturity. He has convenient access to only two option pricing models: a Black-Scholes Merton (BSM OPM), which he presumes is exact due to its analytical solution; and a binomial tree with twelve steps, which he presumes is inaccurate due to the low number of steps. He performs the following calculations:

For the American option, the binomial model returns a price of $8.70 (which he realizes is inaccurate)

He runs the binomial model again, on the same call option, except he assumes it is a European option (i.e., same option parameters but cannot be early exercised); this variant of the binomial returns a price of $8.25.

He prices this European option, also, with his Black-Scholes (which he trusts is exact for the European option) and it gives a price of $8.12.

Based on these findings, he estimates the value of the American option to be equal to $8.70 + ($8.12 - $8.25) = $8.57.

Which of the following methods did Peter use to estimate the price of the American option?

a) Antithetic variable technique b) Control variate technique c) Quasi-random sequences d) None of the above, Peter's method is without merit

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323.3. A Monte Carlo simulation (MCS) assumes that a non-dividend-paying stock follows the process given by the discrete-time version of geometric Brownian motion (gBm):

The initial stock price, S(0), is $100.00. The expected return (mu) is 15.0% per annum. The volatility assumption is 40.0% per annum. Each step in the simulation is five days and the number of trading days per year is 250, such that each time step is 5/250. The random standard normal is generated by inverse transformation of a random uniform variable (with a value from zero to 1.0).

For the first week, the random uniform variable is 0.8980 For the second week, the random uniform variable is 0.2429

Which are nearest to the simulated stock prices, respectively, at the end of the first and second week?

a) $107.48 then $103.55 b) $103.29 then $98.75 c) $97.25 then $101.51 d) $102.78 then $104.14

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Answers:

323.1. C. 38.21% LN(48)+ (0.18 - 0.36^2/2)*1.0 = 3.98640, such that: LN[S(t)] ~ N (3.98640, 0.36^2). Because LN(60) = 4.09434, Prob [S(1.0) > 60] is given by: 1 - N[(4.09434 - 3.98640)/0.36] = 1 - N(0.2984); since N(0.30) = 61.79%, Prob [S(1.0) > 60] ~= 1 - 61.79% ~= 38.21% Sidebar bonus question: Why is the answer different than N(d2) in BSM, or is it the same? N(d2) in BSM employs the risk-free rate, rather then (mu), so it will be lower; for example, if the risk-free rate is 4.0%, then N(d2) in BSM is 24.55% which would represent the probability of exercise in a risk-neutral world. 323.2. B. Control variate technique 323.3. A. $107.48 then $103.55 Per inverse transform, random uniform 0.8980 implies random standard normal of 1.270; and random uniform 0.2420 implies random standard normal of -0.700. S(first week) = 100 + 100*[15%*0.02 + 1.270*40%*sqrt(0.02)] = $107.4842; S(second week) = 107.4842 + 107.4842*[15%*0.02 + (-0.70)*40%*sqrt(0.02)] = $103.5505. Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-323-monte-carlo-simulation-and-gbm-topic-review.7274/

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P1.T2.324. Estimating volatility (Topic Review) 324.1. Consider the following GARCH(1,1) model:

Each of the following is true about this GARCH(1,1) model EXCEPT which is false?

a) If the volatility(n-1) is 5.0% and the return(n-1) is 3.0%, then the updated volatility estimate is about 4.6390%

b) The 10-day forward forecast (n+10) is a daily volatility of about 2.980% c) The model's long-run (unconditional) volatility is 3.3% d) The model's persistence is 0.880

324.2. Analyst Barbara employs an exponentially weighted moving average (EWMA) volatility model to generate a current daily volatility estimate of 2.15%. This EWMA model updates yesterday's volatility of 2.0% with yesterday's daily return of +3.0%. Which is nearest to her lambda parameter?

a) 0.79340 b) 0.87550 c) 0.91810 d) 0.95430

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324.3 Consider the historical series of stock prices below, including the daily returns and the squared daily returns (Return^2). Although only the previous nine days are displayed, the actual horizon includes 55 trading days. The volatility estimates computed for the previous day, t-1, are given for each of the GARCH(1,1), EWMA and moving average (MA) methods. Also given are the lambda for the EWMA model and the GARCH parameters; e.g., the unconditional long-run variance, V(L), is 3.0%^2; the gamma weight applied to this long-run variance is 10%, such that omega = 10%*3.0%^2 = 0.000090:

Today's price drop contributes a dramatic -10.00% return to the series! Which are nearest to the updated volatility estimates given by, respectively, GARCH(1,1) and EWMA? Bonus question: what are the WEIGHTS assigned by each of these models to the (t-5) return^2; i.e., the fifth previous return was also dramatic at -11.12%, what weight is assigned to its square, under each method?

a) GARCH = 4.73% and EWMA = 4.90% b) GARCH = 4.95% and EWMA = 7.00% c) GARCH = 5.25 % and EWMA = 5.25% d) GARCH = 5.25% and EWMA = 5.74%

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Answers:

324.1. C. The model's long-run (unconditional) volatility = SQRT[ 0.0000480 / (1 - 0.06 - 0.82)] = 2.0% In regard to (A), (B) and (D), each is TRUE. 324.2. B. 0.87550 EWMA variance(t) = *variance(t-1) + (1-)*return(t-1)^2, such that: variance(t) = *variance(t-1) + return(t-1)^2 - *return(t-1)^2, and variance(t) - return(t-1)^2 = *variance(t-1) - *return(t-1)^2 = *[variance(t-1) - return(t-1)^2], and = [variance(t) - return(t-1)^2] / [variance(t-1) - return(t-1)^2] = [2.15%^2 - 3.0%^2] / [2.0%^2 - 3.0%^2] = 0.87550 324.3. B. GARCH = 4.95% and EWMA = 7.00% GARCH(1,1) updated volatility estimate (t) = SQRT[omega + alpha*return(t-1)^2 + beta*variance(t-1)] = SQRT[0.000090 + 0.080*0.010 + 0.820*4.3667%^2] = 4.95336% EWMA updates volatility estimate (t) = SQRT[ *variance(t-1) + (1-)*return(t-1)^2 ] = SQRT[ 0.820 *6.1528%^2 + (1 - 0.820) * 0.010 ] = 7.0030% Bonus: In regard to weights, EWMA weight (t-5) = (1-)*^(t-1) = (1-0.820)*0.820^(5-1) = 8.1382%; GARCH weight (t-5) = alpha*beta^(t-1) = 0.080*0.820^(5-1) = 3.6170% Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-324-estimating-volatility-topic-review.7277/

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