t2.review of probability (stock & watson)

L1.T2.Quantitative Analysis:

Review of Probability, Chapter 2 of Stock and Watson

FRM 2012 Practice Questions

By David Harper, CFA FRM CIPM

www.bionicturtle.com

FRM 2012 QUANT: REVIEW OF PROBABILITY 1 www.bionicturtle.com

Table of Contents

Key Ideas from Stock & Watson Chapter 2 ........................................................... 2

Question 201: Random variables ...................................................................... 3

Question 202: Variance of sum of random variables ............................................... 6

Question 203: Skew and kurtosis (Stock & Watson) ................................................ 9

Question 204: Joint, marginal, and conditional probability functions (Stock & Watson) ... 11

Question 205: Sampling distributions (Stock & Watson) ......................................... 13

Question 206: Variance of sample average ........................................................ 15

Question 207: Law of large numbers and Central Limit Theorem (CLT) ....................... 17


Key Ideas from Stock & Watson Chapter 2

Risk measurement is largely the quantification of uncertainty. We quantify uncertainty by characterizing outcomes with random variables. Random variables have distributions which are either discrete or continuous.

In general, we observe samples; and use them to make inferences about a population (in practice, we tend to assume the population exists but it not available to us)

We are concerned with the first four moments of a distribution: o Mean o Variance, the square of the standard deviation.

Annualized standard deviation is called volatility; e.g., 12% volatility per annum. o Skew (a function of the third moment about the mean): a symmetrical

distribution has zero skew or skewness o Kurtosis (a function of the fourth moment about the mean).

The normal distribution has kurtosis = 3.0 As excess kurtosis = 3 – kurtosis, a normal distribution has zero excess

kurtosis Kurtosis > 3.0 refers to a heavy-tailed distribution (a.k.a., leptokurtosis),

which will also tend to have a higher peaked. The concepts of joint, conditional and marginal probability are important. To test a hypothesis about a sample mean (i.e., is the true population mean different than

some value), we use a student t or normal distribution o Student t if the population variance is unknown (it usually is unknown) o If the sample is large, the student t remains applicable, but as it approximates the

normal, for large samples the normal is used since the difference is not material To test a hypothesis about a sample variance, we use the chi-squared To test a joint hypothesis about regression coefficients, we use the F distribution In regard to the normal distribution:

o N(mu, sigma^2) indicates the only two parameters required. For example, N(3,10) connotes a normal distribution with mean of 3 and variance of 10 and, therefore, standard deviation of SQRT(10)

o The standard normal distribution is N(0,1) and therefore requires no parameter specification: by definition it has mean of zero and variance of 1.0.

o Please memorize, with respect to the standard normal distribution: For N(0,1) Pr(Z < -2.33) ~= 1.0% (CDF is one-tailed) For N(0,1) Pr (Z< -1.645)~ = 5.0% (CDF is one-tailed)

The definition of a random sample is technical: the draws (or trials) are independent and identically distributed (i.i.d.)

o Identical: same distribution o Independence: no correlation (in a time series, no autocorrelation)

The assumption of i.i.d. is a precondition for: o Law of large numbers o Central limit theorem (CLT) o Square root rule (SRR) for scaling volatility; e.g., we typically scales a daily

volatility of (V) to an annual volatility with V*SQRT(250). Please note that i.i.d. returns is the unrealistic precondition.


Question 201: Random variables

Define random variables, and distinguish between continuous and

discrete random variables. Define the probability of an event. Define,

calculate, and interpret the mean, standard deviation, and variance of a

random variable.

201.1. Which of the following is most likely to be characterized by a DISCRETE random variable, and consequently, a discrete probability distribution (aka, probability mass function, PMF) and/or a discrete CDF?

a) The future price of a stock under the lognormal assumption (geometric Brownian motion, GBM) that underlies the Black-Scholes-Merton (BSM)

b) The extreme loss tail under extreme value theory (EVT; i.e., GEV or GPD) c) The empirical losses under the simple historical simulation (HS) approach to value at

risk (VaR) d) The sampling distribution of the sample variance

201.2. A model of the frequency of losses (L) per day, for a certain key operational process, assumes the following discrete distribution: zero loss (events per day) with probability (p) = 20%; one loss with p = 30%; two losses with p = 30%; three losses with p = 10%; and four losses with p = 10%. What are, respectively, the expected (average) number of loss events per day, E(L), and the standard deviation of the number of loss events per day, StdDev(L)?

a) E(L) = 1.20 and StdDev(L) = 1.44 b) E(L) = 1.60 and StdDev(L) = 1.20 c) E(L) = 1.80 and StdDev(L) = 2.33 d) E(L) = 2.20 and StdDev(L) = 9.60

201.3. A volatile portfolio produced the following daily returns over the prior five days (in percentage terms, %, for convenience): +5.0, -3.0, +6.0, -1.0, +3.0. Although this is a tiny sample, we have two ways to calculate the daily volatility. The first is to compute a technically proper daily volatility as an unbiased sample standard deviation. The second, a common practice for short-period/daily returns, is to make two simplifying assumptions: assume the mean return is zero since these are daily periods, and divide the sum of squared returns by (n) rather than (n-1). For this sample of only five daily returns, what is respectively (i) the sample daily volatility and (ii) the simplified daily volatility?

a) 1.65 (sample) and 2.55 (simplified) b) 2.96 (sample) and 3.00 (simplified) c) 4.11 (sample) and 3.65 (simplified) d) 3.87 (sample) and 4.00 (simplified)


201.4. Consider the following five random variables:

A standard normal random variable; no parameters needed. A student's t distribution with 10 degrees of freedom; df = 10. A Bernoulli variable that characterizes the probability of default (PD), where PD = 4%; p

= 0.040 A Poisson distribution that characterizes the frequency of operational losses during the

day, where lambda = 5.0 A binomial variable that characterizes the number of defaults in a basket credit default

swap (CDS) of 50 bonds, each with PD = 2%; n = 50, p = 2% Which of the above has, respectively, the lowest value and highest value as its variance

among the set?

a. Standard normal (lowest) and Bernoulli (highest) b. Binomial (lowest) and Student's t (highest) c. Bernoulli (lowest) and Poisson (highest) d. Poisson (lowest) and Binomial (highest)


Answers: 201.1. C. The empirical losses under the simple historical simulation (HS) approach to value at risk (VaR). Historical simulation sorts actual losses (e.g., daily) which informs an empirical and discrete distribution. Put another way, note that identifying the VaR is basically an exercise in identifying the quantile based on a "counting-type" distribution of losses; e.g., -100, -98, -97, .... -40. Another view is that in a discrete distribution the p(X = x) = f(x); contrast with a continuous, where P(X = x) = dxf(x). In a simple historical simulation of 100 losses, the probability of the worst loss, or any loss, is 1/100 = 1.0% = f(x). (note: per Dowd, there are kernel methods to effectively transform a discrete empirical into a continuous pdf, but this question says "simple" HS!) In regard to (A), lognormal is continuous. In regard to (B), EVT approaches are parametric continuous. In regard to (D), the sampling distribution of the sample variance is characterized by the continuous chi-squared distribution; i.e., we use chi-square to test the significance of a sample variance. 201.2. B. E(L) = 1.60 and StdDev(L) = 1.20 E(L) = 20%*0 + 30%*1 + 30%*2 + 10%*3 + 10%*4 = 1.6; Variance(L) = (0 - 1.6)^2*20% + (1 - 1.6)^2*30% + (2 - 1.6)^2*30% + (3 - 1.6)^2*10% + (4 - 1.6)^2*10% = 1.44; Standard deviation (L) = SQRT(1.44) = 1.20. Please note: as we are given ex ante probabilities, and not an empirical sample, there is no application of sample variance concept here; i.e., as this is not a sample and our variance is not an estimate (the value produced by an estimator), we do not need to divide the sum of squared differences by (n-1). 201.3. D. 3.87 (sample) and 4.00 (simplified) The average return = +2; The sum of squared differences = (5-2)^2 + (-3-2)^2 + (6-2)^2 + (-1 - 2)^2 + (3-2)^2 = 60. The sample variance = 60/(n-1) = 15, such that the sample standard deviation = SQRT(15) = 3.8730. The simplified standard deviation = (5^2 + -3^2 + 6^2 + -1^2 + 3^2)/5 = 4.0 While assuming that the mean = 0 is a simplifying assumption, the division by n=5 rather than n=4 is to merely rely on a different but valid estimator (MLE rather than unbiased). 201.4. C. Bernoulli (lowest) and Poisson (highest) In order: Bernoulli has variance = p(1-p) = 4%*96 = 0.0384 Binomial has variance = p(1-p)n = 2%*98%*50 = 0.980 Standard normal has, by definition, mean = 0 and variance = 1.0 Student's t has variance = df/(df-2) = 10/8 = 1.25 Poisson has lambda = variance = mean = 5 <-- easy to remember, yes?!


Question 202: Variance of sum of random variables

AIM: Calculate the mean and variance of sums of random variables.

202.1. A high growth stock has a daily return volatility of 1.60%. The returns are positively autocorrelated such that the correlation between consecutive daily returns is +0.30. What is the two-day volatility of the stock?

a) 1.800% b) 2.263% c) 2.580% d) 3.200%

202.2. A three-bond portfolio contains three par $100 junk bonds with respective default probabilities of 4%, 8% and 12%. Each bond either defaults or repays in full (three Bernoulli variables). The bonds are independent; their default correlation is zero. What is, respectively, the mean value of the three-bond portfolio and the standard deviation of the portfolio's value?

a) mean $276.00 and StdDev $46.65 b) mean $276.00 and StdDev $139.94 c) mean $276.00 and StdDev $2,176.45 d) mean $313.00 and StdDev $94.25

202.3. Assume two random variables X and Y. The variance of Y = 49 and the correlation between X and Y is 0.50. If the variance[2X - 4Y] = 652, which is a solution for the standard deviation of X?

a) 2.0 b) 3.0 c) 6.0 d) 9.0

202.4 A risky bond has a (Bernoulli) probability of default (PD) of 7.0% with loss given default (LGD) of 60.0%. The LGD has a standard deviation of 40.0%. The correlation between LGD and PD is 0.50. What is the bond's expected loss, E[L] = E[PD * LGD]?

a) 3.1% b) 4.2% c) 7.5% d) 9.3%


202.5. Portfolio (P) is equally-weighted in two positions: a 50% position in StableCo (S) plus a 50% position in GrowthCo (G). Volatility of (S) is 9.0% and volatility of (G) is 19.0%. Correlation between (S) and (G) is 0.20. The beta of GrowthCo (G) with respect to the portfolio--denoted Beta (G, P)--is given by the covariance(G,P)/variance(P) where P = 0.5*G + 0.5*S. What is beta(G, P)?

a) 0.45 b) 0.88 c) 1.39 d) 1.55

202.6. Two extremely risky bonds have unconditional probabilities of default (Bernoulli PDs) of 10% and 20%. Their (linear) correlation is 0.35. What is the probability that both bonds default?

a) 2.0% b) 4.6% c) 6.2% d) 9.7%

Answers: 202.1. C. 2.580% variance (X+Y) = variance(X) + variance (Y) + 2*covariance(X,Y), such that volatility (R1 + R2) = SQRT[variance(R1) + variance (R2) + 2*covariance(R1,R2)], such that where R = 1.6%, Two-day volatility (R1 + R2) = SQRT[1.6%^2 + 1.6%^2 + 2*1.6%*1.6%*0.30] = 2.580% Please note, per the positive autocorrelation, this 2.58% is greater than the two-day volatility per the square root rule (SRR) which assumes independence: if independent (i.e., correlation = 0), then 2-day volatility = 1.60% * SQRT(2) = 2.263% Answer (B) is incorrect because it implicitly assumes zero correlation between consecutive returns. 202.2. A. mean $276.00 and StdDev $46.65 The mean = 96%*100 + 92%*100 + 88%*100 = $276. Let B1, B2, and B3 represent the random Bernoulli variables. If independent, variance (100B1 + 100B2 + 100B3) = variance(100*B1) + variance (100*B2) + variance (*B3) = 100^2*variance(B1) + 100^2*variance(B2) + 100^2*variance(B3) = 100^2*[variance(B1) + variance(B2) + variance(B3)] = 10,000 * [4%*96% + 8%*92% + 12%*88%] = 2,176. StdDev = SQRT(2,176) = $46.65


202.3. B. 3.0 Key formulas are: variance (X - Y) = variance(X) + variance(Y) - 2*covariance(X,Y), and varainace (aX) = a^2variance(X). In this case, var(2X - 4Y) = 4var(X) + 16var(Y) - 2*2*4*StdDev(X)*StdDev(Y)*correlation(X,Y). Given var(Y) = 49: var(2X - 4Y) = 4var(X) + 16*49 - 2*2*4*StdDev(X)*7*0.5, and: var(2X - 4Y) = 4var(X) + 784 - 56*StdDev(X), so that: 652 = 4var(X) + 784 - 56*StdDev(X), 0 = 4var(X) - 56StdDev(X) + 132; if we let w = StdDev(X), in order to factor we can express: 0 = 4w^2 - 56w + 132, which factors: 0 = (4w - 12)(w - 11), such that: Either 4w - 12 = 0, and w = 3; or w = 11. So this has two solutions but one is StdDev(X) = 3 such that variance(X) = 9. 202.4. D. 9.3% If PD and LGD are independent, then EL = PD*LGD; In general, E[XY] = E[X]*E[Y] * covariance(X,Y), or E[PD * LGD] = E[PD] * E[LGD] + covariance(PD,LGD) = E[PD * LGD] = E[PD] * E[LGD] + StdDev(PD)*StdDev(LGD)*correlation(PD,LGD) = E[PD * LGD] = 7% * 60% + SQRT(7%*93%)*40%*0.5 = 9.303% 202.5. D. 1.55 variance (P) = 50%^2*9%^2 + 50%^2*19%^2+2*50%*50%*9%*21%*0.2 = 0.0127600 cov(G,P) = cov(G,0.5G+0.5S) = 0.5cov(G,G) + 0.5cov(G,S) = 0.5var(G) + 0.5cov(G,S). In this case, cov(G,P) = 0.5*19%^2 + 0.5*9%*19%*0.2 = 0.019760 beta(G,P) = cov(G,P)/variance(P) = 0.019760 / 0.0127600 = 1.5486 202.6. C. 6.2% Let respective default be represented by X and Y, such that we want: We want E[XY] = E[X]*E[Y] + covariance(X,Y) StdDev(X) = SQRT(10%*90%) = 0.30 and StdDev(Y) = SQRT(20%*80%) = 0.40, so that E[XY] = 10%*20% + 0.30*0.40*0.35 = 6.2%. Please note: if the bonds were independent, then cov(X,Y) = 0, such that E[XY] = 2.0%


Question 203: Skew and kurtosis (Stock & Watson)

AIM: Define, calculate, and interpret the skewness, and kurtosis of a

distribution

203.1. Let random variable W be distributed normally as N(0,10). What are, respectively, the following: i. The fourth moment of W, E[W^4]; and ii. The kurtosis of W?

a) 30.0 (4th moment) and zero (kurtosis) b) 100.0 and 3.0 c) 300.0 and zero d) 300.0 and 3.0

203.2. A random variable (X) has three possible outcomes: 90.0 with 40% probability; 100.0 with 50% probability; and 110.0 with 10% probability. What is the skewness of the variable's distribution?

a) -1.82 b) -0.95 c) 0.37 d) 0.74

203.3. An analyst gather information about the return distribution for two portfolios during the same period: Portfolio A shows skewness of 0.9 and kurtosis of 3.7; Portfolio B shows skewness of 1.3 and kurtosis of 2.1. The analyst asserts "Portfolio A is more peaked--that is, has a higher peak--than a normal distribution and Portfolio B has a long right tail."

a) The analyst is correct about both portfolios b) The analyst is correct about A but incorrect about B c) The analyst is correct about B but incorrect about A d) The analyst is wrong about both portfolios


Answers: 203.1. D. 300.0 (fourth moment) and 3.0 (kurtosis) N(0, 10) signifies a normal distribution with mean = 0, variance = 10; and, by definition, skew = 0, and kurtosis = 3.0. Kurtosis = E[(W-E[W])^4]/sigma(W)^4 = E[(W-E[W])^4]/variance(W)^2. In this case, therefore, 3 = E[(W-0)^4]/10^2 = E[W^4]/100. The fourth moment E[W^4] = 3*100 = 300. By the way, the fourth moment about the mean, E[(W-E[W])^4], is therefore also equal to 300, since the mean is zero. In a normal distribution, "excess kurtosis" is equal to zero as excess kurtosis = kurtosis - 3.0. 203.2. C. 0.37 Skewness = E[(X - E[X])^3]/variance(X)^(3/2). Average (X) = 97.0; E[(X - E[X])^3] = 40%*(90-97)^3 + 50%*(100-97)^3 + 10%*(110-97)^3 = 40%*-343 + 50%*27 + 10%*2197 = -137.2 + 13.5 + 219.7 = 96.0; Variance (i.e., second moment about the mean) = 40%*7^2 + 50%*3^2 + 10%*13^2 = 41.0; Skewness = 96.0/41.0^1.5 = 0.3657 203.3. A. The analyst is correct about both portfolios Portfolio A has kurtosis greater than 3.0 and, most importantly, is heavy-tailed but also leptokurtic (higher peaked). Portfolio B has positive skew and therefore has a longer right tail.


Question 204: Joint, marginal, and conditional

probability functions (Stock & Watson)

AIM: Describe Joint, marginal, and conditional probability functions

204.1. X and Y are discrete random variables with the following joint distribution; e.g., Pr (X = 4, Y = 30) = 0.07.

What is the standard deviation of the conditional probability Pr (Y | X = 7)?

a) 10.3 b) 14.7 c) 21.2 d) 29.4

204.2. Sally's commute (C) is either long (L) or short (S). While commuting, it either rains (R = Y) or it does not (R = N). Today, the marginal (aka, unconditional) probability of no rain is 75%; P(R = N) = 75%. The joint probability of rain and a short commute is 10%; i.e., P(R = Y, C = S) = 10%. What is the probability of a short commute conditional on it being rainy, P (C = S | R = Y)?

a) 10% b) 25% c) 40% d) 68%

204.3. Economists predict the economy has a 40% of experiencing a recession in 2012; marginal P(R) = 40% and therefore the marginal probability of no recession P(R') = 60%. Let P(S) be the probability the S&P 500 index ends the year above 1400, such that P(S') is the probability the index does not end the year above 1400. If there is a recession, the probability of the index ending the year above 1400 is only 30%; P(S|R) = 30%. If there is not a recession, the probability of the index ending above 1400 is 50%; P(S|R') = 50%. Bayes' Theorem tells us that the conditional probability, P(R|S), is equal to the joint probability P(R,S) divided by the marginal probability, P(S). At the end of the year, the index does end above 1400, such that we observe (S) not (S'). What is the probability of a recession conditional on the the index ending above 1400; i.e., P(R|S)?

a) 12.0% b) 28.6% c) 40.0% d) 42.0%


Answers:

204.1. A. 10.3 E(Y|X=7) = 10*(0.05/0.32) + 20*(0.03/0.32) +30*(0.13/0.32) + 40*(0.11/0.32)= 29.375. E(Y^2|X) = 10^2*(0.05/0.32) + 20^2*(0.03/0.32) +30^2*(0.13/0.32) + 40^2*(0.11/0.32)= 968.75 Variance(Y|X=7) = 968.75 - 29.375^2 = 105.8594. StdDev(Y|X=7) = SQRT(105.8594) = 10.289. 204.2. C. 40% The conditional probability Pr(C = S | R = Y ) = Pr(C = S, R = Y ) / Pr (R = Y). The marginal probability of rain Pr (R = Y) = 1 - 75% = 25%; such that The conditional probability Pr(C = S | R = Y ) = 10% / 25% = 40%. 204.3. B. 28.6% According to Bayes, P(R|S) = P(R,S) / P(S). In this case, P(R,S) = P(R)*P(S|R) = 40%*30% = 12%. P(S) = P(R)*P(S|R) + P(R')*P(S|R') = 12% + 60%*50% = 12% + 30% = 42%. Such that, P(R|S) = 12%/42% = 28.6%; i.e., the ex post knowledge of (S) decreases the conditional probability of recession from its marginal probability of 40%.


Question 205: Sampling distributions (Stock & Watson)

AIM: Describe the key properties of the normal, standard normal,

multivariate normal, Chi-squared, Student t, and F distributions.

205.1. Two variables each have a normal distribution: X = N(20,4) and Y = N(40,9). The correlation between X and Y is 0.20. Let the (J) be a bivariate normal distribution such that J = 20*X + 28*Y. What is the Pr(1,355 < J < 1,753)?

a) 90.0% b) 93.0% c) 94.0% d) 96.0%

205.2. Each of the following is true about the student t distribution EXCEPT:

a) The student t distribution has skew equal to zero; variance equal to df/(df - 2) where (df) is degrees of freedom; and kurtosis greater than 3.0 (leptokurtosis with heavier tail and higher peak compared to the normal)

b) To test of significance of a single partial slope coefficient in a (sample) multiple regression with three independent variables (aka, regressors), we use a critical t with degrees of freedom (d.f) equal to the sample size minus four (n - 4)

c) The student's t distribution is the distribution of the ratio of a standard normal random variable divided by the square root of an independently distributed chi-squared random variable with (m) degrees of freedom divided by (m)

d) For asset returns involving large sample sizes (for example, n = 1,000), the student t should be used to simulate heavy tails as asset returns exhibit heavy tails

205.3. Each of the following is true about the chi-square and F distributions EXCEPT:

a) The chi-square distribution is used to test a hypothesis about a sample variance; i.e., given an observed sample variance, is the true population variance different than a specified value?

b) As degrees of freedom increase, the chi-square approaches a lognormal distribution and the F distribution approaches a gamma distribution

c) The F distribution is used to test the joint hypothesis that the partial slope coefficients in a multiple regression are significant; i.e., is the overall multiple regression significant?

d) Given a computed F ratio, where F ratio = (ESS/df)/(SSR/df), and sample size (n), we can compute the coefficient of determination (R^2) in a multiple regression with (k) independent variabels (regressors)


Answers: 205.1. C. 94.0% Please note N(20,4) signifies mean of 20 and variance of 4. In this case, Variance (J) = 20^2*4 + 28^2*9 + 2*20*28*SQRT(4)*SQRT(9)*0.2 = 10,000; such that: standard deviation (J) = 100 Expected value of J, E[J] = 20*20 + 28*40 = 1,520 As standardized variables are (1,355 - 1,520)/100 = -1.65 and (1,753 - 1,520)/100 = 2.33, we have: Pr (1,355 < J < 1,753) in standardized terms is Pr (-1.65 < N < 2.33) = 99.0% - 5.0% = 94.0% Please note: for the exam, you cannot be expected to know the normal CDF, N(.), with two exceptions due to their common usage. You simply must know:

N(-2.33) = 1% and N(-1.645) = 5%, and due to the symmetry of the normal of course, N(1.645) = 95% and N(2.33) = 99%.

205.2. D. For large samples, the student t approximates the normal. For example, if n = 1,000, 99% one-tailed = 2.330083 compared to z = 2.32634. For large samples, the kurtosis is merely technical and the student t is NOT useful to approximate heavy tails! In regard to (A), (B), and (C), each is true. 205.3. B. False, both approach a normal distribution; all of the so-called sampling distributions (student t, chi-square, F) approach normal as d.f. --> inf. In regard to (A), (C) and (D), each is true.


Question 206: Variance of sample average

AIMs: Define, calculate, and interpret the mean and variance of the

sample average (Stock & Watson)

206.1. A portfolio is equally weighted among nine pairwise independent assets, each with identical volatility of 14.0%; i.e., n = 9, sigma (i) = 14%, weight of asset (i) = 1/9, and all correlations (i,j) = 0. If we add another independent asset with same volatility of 14.0%, such that that (n) increases to from 9 to 10, and each asset weight dilutes to 1/10, what is absolute change to portfolio volatility?

a) Zero b) Reduced by 0.24% (absolute) c) Reduced by 1.18% (absolute) d) Reduced by 2.48% (absolute)

206.2. A stock has an expected (i.i.d.) return of 9.0% per annum and volatility of 10% per annum. The distribution of the average (continuously compounded) rate of return has a mean of 8.5% per annual as 9.0% - 10.0%^2/2 = 8.5%; i.e., over several years, the average realized return is expected to be 8.5% per year. Over a five-year horizon, we can be 95% confident that the realized average (i.e. per year) per annum return will exceed what level?

a) -7.95% b) 0.85% c) 1.14% d) 4.47%

206.3. A basket credit default swap (basket CDS) references one hundred (100) very risky credits. Each credit is characterized by a random Bernoulli variable and either defaults with probability of 9.0% or does not default. Further, the credits are uncorrelated; note these two assumptions satisfy i.i.d. as the distributions are identical and independent. In this way, the basket's expected average default rate is 9.0%. What is the 95% confidence interval, around this expected mean, for basket's average default rate?

a) 3.4% < E[average default rate] < 14.6% b) 4.3% < E[average default rate] < 13.7% c) 5.2% < E[average default rate] < 12.6% d) 6.1% < E[average default rate] < 11.9%


Answers: 206.1. B. Reduced by 0.24% Variance (n = 9) = 14%^2/9, such that portfolio volatility = SQRT(14%^2/9) = 4.6667% portfolio volatility (n = 10) = SQRT(14%^2/10) = 4.4272% Improvement = 4.4272% - 4.6667% = -0.2395%. 206.2. C. 1.14% The AVERAGE return has a variance of 10%^2/5 = 0.0020; such that the AVERAGE return has volatility of SQRT(10%^2/5) = 4.47% The mean is 8.5% per year, such that 95% confidence bound is given by 8.5% - 1.645*4.47% = 1.143% 206.3. A. 3.4% < E[average default rate] < 14.6% The variance of a single credit = p*(1-p) = 9%*(1-9%) such that the variance of the AVERAGE default rate of 100 i.i.d. credits is (9%*91%)/100 = 0.000819; The standard deviation of the average default rate = SQRT[(9%*91%)/100] = 2.86%. The 95% confidence interval is: 9.0% +/- 1.96*2.86%: 3.39% < average default rate < 14.61% Please note: here the individual variable is non normal (Bernoulli). But that doesn't matter, CLT only requires i.i.d. and tells us that the average will tend toward (converge) a normal distribution as the sample increases.


Question 207: Law of large numbers and Central Limit Theorem (CLT)

AIMs: Define and describe random sampling and what is meant by i.i.d.

Describe, interpret, and apply the Law of Large Numbers and the Central

Limit Theorem (Stock & Watson)

207.1. Analyst Joe wants to apply the square root rule (SRR) to scale daily asset volatility into monthly asset volatility. For example, if the daily volatility is (D), then the scaled monthly volatility will be given by M = D*SQRT(20). Consider the following possible assumptions:

I. Each daily return has a normal distribution, although the mean and variance varies II. Knowledge of today's return gives no information about tomorrows return

III. Daily returns are autocorrelated (positive serial correlation) IV. Each daily return is non-normal, with heavy tail, although distributional moments are

constant Joe is informed that application of the square root rule (SRR) requires that returns are i.i.d. Therefore, which of the above assumptions is (are) necessary to legitimately scale the volatility?

a) I. only b) I. and II. c) II. and III. d) II. and IV.

207.2. Unrealistically but somehow, Analyst Sally has determined that the true population default rate of a single BB-rated bond is 1.0%; i.e., default is a Bernoulli with p(default) = 1.0%. She is analyzing a collateralized debt obligation (CDO) which references a sample size of (N) of these BB-rated bonds. She is interested in the default rate of the entire sample referenced by the CDO. Which of the following most nearly summarizes the law of large numbers?

a) As N increases, the number of defaults (D) must also increases b) As N increases, the CDO's default rate of D/N will exceed 1.0% with a probability that

tends toward 1.0. c) As N increases, the CDO's default rate of D/N will converge to 1.0%, but only if bond

defaults are i.i.d. d) As N increases, the CDO's default rate of D/N will converge to 1.0% and no assumptions

are required


207.3 Portfolio Manager Roger is analyzing a universe of potential stocks; unrealistically, it happens to be that each stock offers identical expected returns (mu) of 6.0% and identical standard deviations (sigma) of 18.0%. Roger wants to invest in a sample size of (N) of the stocks. Let (R) equal the average return of the invested sample. Roger argues that, according to the Central Limit Theorem, as (N) increases, the distribution of [(R - 6.0%)/SQRT(18%^2/N)] becomes increasingly well approximated by the standard normal distribution. Which of the following best characterizes Roger's argument about the CLT?

a) Roger is correct and no further conditions are required b) Roger is correct, but only if he can assume that the stock returns are characterized by a

normal distribution c) Roger is correct, if he can also assume independence among returns, but the stock

returns do NOT need to be normally distributed d) Roger is plainly incorrect, his argument is unrelated to the CLT

Answers: 207.1. D. II. and IV. Normality is not required. The square root rule (SRR) assumes i.i.d. returns. "Knowledge of today's return gives no information about tomorrows return" signifies INDEPENDENCE. "Distributional moments are constant" signifies IDENTICAL DISTRIBUTIONS. 207.2. C. As N increases, the CDO's default rate of D/N will converge to 1.0%, if bond defaults are i.i.d. 207.3 C. Roger is correct, if he can also assume independence among returns, but the stock returns do NOT need to be normally distributed The CLT does not require the population distribution to be normal. Rather, it requires i.i.d. and finite variance.

t2.review of probability (stock & watson)

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