## LO 1.2: Estimate VaR using a parametric approach for both normal and lognormal

LO 1.2: Estimate VaR using a parametric approach for both normal and lognormal return distributions.
In contrast to the historical simulation method, the parametric approach (e.g., the delta- normal approach) explicitly assumes a distribution for the underlying observations. For this LO, we will analyze two cases: (1) VaR for returns that follow a normal distribution, and (2) VaR for returns that follow a lognormal distribution.
Norm al VaR
Intuitively, the VaR for a given confidence level denotes the point that separates the tail losses from the remaining distribution. The VaR cutoff will be in the left tail of the returns distribution. Flence, the calculated value at risk is negative, but is typically reported as a positive value since the negative amount is implied (i.e., it is the value that is at risk). In equation form, the VaR at significance level a is:
VaR(a% ) = pp/L + CTP/L x za
where p and a denote the mean and standard deviation of the profit/loss distribution and 2: denotes the critical value (i.e., quantile) of the standard normal. In practice, the population parameters (i and a are not likely known, in which case the researcher will use the sample mean and standard deviation.
Example: Computing VaR (normal distribution)
Assume that the profit/loss distribution for XYZ is normally distributed with an annual mean of \$ 13 million and a standard deviation of \$ 10 million. Calculate the VaR at the 93% and 99% confidence levels using a parametric approach.
VaR(5%) = -\$15 million + \$10 million x 1.65 = \$1.5 million. Therefore, XYZ expects to lose at most \$1.5 million over the next year with 95% confidence. Equivalently, XYZ expects to lose more than \$1.5 million with a 5% probability.
VaR(l%) = -\$15 million + \$10 million x 2.33 = \$8.3 million. Note that the VaR (at 99% confidence) is greater than the VaR (at 95% confidence) as follows from the definition of value at risk.
Now suppose that the data you are using is arithmetic return data rather than profit/loss data. The arithmetic returns follow a normal distribution as well. As you would expect, because of the relationship between prices, profits/losses, and returns, the corresponding VaR is very similar in format:
VaR(a%) = (pr + CTr x za ) x Pt-1
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Example: Computing VaR (arithmetic returns)
A portfolio has a beginning period value of \$100. The arithmetic returns follow a normal distribution with a mean of 10% and a standard deviation of 20%. Calculate VaR at both the 95% and 99% confidence levels.
VaR(l%) = (-10% + 2.33 x 20%) x 100 = \$36.6
Lognormal VaR
The lognormal distribution is right-skewed with positive outliers and bounded below by zero. As a result, the lognormal distribution is commonly used to counter the possibility of negative asset prices (P ). Technically, if we assume that geometric returns follow a normal distribution (jiR, crR), then the natural logarithm of asset prices follows a normal distribution and P follows a lognormal distribution. After some algebraic manipulation, we can derive the following expression for lognormal VaR:
VaR(a%) = Pt1 x (l – e^R -^R *^)
Example: Computing VaR (lognormal distribution)
A diversified portfolio exhibits a normally distributed geometric return with mean and standard deviation of 10% and 20%, respectively. Calculate the 5% and 1% lognormal VaR assuming the beginning period portfolio value is \$100.
Lognormal VaR(5%) = 100 x (1 exp[0.1 0.2 x 1.65])
= 100 x (1 – exp[-0.23]) = \$20.55
Lognormal VaR(l%) = 100 x (1 exp[0.1 0.2 x 2.33])
= 100 x (1 exp[0.366]) = \$30.65
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Note that the calculation of lognormal VaR (geometric returns) and normal VaR (arithmetic returns) will be similar when we are dealing with short-time periods and practical return estimates.
E x p e c t e d S h o r t f a l l

## LO 1.1: Estimate VaR using a historical simulation approach.

LO 1.1: Estimate VaR using a historical simulation approach.
Estimating VaR with a historical simulation approach is by far the simplest and most straightforward VaR method. To make this calculation, you simply order return observations from largest to smallest. The observation that follows the threshold loss level denotes the VaR limit. We are essentially searching for the observation that separates the tail from the body of the distribution. More generally, the observation that determines VaR for n observations at the (1 a) confidence level would be: (a x n) + 1.
Professors Note: Recall that the confidence level, (1 a), is typically a large value (e.g., 95% ) whereas the significance level, usually denoted as a , is much smaller (e.g., 5%).
To illustrate this VaR method, assume you have gathered 1,000 monthly returns for a security and produced the distribution shown in Figure 1. You decide that you want to compute the monthly VaR for this security at a confidence level of 93%. A ta95% confidence level, the lower tail displays the lowest 3% of the underlying distributions returns. For this distribution, the value associated with a 95% confidence level is a return of
15.5%. If you have \$1,000,000 invested in this security, the one-month VaR is \$155,000 (-15.5% x \$1,000,000).
Figure 1: Histogram of Monthly Returns – u C a j 3 crv U – i
70 –
60 –
50 –
40 –
30 –
20-
10-
0
/
5%
ProbabilityJ of Loss ———-

\
6
/
N ‘8 V
4
/
& 4 ”
q
/
/
b
.v \
Monthly Return
Q ni r i ‘i i \o o
lo
\O
4 V5
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Example: Identifying the VaR limit
Identify the ordered observation in a sample of 1,000 data points that corresponds to VaR at a 93% confidence level.
Since VaR is to be estimated at 93% confidence, this means that 5% (i.e., 50) of the ordered observations would fall in the tail of the distribution. Therefore, the 51st ordered loss observation would separate the 5% of largest losses from the remaining 95% of returns.
Professors Note: VaR is the quantile that separates the tail from the body of the distribution. With 1,000 observations at a 95% confidence level, there is a certain level o f arbitrariness in how the ordered observations relate to VaR. In other words, should VaR be the 50th observation (i.e., a x n), the 51st observation [i.e., (a x n) + 1], or some combination o f these observations? In this example, using the 51st observation was the approximation for VaR, and the method used in the assigned reading. However, on past FRM exams, VaR using the historical simulation method has been calculated as just: (a x n), in this case, as the 50th observation.
Example: Computing VaR
A long history of profit/loss data closely approximates a standard normal distribution (mean equals zero; standard deviation equals one). Estimate the 5% VaR using the historical simulation approach.
The VaR limit will be at the observation that separates the tail loss with area equal to 5% from the remainder of the distribution. Since the distribution is closely approximated by the standard normal distribution, the VaR is 1.65 (5% critical value from the stable). Recall that since VaR is a one-tailed test, the entire significance level of 5% is in the left tail of the returns distribution.
>From a practical perspective, the historical simulation approach is sensible only if you expect future performance to follow the same return generating process as in the past. Furthermore, this approach is unable to adjust for changing economic conditions or abrupt shifts in parameter values.
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Pa r a m e t r i c E s t i m a t i o n A p p r o a c h e s

## LO 1.1: Estimate VaR using a historical simulation approach.

LO 1.1: Estimate VaR using a historical simulation approach.
Estimating VaR with a historical simulation approach is by far the simplest and most straightforward VaR method. To make this calculation, you simply order return observations from largest to smallest. The observation that follows the threshold loss level denotes the VaR limit. We are essentially searching for the observation that separates the tail from the body of the distribution. More generally, the observation that determines VaR for n observations at the (1 a) confidence level would be: (a x n) + 1.
Professors Note: Recall that the confidence level, (1 a), is typically a large value (e.g., 95% ) whereas the significance level, usually denoted as a , is much smaller (e.g., 5%).
To illustrate this VaR method, assume you have gathered 1,000 monthly returns for a security and produced the distribution shown in Figure 1. You decide that you want to compute the monthly VaR for this security at a confidence level of 93%. A ta95% confidence level, the lower tail displays the lowest 3% of the underlying distributions returns. For this distribution, the value associated with a 95% confidence level is a return of
15.5%. If you have \$1,000,000 invested in this security, the one-month VaR is \$155,000 (-15.5% x \$1,000,000).
Figure 1: Histogram of Monthly Returns – u C a j 3 crv U – i
70 –
60 –
50 –
40 –
30 –
20-
10-
0
/
5%
ProbabilityJ of Loss ———-

\
6
/
N ‘8 V
4
/
& 4 ”
q
/
/
b
.v \
Monthly Return
Q ni r i ‘i i \o o
lo
\O
4 V5
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Example: Identifying the VaR limit
Identify the ordered observation in a sample of 1,000 data points that corresponds to VaR at a 93% confidence level.
Since VaR is to be estimated at 93% confidence, this means that 5% (i.e., 50) of the ordered observations would fall in the tail of the distribution. Therefore, the 51st ordered loss observation would separate the 5% of largest losses from the remaining 95% of returns.
Professors Note: VaR is the quantile that separates the tail from the body of the distribution. With 1,000 observations at a 95% confidence level, there is a certain level o f arbitrariness in how the ordered observations relate to VaR. In other words, should VaR be the 50th observation (i.e., a x n), the 51st observation [i.e., (a x n) + 1], or some combination o f these observations? In this example, using the 51st observation was the approximation for VaR, and the method used in the assigned reading. However, on past FRM exams, VaR using the historical simulation method has been calculated as just: (a x n), in this case, as the 50th observation.
Example: Computing VaR
A long history of profit/loss data closely approximates a standard normal distribution (mean equals zero; standard deviation equals one). Estimate the 5% VaR using the historical simulation approach.
The VaR limit will be at the observation that separates the tail loss with area equal to 5% from the remainder of the distribution. Since the distribution is closely approximated by the standard normal distribution, the VaR is 1.65 (5% critical value from the stable). Recall that since VaR is a one-tailed test, the entire significance level of 5% is in the left tail of the returns distribution.
>From a practical perspective, the historical simulation approach is sensible only if you expect future performance to follow the same return generating process as in the past. Furthermore, this approach is unable to adjust for changing economic conditions or abrupt shifts in parameter values.
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Topic 1 Cross Reference to GARP Assigned Reading – Dowd, Chapter 3
Pa r a m e t r i c E s t i m a t i o n A p p r o a c h e s

## LO 16.9: Explain the impact of a single asset price jump on a volatility smile.

LO 16.9: Explain the impact of a single asset price jump on a volatility smile.
Price jumps can occur for a number of reasons. One reason may be the expectation of a significant news event that causes the underlying asset to move either up or down by a large amount. This would cause the underlying distribution to become bimodal, but with the same expected return and standard deviation as a unimodal, or standard, price-change distribution.
Implied volatility is affected by price jumps and the probabilities assumed for either a large up or down movement. The usual result, however, is that at-the-money options tend to have a higher implied volatility than either out-of-the-money or in-the-money options. Away-from-the-money options exhibit a lower implied volatility than at-the-money options. Instead of a volatility smile, price jumps would generate a volatility frown, as in Figure 3.
Figure 3: Volatility Smile (Frown) W ith Price Jump Implied volatility
Strike price
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
K e y C o n c e p t s
LO 16.1 When option traders allow implied volatility to depend on strike price, patterns of implied volatility resemble volatility smiles.
LO 16.2 Put-call parity indicates that the deviation between market prices and Black-Scholes-Merton prices will be equivalent for calls and puts. Hence, implied volatility will be the same for calls and puts.
LO 16.3 Currency traders believe there is a greater chance of extreme price movements than predicted by a lognormal distribution. Equity traders believe the probability of large down movements in price is greater than large up movements in price, as compared with a lognormal distribution.
LO 16.4 The volatility pattern used by traders to price currency options generates implied volatilities that are higher for deep in-the-money and deep out-of-the-money options, as compared to the implied volatility for at-the-money options.
LO 16.5 The volatility smile exhibited by equity options is more of a smirk, with implied volatility higher for low strike prices. This has been attributed to leverage and crashophobia effects.
LO 16.6 Alternative methods to studying volatility patterns include: replacing strike price with strike price divided by stock price, replacing strike price with strike price divided by the forward price for the underlying asset, and replacing strike price with option delta.
LO 16.7 Volatility term structures and volatility surfaces are used by traders to judge consistency in model-generated option prices.
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LO 16.8 Volatility smiles that are not flat require the use of implied volatility functions or trees to correctly calculate option Greeks.
LO 16.9 Price jumps may generate volatility frowns instead of smiles.
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
C o n c e p t C h e c k e r s
1.
2.
3.
4.
3.
The market price deviations for puts and calls from Black-Scholes-Merton prices indicate: A. equivalent put and call implied volatility. B. equivalent put and call moneyness. C. unequal put and call implied volatility. D. unequal put and call moneyness.
An empirical distribution that exhibits a fatter right tail than that of a lognormal distribution would indicate: A. equal implied volatilities across low and high strike prices. B. greater implied volatilities for low strike prices. C. greater implied volatilities for high strike prices. D. higher implied volatilities for mid-range strike prices.
the same across maturities for given strike prices. the same for short time periods. The sticky strike rule assumes that implied volatility is: A. B. C. the same across strike prices for given maturities. D. different across strike prices for given maturities.
Compared to at-the-money currency options, out-of-the-money currency options exhibit which of the following volatility traits? A. Lower implied volatility. B. A frown. C. A smirk. D. Higher implied volatility.
Which of the following regarding equity option volatility is true? A. There is higher implied price volatility for away-from-the-money equity options. B. Crashophobia suggests actual equity volatility increases when stock prices
decline.
C. Compared to the lognormal distribution, traders believe the probability of large
down movements in price is similar to large up movements.
D. Increasing leverage at lower equity prices suggests increasing volatility.
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
C o n c e p t C h e c k e r A n s w e r s
1. A Put-call parity indicates that the implied volatility of a call and put will be equal for the same
strike price and time to expiration.
2. C An empirical distribution with a fat right tail generates a higher implied volatility for higher
strike prices due to the increased probability of observing high underlying asset prices. The pricing indication is that in-the-money calls and out-of-the-money puts would be expensive.
3. B The sticky strike rule, when applied to calculating option sensitivity measures, assumes
implied volatility is the same over short time periods.
4. D Away-from-the-money currency options have greater implied volatility than at-the-money
options. This pattern results in a volatility smile.
5. D There is higher implied price volatility for low strike price equity options. Crashophobia is based on the idea that large price declines are more likely than assumed in Black-Scholes- Merton prices, not that volatility increases when prices decline. Compared to the lognormal distribution, traders believe the probability of large down movements in price is higher than large up movements. Increasing leverage at lower equity prices suggests increasing volatility.
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10 Questions: 30 Minutes
1.
2.
3.
4.
An analyst for Z Corporation is determining the value at risk (VaR) for the corporations profit/loss distribution that is assumed to be normally distributed. The profit/loss distribution has an annual mean of \$3 million and a standard deviation of \$3.5 million. Using a parametric approach, what is the VaR with a 99% confidence level? A. \$0,775 million. B. \$3,155 million. C. \$5,775 million. D. \$8,155 million.
The Basel Committee requires backtesting of actual losses to VaR calculations. How many exceptions would need to occur in a 250-day trading period for the capital multiplier to increase from three to four? two to five. A. B. five to seven. C. seven to nine. D. ten or more.
The top-down approach to risk aggregation assumes that a banks portfolio can be cleanly subdivided according to market, credit, and operational risk measures. In contrast, a bottom-up approach attempts to account for interactions among various risk factors. In order to assess which approach is more appropriate, academic studies evaluate the ratio of integrated risks to separate risks. Regarding studies of top-down and bottom-up approaches, which of the following statements is incorrect? A. Top-down studies suggest that risk diversification is present. B. Bottom-up studies sometimes calculate the ratio of integrated risks to separate
C. Bottom-up studies suggest that risk diversification should be questioned. D. Top-down studies calculate the ratio of integrated risks to separate risks to be
risks to be less than one.
greater than one.
Commercial Bank Z has a \$3 million loan to company A and a \$3 million loan to company B. Companies A and B each have a 5% and 4% default probability, respectively. The default correlation between companies A and B is 0.6. What is the expected loss (EL) for the commercial bank under the worst case scenario? a. b. c. d.
\$83,700. \$133,900. \$165,600. \$233,800.
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5.
6.
7.
8.
Book 1 Self-Test: Market Risk Measurement and Management
A risk manager should always pay careful attention to the limitations and advantages of applying financial models such as the value at risk (VaR) and Black-Scholes- Merton (BSM) option pricing model. Which of the following statements regarding financial models is correct? a. Financial models should always be calibrated using most recent market data
because it is more likely to be accurate in extrapolating trends.
b. When applying the VaR model, empirical studies imply asset returns closely
c. The Black-Scholes-Merton option pricing model is a good example of
the advantage of using financial models because the model eliminates all mathematical inconsistences that can occur with human judgment.
d. A good example of a limitation of a financial model is the assumption of
constant volatility when applying the Black-Scholes-Merton (BSM) option pricing model.
Assume that a trader wishes to set up a hedge such that he sells \$100,000 of a Treasury bond and buys TIPS as a hedge. Using a historical yield regression framework, assume the DV01 on the T-bond is 0.072, the DV01 on the TIPS is 0.051, and the hedge adjustment factor (regression beta coefficient) is 1.2. What is the face value of the offsetting TIPS position needed to carry out this regression hedge? A. \$138,462. B. \$169,412. C. \$268,499. D. \$280,067.
A constant maturity Treasury (CMT) swap pays (\$1,000,000 / 2) x (y^-p 9%) every six months. There is a 70% probability of an increase in the 6-month spot rate and a 60% probability of an increase in the 1-year spot rate. The rate change in all cases is 0.50% per period, and the initial yCMT is 9%. What is the value of this CMT swap? A. \$2,325. B. \$2,229. C. \$2,429. D. \$905.
Suppose the market expects that the current 1-year rate for zero-coupon bonds with a face value of \$1 will remain at 5%, but the 1-year rate in one year will be 3%. What is the 2-year spot rate for zero-coupon bonds? A. 3.995%. B. 4.088%. C. 4.005%. D. 4.115%.
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9.
10.
An analyst is modeling spot rate changes using short rate term structure models. The current short-term interest rate is 5% with a volatility of 80bps. After one month passes the realization of dw, a normally distributed random variable with mean 0 and standard deviation Vdt, is -0.5. Assume a constant interest rate drift, X, of 0.36%. What should the analyst compute as the new spot rate? A. 5.37%. B. 4.63%. C. 5.76%. D. 4.24%.
Which of the following statements is incorrect regarding volatility smiles? A. Currency options exhibit volatility smiles because the at-the-money options have
higher implied volatility than away-from-the-money options.
B. Volatility frowns result when jumps occur in asset prices. C. Equity options exhibit a volatility smirk because low strike price options have
greater implied volatility.
D. Relative to currency traders, it appears that equity traders expectations of
extreme price movements are more asymmetric.
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Se l f -Te st A n s w e r s: M a r k e t Ri s k M e a su r e me n t a n d Ma n a g e me n t
1. B The population mean and standard deviations are unknown; therefore, the standard normal
z-value of 2.33 is used for a 99% confidence level.
VaR(l%) = -5.0 million + (\$3.5 million)(2.33) = -5.0 million + 8.155 million = 3.155 million (See Topic 1)
2. D Ten or more backtesting violations require the institution to use a capital multiplier of four.
(See Topic 3)
3. D Top-down studies calculate this ratio to be less than one, which suggests that risk
diversification is present and ignored by the separate approach. Bottom-up studies also often calculate this ratio to be less than one; however, this research has not been conclusive, and has recently found evidence of risk compounding, which produces a ratio greater than one. Thus, bottom-up studies suggests that risk diversification should be questioned. (See Topic 5)
4. C The default probability of company A is 5%. Thus, the standard deviation for company A is:
V0.05(l 0.05) – 0.2179
Company B has a default probability of 4% and, therefore, will have a standard deviation of 0.1960. We can now calculate the expected loss under the worst case scenario where both companies A and B are in default. Assuming that the default correlation between A and B is 0.6, the joint probability of default is:
P(AB) = 0.6^0.05(0.95) x 0.04(0.96) + 0.05 x 0.04 = 0.6V0.001824 + 0.002 = 0.0276
Thus, the expected loss for the commercial bank is \$165,600 (= 0.0276 x \$6,000,000). (See Topic 6)
5. D The Black-Scholes-Merton (BSM) option pricing model assumes strike prices have a
constant volatility. However, numerous empirical studies find higher volatility for out-of- the-money options and a volatility skew in equity markets. Thus, this is a good example of a limitation of financial models. The choice of time period used to calibrate the parameter inputs for the model can have a big impact on the results. Risk managers used volatility and correlation estimates from pre-crisis periods during the recent financial crisis, and this resulted in significantly underestimating the risk for financial models. All financial models should be stress tested using scenarios of extreme economic conditions. VaR models often assume asset returns have a normal distribution. However, empirical studies find higher kurtosis in return distributions. High kurtosis implies a distribution with fatter tails than the normal distribution. Thus, the normal distribution is not the best assumption for the underlying distribution. Financial models contain mathematical inconsistencies. For example, in applying the BSM option pricing model for up-and-out calls and puts and down-and-out calls and puts, there are rare cases where the inputs make the model insensitive to changes in implied volatility and option maturity. (See Topic 8)
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Book 1 Self-Test Answers: Market Risk Measurement and Management
6 . B Defining JF^ and P* as the face amounts of the real and nominal bonds, respectively, and
their corresponding DVO Is as DV01R and DV01R, a DV01 hedge is adjusted by the hedge adjustment factor, or beta, as follows:
F r = FN x
DV01 N DV01 R x(3 0.072 10.051 J
(cid:31)R F = 1 0 0 ,0 0 0 x
x 1.2 = 169,412
(See Topic 10)
7 . A The payoff in each period is (\$1,000,000 / 2) x (yCMT – 9 % ). For example, the 1-year payoff of \$5,000 in the figure below is calculated as (\$1,000,000 / 2) x (10% 9% ) = \$5,000. The other numbers in the year one cells are calculated similarly.
In six months, the payoff if interest rates increase to 9.50% is (\$1,000,000 / 2 ) x (9.5% 9.0%) = \$2,500. Note that the price in this cell equals the present value of the probability weighted 1 -year values plus the 6-month payoff:
months, U
(\$5,000×0.6)+ (\$0x0.4)
+ 0.095 1
+ \$2,500 = \$5,363.96
The other cell value in six months is calculated similarly and results in a loss of \$4,418.47.
The value of the CMT swap today is the present value of the probability weighted 6-month values:
(\$5,363.96 x 0.7) + ( – \$4,418.47 x 0.3)
+ 0.09 1
\$2,324.62
1 0 % yCMT Price = \$5,000
Tc mt
8 .5 %
Price = -\$ 4 ,4 18
7 c m t _ 8 %
Price = -\$5,000
Today
6 months year 1 year
Thus the correct response is A. The other answers are incorrect because they do not correctly discount the future values or omit the 6-month payoff from the 6-month values.
(See Topic 11)
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Book 1 Self-Test Answers: Market Risk Measurement and Management
8. A The 2-year spot rate is computed as follows:
f (2 ) = 2/(1 .05)(1.03) – 1 = 3.995%
(See Topic 12)
9. B This short rate process has an annualized drift of 0.36%, so it requires the use of Model 2
(with constant drift). The change in the spot rate is computed as:
dr = Xdt + crdw
dr = (0.36% / 12) + (0.8% x -0.5) = -0.37% = -3 7 basis points
Since the initial short-term rate was 5% and d r is 0.37%, the new spot rate in one month is:
5% – 0.37% = 4.63% (See Topic 13)
10. A Currency options exhibit volatility smiles because the at-the-money options have lower
implied volatility than away-from-the-money options.
Equity traders believe that the probability of large price decreases is greater than the probability of large price increases. Currency traders beliefs about volatility are more symmetric as there is no large skew in the distribution of expected currency values (i.e., there is a greater chance of large price movements in either direction).
(See Topic 16)
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Market Risk Measurement and Management
F o r m u l a s
Topic 1
profit/loss data: P/Lt = Pt + Dt Pt l
arithmetic return: rt
geometric return: Rt =
+ Dt ~ Pt-l
P t-1
P . + D . P -l
J
delta-normal VaR: VaR(a%) = (p,r + crr X za ) x Pt_1
lognormal VaR: VaR(a%) = Pt_1 x |l e^R CTRXZa j
standard error of a quantile: se (q) =
V p(l p)/n
f(q)
Topic 2 age-weighted historical simulation: w(i) = XM (1-X )
1-X ”
Topic 3
model accuracy test:
x pT
V p(l p)T
unconditional coverage test statistic:
LR = -2ln[(l – p)T-NpN] + 2ln{[l – (N/T)]t “n (N/T)n}
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Topic 6 portfolio mean return: pp = wxpx + wYpY
Book 1 Formulas
portfolio standard deviation: Op = V W X X + w y a y + 2 w x w y c o v x y 2 2 j 2 2
2 2
n ( X t -|j,x )(Yt – PY) t1 covariance: cov^y —-

n 1
covXY correlation: PXY = a x o Y
realized correlation: Prealized 2n n E ft,;
>)
correlation swap payoff: notional amount x (preajizej pfixeci)
joint probability of default: P(AB) = Pa b ^PDa CI PDa ) x PDb (1 PDB) + PDA X PDg
Topic 7 mean reversion rate: St – S j = a(p, S j)
autocorrelation: AC(pt,pt
) _ cov(Pt>Pt-l) a(pt)xa(P[_,)
t_1
Topic 8
correlation w ith expectation values: PXY =
E(XY) – E(X)E(Y)
Ve (X2)-(E (X ))2 x Ve (Y2)-(E (Y ))2
Spearmans rank correlation: Ps 1 ~ i = l
n(n2 -1 )
n
6 E d ?
Kendall s r: r n(n 1) / 2
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Book 1 Formulas
Topic 12
2- year spot rate: r (2) = %J(l + tj )(l + r2) 1
3- year spot rate: r (3) = %J(l + q ) (l + r2) (l + r3) 1
Jensens inequality: E
1
(l + r)
1
Topic 13
Model 1:
dr = crdw
where: dr = change in interest rates over small time interval, dt dt = small time interval (measured in years) a = annual basis-point volatility of rate changes dw = normally distributed random variable with mean 0 and standard deviation \fdt
Model 2: dr = Xdt + crdw
Vasicek model:
dr = k(9 – r)dt + crdw
where: k = a parameter that measures the speed of reversion adjustment 9 = long-run value of the short-term rate assuming risk neutrality r = current interest rate level
long-run value of short-term rate: X A
9 q -| k
where: the long-run true rate of interest = the long-run true rate of interest
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Book 1 Formulas
Topic 14
Model 3:
dr = \(t)dt + cre-atdw
where: cr = volatility at t = 0, which decreases exponentially to 0 for a > 0
CIR model: dr = k(0 r)dt + cr Vr dw
Model 4: dr = ardt + crdw
Topic 16
put-call parity: c – p = S – PV(X)
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Probability Example
Assume that the annual earnings per share (EPS) for a large sample of firms is normally distributed with a mean of \$5.00 and a standard deviation of \$1.50. What is the approximate probability of an observed EPS value falling between \$3.00 and \$7.25?
If EPS = x = \$7.25, then z = (x – p)/a = (\$7.25 – \$5.00)/\$ 1.50 – +1.50
If EPS = x = \$3.00, then z = (x – p)/a = (\$3.00 – \$5.00)/\$1.50 – -1.33
Forz-value o f 1.50: Use the row headed 1.5 and the column headed 0 to find the value 0.9332. This represents the area under the curve to the left of the critical value 1.50.
Forz-value o f 1.33: Use the row headed 1.3 and the column headed 3 to find the value 0.9082. This represents the area under the curve to the left of the critical value +1.33. The area to the left of1.33 is 1 0.9082 = 0.0918.
The area between these critical values is 0.9332 0.0918 = 0.8414, or 84.14%.
Hypothesis Testing One-Tailed Test Example
A sample of a stocks returns on 36 non-consecutive days results in a mean return of 2.0%. Assume the population standard deviation is 20.0%. Can we say with 95% confidence that the mean return is greater than 0%? (2.0 – 0.0) / (20.0 / 6) = 0.60. Hq: p < 0.0%, Ha : p > 0.0%. The test statistic = ^-statistic = = (2.0 – 0.0) / (20.0 / 6) = 0.60.
x – p o
The significance level = 1.0 – 0.95 = 0.05, or 5%.
Since this is a one-tailed test with an alpha of 0.05, we need to find the value 0.95 in the cumulative z-table. The closest value is 0.9505, with a corresponding critical z-value of 1.65. Since the test statistic is less than the critical value, we fail to reject HQ.
Hypothesis Testing Two-Tailed Test Example
Using the same assumptions as before, suppose that the analyst now wants to determine if he can say with 99% confidence that the stocks return is not equal to 0.0%.
Hq: p = 0.0%, Ha: p ^ 0.0%. The test statistic (z-value) = (2.0 0.0) / (20.0 / 6) = 0.60. The significance level = 1.0 – 0.99 = 0.01, or 1%.
Since this is a two-tailed test with an alpha of 0.01, there is a 0.005 rejection region in both tails. Thus, we need to find the value 0.995 (1.0 0.005) in the table. The closest value is 0.9951, which corresponds to a critical z-value of 2.58. Since the test statistic is less than the critical value, we fail to reject HQ and conclude that the stocks return equals 0.0%.
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## LO 16.9: Explain the impact of a single asset price jump on a volatility smile.

LO 16.9: Explain the impact of a single asset price jump on a volatility smile.
Price jumps can occur for a number of reasons. One reason may be the expectation of a significant news event that causes the underlying asset to move either up or down by a large amount. This would cause the underlying distribution to become bimodal, but with the same expected return and standard deviation as a unimodal, or standard, price-change distribution.
Implied volatility is affected by price jumps and the probabilities assumed for either a large up or down movement. The usual result, however, is that at-the-money options tend to have a higher implied volatility than either out-of-the-money or in-the-money options. Away-from-the-money options exhibit a lower implied volatility than at-the-money options. Instead of a volatility smile, price jumps would generate a volatility frown, as in Figure 3.
Figure 3: Volatility Smile (Frown) W ith Price Jump Implied volatility
Strike price
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
K e y C o n c e p t s
LO 16.1 When option traders allow implied volatility to depend on strike price, patterns of implied volatility resemble volatility smiles.
LO 16.2 Put-call parity indicates that the deviation between market prices and Black-Scholes-Merton prices will be equivalent for calls and puts. Hence, implied volatility will be the same for calls and puts.
LO 16.3 Currency traders believe there is a greater chance of extreme price movements than predicted by a lognormal distribution. Equity traders believe the probability of large down movements in price is greater than large up movements in price, as compared with a lognormal distribution.
LO 16.4 The volatility pattern used by traders to price currency options generates implied volatilities that are higher for deep in-the-money and deep out-of-the-money options, as compared to the implied volatility for at-the-money options.
LO 16.5 The volatility smile exhibited by equity options is more of a smirk, with implied volatility higher for low strike prices. This has been attributed to leverage and crashophobia effects.
LO 16.6 Alternative methods to studying volatility patterns include: replacing strike price with strike price divided by stock price, replacing strike price with strike price divided by the forward price for the underlying asset, and replacing strike price with option delta.
LO 16.7 Volatility term structures and volatility surfaces are used by traders to judge consistency in model-generated option prices.
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
LO 16.8 Volatility smiles that are not flat require the use of implied volatility functions or trees to correctly calculate option Greeks.
LO 16.9 Price jumps may generate volatility frowns instead of smiles.
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
C o n c e p t C h e c k e r s
1.
2.
3.
4.
3.
The market price deviations for puts and calls from Black-Scholes-Merton prices indicate: A. equivalent put and call implied volatility. B. equivalent put and call moneyness. C. unequal put and call implied volatility. D. unequal put and call moneyness.
An empirical distribution that exhibits a fatter right tail than that of a lognormal distribution would indicate: A. equal implied volatilities across low and high strike prices. B. greater implied volatilities for low strike prices. C. greater implied volatilities for high strike prices. D. higher implied volatilities for mid-range strike prices.
the same across maturities for given strike prices. the same for short time periods. The sticky strike rule assumes that implied volatility is: A. B. C. the same across strike prices for given maturities. D. different across strike prices for given maturities.
Compared to at-the-money currency options, out-of-the-money currency options exhibit which of the following volatility traits? A. Lower implied volatility. B. A frown. C. A smirk. D. Higher implied volatility.
Which of the following regarding equity option volatility is true? A. There is higher implied price volatility for away-from-the-money equity options. B. Crashophobia suggests actual equity volatility increases when stock prices
decline.
C. Compared to the lognormal distribution, traders believe the probability of large
down movements in price is similar to large up movements.
D. Increasing leverage at lower equity prices suggests increasing volatility.
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Topic 16 Cross Reference to GARP Assigned Reading – Hull, Chapter 20
C o n c e p t C h e c k e r A n s w e r s
1. A Put-call parity indicates that the implied volatility of a call and put will be equal for the same
strike price and time to expiration.
2. C An empirical distribution with a fat right tail generates a higher implied volatility for higher
strike prices due to the increased probability of observing high underlying asset prices. The pricing indication is that in-the-money calls and out-of-the-money puts would be expensive.
3. B The sticky strike rule, when applied to calculating option sensitivity measures, assumes
implied volatility is the same over short time periods.
4. D Away-from-the-money currency options have greater implied volatility than at-the-money
options. This pattern results in a volatility smile.
5. D There is higher implied price volatility for low strike price equity options. Crashophobia is based on the idea that large price declines are more likely than assumed in Black-Scholes- Merton prices, not that volatility increases when prices decline. Compared to the lognormal distribution, traders believe the probability of large down movements in price is higher than large up movements. Increasing leverage at lower equity prices suggests increasing volatility.
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Se l f -Te s t : M a r k e t Ri s k M e a su r e me n t a n d Ma n a g e me n t
10 Questions: 30 Minutes
1.
2.
3.
4.
An analyst for Z Corporation is determining the value at risk (VaR) for the corporations profit/loss distribution that is assumed to be normally distributed. The profit/loss distribution has an annual mean of \$3 million and a standard deviation of \$3.5 million. Using a parametric approach, what is the VaR with a 99% confidence level? A. \$0,775 million. B. \$3,155 million. C. \$5,775 million. D. \$8,155 million.
The Basel Committee requires backtesting of actual losses to VaR calculations. How many exceptions would need to occur in a 250-day trading period for the capital multiplier to increase from three to four? two to five. A. B. five to seven. C. seven to nine. D. ten or more.
The top-down approach to risk aggregation assumes that a banks portfolio can be cleanly subdivided according to market, credit, and operational risk measures. In contrast, a bottom-up approach attempts to account for interactions among various risk factors. In order to assess which approach is more appropriate, academic studies evaluate the ratio of integrated risks to separate risks. Regarding studies of top-down and bottom-up approaches, which of the following statements is incorrect? A. Top-down studies suggest that risk diversification is present. B. Bottom-up studies sometimes calculate the ratio of integrated risks to separate
C. Bottom-up studies suggest that risk diversification should be questioned. D. Top-down studies calculate the ratio of integrated risks to separate risks to be
risks to be less than one.
greater than one.
Commercial Bank Z has a \$3 million loan to company A and a \$3 million loan to company B. Companies A and B each have a 5% and 4% default probability, respectively. The default correlation between companies A and B is 0.6. What is the expected loss (EL) for the commercial bank under the worst case scenario? a. b. c. d.
\$83,700. \$133,900. \$165,600. \$233,800.
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5.
6.
7.
8.
Book 1 Self-Test: Market Risk Measurement and Management
A risk manager should always pay careful attention to the limitations and advantages of applying financial models such as the value at risk (VaR) and Black-Scholes- Merton (BSM) option pricing model. Which of the following statements regarding financial models is correct? a. Financial models should always be calibrated using most recent market data
because it is more likely to be accurate in extrapolating trends.
b. When applying the VaR model, empirical studies imply asset returns closely
c. The Black-Scholes-Merton option pricing model is a good example of
the advantage of using financial models because the model eliminates all mathematical inconsistences that can occur with human judgment.
d. A good example of a limitation of a financial model is the assumption of
constant volatility when applying the Black-Scholes-Merton (BSM) option pricing model.
Assume that a trader wishes to set up a hedge such that he sells \$100,000 of a Treasury bond and buys TIPS as a hedge. Using a historical yield regression framework, assume the DV01 on the T-bond is 0.072, the DV01 on the TIPS is 0.051, and the hedge adjustment factor (regression beta coefficient) is 1.2. What is the face value of the offsetting TIPS position needed to carry out this regression hedge? A. \$138,462. B. \$169,412. C. \$268,499. D. \$280,067.
A constant maturity Treasury (CMT) swap pays (\$1,000,000 / 2) x (y^-p 9%) every six months. There is a 70% probability of an increase in the 6-month spot rate and a 60% probability of an increase in the 1-year spot rate. The rate change in all cases is 0.50% per period, and the initial yCMT is 9%. What is the value of this CMT swap? A. \$2,325. B. \$2,229. C. \$2,429. D. \$905.
Suppose the market expects that the current 1-year rate for zero-coupon bonds with a face value of \$1 will remain at 5%, but the 1-year rate in one year will be 3%. What is the 2-year spot rate for zero-coupon bonds? A. 3.995%. B. 4.088%. C. 4.005%. D. 4.115%.
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Book 1 Self-Test: Market Risk Measurement and Management
9.
10.
An analyst is modeling spot rate changes using short rate term structure models. The current short-term interest rate is 5% with a volatility of 80bps. After one month passes the realization of dw, a normally distributed random variable with mean 0 and standard deviation Vdt, is -0.5. Assume a constant interest rate drift, X, of 0.36%. What should the analyst compute as the new spot rate? A. 5.37%. B. 4.63%. C. 5.76%. D. 4.24%.
Which of the following statements is incorrect regarding volatility smiles? A. Currency options exhibit volatility smiles because the at-the-money options have
higher implied volatility than away-from-the-money options.
B. Volatility frowns result when jumps occur in asset prices. C. Equity options exhibit a volatility smirk because low strike price options have
greater implied volatility.
D. Relative to currency traders, it appears that equity traders expectations of
extreme price movements are more asymmetric.
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Se l f -Te st A n s w e r s: M a r k e t Ri s k M e a su r e me n t a n d Ma n a g e me n t
1. B The population mean and standard deviations are unknown; therefore, the standard normal
z-value of 2.33 is used for a 99% confidence level.
VaR(l%) = -5.0 million + (\$3.5 million)(2.33) = -5.0 million + 8.155 million = 3.155 million (See Topic 1)
2. D Ten or more backtesting violations require the institution to use a capital multiplier of four.
(See Topic 3)
3. D Top-down studies calculate this ratio to be less than one, which suggests that risk
diversification is present and ignored by the separate approach. Bottom-up studies also often calculate this ratio to be less than one; however, this research has not been conclusive, and has recently found evidence of risk compounding, which produces a ratio greater than one. Thus, bottom-up studies suggests that risk diversification should be questioned. (See Topic 5)
4. C The default probability of company A is 5%. Thus, the standard deviation for company A is:
V0.05(l 0.05) – 0.2179
Company B has a default probability of 4% and, therefore, will have a standard deviation of 0.1960. We can now calculate the expected loss under the worst case scenario where both companies A and B are in default. Assuming that the default correlation between A and B is 0.6, the joint probability of default is:
P(AB) = 0.6^0.05(0.95) x 0.04(0.96) + 0.05 x 0.04 = 0.6V0.001824 + 0.002 = 0.0276
Thus, the expected loss for the commercial bank is \$165,600 (= 0.0276 x \$6,000,000). (See Topic 6)
5. D The Black-Scholes-Merton (BSM) option pricing model assumes strike prices have a
constant volatility. However, numerous empirical studies find higher volatility for out-of- the-money options and a volatility skew in equity markets. Thus, this is a good example of a limitation of financial models. The choice of time period used to calibrate the parameter inputs for the model can have a big impact on the results. Risk managers used volatility and correlation estimates from pre-crisis periods during the recent financial crisis, and this resulted in significantly underestimating the risk for financial models. All financial models should be stress tested using scenarios of extreme economic conditions. VaR models often assume asset returns have a normal distribution. However, empirical studies find higher kurtosis in return distributions. High kurtosis implies a distribution with fatter tails than the normal distribution. Thus, the normal distribution is not the best assumption for the underlying distribution. Financial models contain mathematical inconsistencies. For example, in applying the BSM option pricing model for up-and-out calls and puts and down-and-out calls and puts, there are rare cases where the inputs make the model insensitive to changes in implied volatility and option maturity. (See Topic 8)
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Book 1 Self-Test Answers: Market Risk Measurement and Management
6 . B Defining JF^ and P* as the face amounts of the real and nominal bonds, respectively, and
their corresponding DVO Is as DV01R and DV01R, a DV01 hedge is adjusted by the hedge adjustment factor, or beta, as follows:
F r = FN x
DV01 N DV01 R x(3 0.072 10.051 J
(cid:31)R F = 1 0 0 ,0 0 0 x
x 1.2 = 169,412
(See Topic 10)
7 . A The payoff in each period is (\$1,000,000 / 2) x (yCMT – 9 % ). For example, the 1-year payoff of \$5,000 in the figure below is calculated as (\$1,000,000 / 2) x (10% 9% ) = \$5,000. The other numbers in the year one cells are calculated similarly.
In six months, the payoff if interest rates increase to 9.50% is (\$1,000,000 / 2 ) x (9.5% 9.0%) = \$2,500. Note that the price in this cell equals the present value of the probability weighted 1 -year values plus the 6-month payoff:
months, U
(\$5,000×0.6)+ (\$0x0.4)
+ 0.095 1
+ \$2,500 = \$5,363.96
The other cell value in six months is calculated similarly and results in a loss of \$4,418.47.
The value of the CMT swap today is the present value of the probability weighted 6-month values:
(\$5,363.96 x 0.7) + ( – \$4,418.47 x 0.3)
+ 0.09 1
\$2,324.62
1 0 % yCMT Price = \$5,000
Tc mt
8 .5 %
Price = -\$ 4 ,4 18
7 c m t _ 8 %
Price = -\$5,000
Today
6 months year 1 year
Thus the correct response is A. The other answers are incorrect because they do not correctly discount the future values or omit the 6-month payoff from the 6-month values.
(See Topic 11)
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Book 1 Self-Test Answers: Market Risk Measurement and Management
8. A The 2-year spot rate is computed as follows:
f (2 ) = 2/(1 .05)(1.03) – 1 = 3.995%
(See Topic 12)
9. B This short rate process has an annualized drift of 0.36%, so it requires the use of Model 2
(with constant drift). The change in the spot rate is computed as:
dr = Xdt + crdw
dr = (0.36% / 12) + (0.8% x -0.5) = -0.37% = -3 7 basis points
Since the initial short-term rate was 5% and d r is 0.37%, the new spot rate in one month is:
5% – 0.37% = 4.63% (See Topic 13)
10. A Currency options exhibit volatility smiles because the at-the-money options have lower
implied volatility than away-from-the-money options.
Equity traders believe that the probability of large price decreases is greater than the probability of large price increases. Currency traders beliefs about volatility are more symmetric as there is no large skew in the distribution of expected currency values (i.e., there is a greater chance of large price movements in either direction).
(See Topic 16)
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Market Risk Measurement and Management
F o r m u l a s
Topic 1
profit/loss data: P/Lt = Pt + Dt Pt l
arithmetic return: rt
geometric return: Rt =
+ Dt ~ Pt-l
P t-1
P . + D . P -l
J
delta-normal VaR: VaR(a%) = (p,r + crr X za ) x Pt_1
lognormal VaR: VaR(a%) = Pt_1 x |l e^R CTRXZa j
standard error of a quantile: se (q) =
V p(l p)/n
f(q)
Topic 2 age-weighted historical simulation: w(i) = XM (1-X )
1-X ”
Topic 3
model accuracy test:
x pT
V p(l p)T
unconditional coverage test statistic:
LR = -2ln[(l – p)T-NpN] + 2ln{[l – (N/T)]t “n (N/T)n}
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Topic 6 portfolio mean return: pp = wxpx + wYpY
Book 1 Formulas
portfolio standard deviation: Op = V W X X + w y a y + 2 w x w y c o v x y 2 2 j 2 2
2 2
n ( X t -|j,x )(Yt – PY) t1 covariance: cov^y —-

n 1
covXY correlation: PXY = a x o Y
realized correlation: Prealized 2n n E ft,;
>)
correlation swap payoff: notional amount x (preajizej pfixeci)
joint probability of default: P(AB) = Pa b ^PDa CI PDa ) x PDb (1 PDB) + PDA X PDg
Topic 7 mean reversion rate: St – S j = a(p, S j)
autocorrelation: AC(pt,pt
) _ cov(Pt>Pt-l) a(pt)xa(P[_,)
t_1
Topic 8
correlation w ith expectation values: PXY =
E(XY) – E(X)E(Y)
Ve (X2)-(E (X ))2 x Ve (Y2)-(E (Y ))2
Spearmans rank correlation: Ps 1 ~ i = l
n(n2 -1 )
n
6 E d ?
Kendall s r: r n(n 1) / 2
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Book 1 Formulas
Topic 12
2- year spot rate: r (2) = %J(l + tj )(l + r2) 1
3- year spot rate: r (3) = %J(l + q ) (l + r2) (l + r3) 1
Jensens inequality: E
1
(l + r)
1
Topic 13
Model 1:
dr = crdw
where: dr = change in interest rates over small time interval, dt dt = small time interval (measured in years) a = annual basis-point volatility of rate changes dw = normally distributed random variable with mean 0 and standard deviation \fdt
Model 2: dr = Xdt + crdw
Vasicek model:
dr = k(9 – r)dt + crdw
where: k = a parameter that measures the speed of reversion adjustment 9 = long-run value of the short-term rate assuming risk neutrality r = current interest rate level
long-run value of short-term rate: X A
9 q -| k
where: the long-run true rate of interest = the long-run true rate of interest
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Book 1 Formulas
Topic 14
Model 3:
dr = \(t)dt + cre-atdw
where: cr = volatility at t = 0, which decreases exponentially to 0 for a > 0
CIR model: dr = k(0 r)dt + cr Vr dw
Model 4: dr = ardt + crdw
Topic 16
put-call parity: c – p = S – PV(X)
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U si n g t h e C u mu l a t i v e Z-Ta b l e
Probability Example
Assume that the annual earnings per share (EPS) for a large sample of firms is normally distributed with a mean of \$5.00 and a standard deviation of \$1.50. What is the approximate probability of an observed EPS value falling between \$3.00 and \$7.25?
If EPS = x = \$7.25, then z = (x – p)/a = (\$7.25 – \$5.00)/\$ 1.50 – +1.50
If EPS = x = \$3.00, then z = (x – p)/a = (\$3.00 – \$5.00)/\$1.50 – -1.33
Forz-value o f 1.50: Use the row headed 1.5 and the column headed 0 to find the value 0.9332. This represents the area under the curve to the left of the critical value 1.50.
Forz-value o f 1.33: Use the row headed 1.3 and the column headed 3 to find the value 0.9082. This represents the area under the curve to the left of the critical value +1.33. The area to the left of1.33 is 1 0.9082 = 0.0918.
The area between these critical values is 0.9332 0.0918 = 0.8414, or 84.14%.
Hypothesis Testing One-Tailed Test Example
A sample of a stocks returns on 36 non-consecutive days results in a mean return of 2.0%. Assume the population standard deviation is 20.0%. Can we say with 95% confidence that the mean return is greater than 0%? (2.0 – 0.0) / (20.0 / 6) = 0.60. Hq: p < 0.0%, Ha : p > 0.0%. The test statistic = ^-statistic = = (2.0 – 0.0) / (20.0 / 6) = 0.60.
x – p o
The significance level = 1.0 – 0.95 = 0.05, or 5%.
Since this is a one-tailed test with an alpha of 0.05, we need to find the value 0.95 in the cumulative z-table. The closest value is 0.9505, with a corresponding critical z-value of 1.65. Since the test statistic is less than the critical value, we fail to reject HQ.
Hypothesis Testing Two-Tailed Test Example
Using the same assumptions as before, suppose that the analyst now wants to determine if he can say with 99% confidence that the stocks return is not equal to 0.0%.
Hq: p = 0.0%, Ha: p ^ 0.0%. The test statistic (z-value) = (2.0 0.0) / (20.0 / 6) = 0.60. The significance level = 1.0 – 0.99 = 0.01, or 1%.
Since this is a two-tailed test with an alpha of 0.01, there is a 0.005 rejection region in both tails. Thus, we need to find the value 0.995 (1.0 0.005) in the table. The closest value is 0.9951, which corresponds to a critical z-value of 2.58. Since the test statistic is less than the critical value, we fail to reject HQ and conclude that the stocks return equals 0.0%.
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