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1 CF4103 Financial Time Series Analysis Suggested Solutions of Tutorial 10 (Semester 2/06-07) Questions and Answers 1. Explain the exponentially weighted moving average (EWMA) model for estimating volatility from historical data. Solution. Define u i as (S i S i 1 )/S i 1, where S i is the value of a market variable on day i. In the EWMA model, the variance rate of the market variable (i.e., the square of its volatility) calculated for day n is a weighted average of the u 2 n i s (i = 1, 2, 3, ). For some constant λ (0 < λ < 1) the weight given to u 2 n i 1 is λ times the weight given to u 2 n i. The volatility estimated for day n, σ n, is related to the volatility estimated for day n 1, σ n 1, by σn 2 = λσn (1 λ)u 2 n 1. This formula shows that the EWMA model has one very attractive property. To calculate the volatility estimate for day n, it is sufficient to know the volatility estimate for day n 1 and u n 1. 1
2 2. What is the difference between the exponentially weighted moving average model and the GARCH(1, 1) model for updating volatilities? Solution. The EWMA model produces a forecast of the daily variance rate for day n which is a weighted average of (i) the forecast for day n 1, and (ii) the square of the proportional change on day n 1. The GARCH(1, 1) model produces a forecast of the daily variance for day n which is a weighted average of (i) the forecast for day n 1, (ii) The square of the proportional change on day n 1. and (iii) a long run average variance rate. GARCH (1, 1) adapts the EWMA model by giving some weight to a long run average variance rate. Whereas the EWMA has no mean reversion. GARCH(1, 1) is consistent with a mean-reverting variance rate model. 3. The most recent estimate of the daily volatility of an asset is 1.5% and the price of the asset at the close of trading yesterday was $ The parameter λ in the EWMA model is Suppose that the price of the asset at the close of trading today is $ How will this cause the volatility to be updated by the EWMA model? Solution. In the case σ n 1 = and u n = 0.5/30 = 2
3 , so that the equation σn 2 = λσn (1 λ)u 2 n 1 gives σn 2 = = The volatility estimate on day n is therefore = or %. 4. A company uses an EWMA model for forecasting volatility. It decides to change the parameter λ from 0.95 to Explain the likely impact on the forecasts. Solution. Reducing λ from 0.95 to 0.85 means that more weight is put on recent observations of u 2 i and less weight is given to older observations. Volatilities calculated with λ = 0.85 will react more quickly to new information and will bounce around much more than volatilities calculated with λ = The volatility of a certain market variable is 30% per annum. Calculate a 99% confidence interval for the size of the percentage daily change in the variable. Solution. The volatility per day is 30/ 252 = 1.89%. There is a 99% chance that a normally distributed variable will lie within 2.57 standard deviations. We are therefore 99% 3
4 confident that the daily change will be less than % = 4.86%. 6. A company uses the GARCH(1, 1) model for updating volatility. The three parameters are ω, α and β. Describe the impact of making a small increase in each of the parameters while keeping the others fixed. Solution. The weight given to the long-run average variance rate is 1 α β and the long-run average variance rate is ω/(1 α β). Increasing ω increases the long-run average variance rate. Increasing α increases the weight given to the most recent data item, reduces the weight given to the long-run average variance rate, and increases the level of the long-run average variance rate. Increasing β increases the weight given to the previous variance estimate, reduces the weight given to the long-run average variance rate, and increases the level of the long-run average variance rate. 7. The most recent estimate of the daily volatility of the US dollar/sterling exchange rate is 0.6% and the exchange rate at 4 p.m. yesterday was The parameter λ in the EWMA model is 0.9. Suppose that the exchange rate at 4 p.m. today proves to be How would the estimate of the daily volatility be updated? 4
5 Solution. The proportional daily change is 0.005/ = The current daily variance estimate is = The new daily variance estimate is = The new volatility is the square root of this. It is or 0.579%. 8. Assume that S&P 500 at the close of trading yesterday was 1, 040 and the daily volatility of the index was estimated as 1% per day at that time. The parameters in a GARCH(1, 1) model are ω = , α = 0.06 and β = If the level of the index at close of trading today is 1, 060, what is the new volatility estimate? Solution. With the usual notation u n 1 = 20/1040 = so that σn 2 = = so that σ n = The new volatility estimate is therefore 1.078% per day. 9. Suppose that the daily volatilities of asset A and asset B, calculated at the close of trading yesterday, are 1.6% and 2.5%, respectively. The prices of the assets at close of trading yesterday were $20 and $40, and the estimate of the 5
6 coefficient of correlation between the returns on the two assets was The parameter λ used in the EWMA model is (a) Calculate the current estimate of the covariance between the assets. (b) On the assumption that the prices of the assets at close of trading today are $20.5 and $40.5, update the correlation estimate. Solution. (a) The volatilities and correlation imply that the current estimate of the covariance is = (b) If the prices of the assets at close of trading are $20.5 and $40.5, the proportional changes are 0.5/20 = and 0.5/40 = The new covariance estimate is = The new variance estimate for asset A is = so that the new volatility is The new variance estimate for asset B is =
7 so that the new volatility is The new correlation estimate is = The parameters of a GARCH(1, 1) model are estimated as ω = , α = 0.05 and β = What is the longrun average volatility and what is the equation describing the way that the variance rate reverts to its long-run average? If the current volatility is 20% per year, what is the expected volatility in 20 days? Solution. The long-run average variance rate is ω/(1 α β) or /0.03 = The long-run average volatility is or 1.155%. The equation describing the way the variance rate reverts to its long-run average is In this case, E[σ 2 n+k] = V L + (α + β) k (σ 2 n V L ). E[σ 2 n+k] = k (σ 2 n ) If the current volatility is 20% per year,σ n = 0.2/ 252 = The expected variance rate in 20 days is ( ) = The expected variance rate in 20 days is therefore = or 1.21% per day. 7
8 11. Suppose that the current daily volatilities of asset X and asset Y are 1.0% and 1.2%, respectively. The prices of these assets at the close of trading yesterday were $30 and $50, and the estimate of the coefficient of correlation between the returns on the two assets made at this time was Correlations and volatilities are updated using a GARCH(1, 1) model. The estimates of the model s parameters are α = 0.04 and β = For the correlation ω = , and for the volatilities ω = If the prices of the two assets at the close of trading today are $31 and $51, how is the correlation estimate updated? Solution. Using the notation in the text σ u,n 1 = 0.01 and σ v,n 1 = and the most recent estimate of the covariance between the asset returns is Cov n 1 = = The variable u n 1 = 1/30 = and the variable v n 1 = 1/50 = The new estimate of the covariance, Cov n, is = The new estimate of the variance of the first asset, σu,n 2 is = so that σ u,n = = or 1.189%. The new 8
9 estimate of the variance of the second asset, σv,n 2 is = so that σ v,n = = pr 1.242%. The new estimate of the correlation between the assets is therefore /( ) = Suppose that the daily volatility of FTSE 100 stock index (measured in pounds sterling) is 1.8% and the daily volatility of the dollar/sterling exchange rate is 0.9%. Suppose further that the correlation between the FTSE 100 and the dollar/sterling exchange rate is 0.4. What is the volatility of the FTSE 100 when it is translated to US dollars? Assume that the dollar/sterling exchange rate is expressed as the number of US dollars per pound sterling. Solution. The FTSE expressed in dollars is XY where X is the FTSE expressed in sterling and Y is the exchange rate (value of one pound in dollars). Define x i as the proportional change in X on day i and y i as the proportional change in Y on day i. The proportional change in XY is approximately x i + y i. The standard deviation of x i is and the standard deviation of y i is The correlation between the two is 0.4. The variance of x i + y i is therefore =
10 so that the volatility of x i +y i is or 2.31%. This is the volatility of the FTSE expressed in dollars. Note that it is greater than the volatility of the FTSE expressed in sterling. This is the impact of the positive correlation. When the FTSE increases the value of sterling measured in dollars also tends to increase. The created an even bigger increase in the value of FT-SE measured in dollars. Similarly for a decrease in the FT-SE. 13. Suppose that in the above problem the correlation between the S&P 500 Index (measured in dollars) and the FTSE 100 Index (measured in sterling) is 0.7, the correlation between the S&P 500 Index (measured in dollars) and the dollar/sterling exchange rate is 0.3, and the daily volatility of S&P 500 index is 1.6%. What is the correlation between the S&P 500 index (measured in dollars) and the FTSE 100 index when it is translated to dollars? Solution. Continuing with the notation in the above problem, define z i as the proportional change in the value of the S&P 500 on day i. The covariance between x i and z i is = The covariance between y i and z i is = The covariance between x i + y i and z i equals the covariance between 10
11 x i and z i plus the covariance between y i and z i. It is = The correlation between x i + y i and z i is = Note that the volatility of the S&P 500 drops out in this calculation. 14. Show that the GARCH(1, 1) model σ 2 n = ω+αu 2 n 1+βσ 2 n 1 is equivalent to the stochastic volatility model dv = a(v L V )dt + ξv dz, where time is measured in days, V is the square of the volatility of the asset price, and ω a = 1 α β, V L = 1 α β, ξ = α 2. What is the stochastic volatility model when time is measured in years? Solution. so that σ 2 n = ω + αu 2 n 1 + β 2 n 1 σn 2 σn 1 2 = ω + (β 1)σn αu 2 n 1. The variance u 2 n 1 has a mean of σn 1 2 and a variance of E(u n 1 ) 4 [E(u 2 n 1)] 2 = 2σn
12 The standard deviation of u 2 n 1 is 2σn 1. 2 Assuming the u i are generated by a Wiener process, dz, we can therefore write u 2 n 1 = σn σn 1ε, 2 where ε is a random sample form a standard normal distribution. Substituting this into a equation for σn 2 σn 1 2 we get σn 2 σn 1 2 = ω + (α + β 1)σn α 2σn 1ε. 2 We can write V = σn 2 σn 1 2 and V = σn 1. 2 Also a = 1 α β, av L = ω and ξ = α 2 so that V = a(v L V ) + ξεv. Because time is measured in days, t = 1 and V = a(v L V ) t + ξv ε t. The result follows. When time is measured in years t = 1/252 so that V = a(v L V )252 t + ξv ε 252 t and the process for V is dv = 252a(V L V )dt + ξv 252dz. 12
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