Type:
Solution Manual
Resource:
Analysis of Financial Time Series
Edition:
3rd Edition
Author(s):
Ruey S. Tsay
Solutions to Exercises of Chapter 1 Problem 1: Summary statistics of daily returns from January 1999 to December 2008. Stock
mean
St. Dev. Skew Ex. Kurt. Min. Max. t-ratio(p-value) (a) Percentage simple returns: 2515 data points AXP 0.015 2.446 -0.035 6.048 −17.595 17.927 0.299(0.77) CAT 0.060 2.170 0.012 4.453 −14.518 14.723 1.375(0.17) SBUX 0.048 2.683 −0.082 8.746 −28.286 14.635 0.898(0.37) (b) Percentage log returns: 2515 data points AXP −0.015 2.453 −0.336 6.486 −19.352 16.489 −0.316(0.75) CAT 0.036 2.171 −0.202 4.695 −15.686 13.735 0.830(0.41) SBUX 0.012 2.696 −0.597 12.90 −33.249 13.659 0.221(0.83) The sample means of the log returns are not significantly different from zero at the 5% level. Problem 2: Summary statistics of monthly returns from 1975 to 2008. Stock
mean
St. Dev. Skew Ex. Kurt. Min. Max. t-ratio(p-value) (a) Percentage simple returns: 408 data points GM 0.557 9.273 −0.383 2.048 −38.931 27.662 1.213(0.23) VW 1.012 4.507 −0.743 2.666 −22.536 14.160 4.534(0.00) EW 1.331 5.596 −0.300 4.334 −27.225 29.926 4.806(0.00) SP 0.730 4.360 −0.571 2.269 −21.763 13.177 3.382(0.00) (b) Percentage log returns: 408 data points GM 0.110 9.950 −1.024 4.021 −49.317 24.422 0.232(0.82) VW 0.905 4.561 −1.051 3.938 −25.536 13.243 4.006(0.00) EW 1.167 5.626 −0.836 5.242 −31.779 26.179 4.190(0.00) SP 0.632 4.402 −0.855 3.335 −24.543 12.378 2.900(0.00) The sample mean of the monthly GM stock return is not significantly different from zero, but those of the monthly log returns of indices are all significantly different from zero at the 5% level. Problem 3: Focus on monthly S&P composite log returns. • Average annual return = sample mean * 12 = (sum of log returns)/(number of years) = 7.583%. P • V = exp(0.07583 ∗ (2008 − 1975 + 1)) ≈ $13.17. (or V = exp( 408 t=1 rt ), where rt is the monthly log return). 3
4 Problem 4: Daily log returns of American Express stock from 1999 to 2008. p • Skewness: test-statistic is t = −0.336/ 6/2515 = −6.888 with p value 5.66 × 10−12 . Thus, we reject the null hypothesis of no skewness at the 5% level. • Kurtosis: test-statistic is t = √ 6.486
24/2515
= 66.40, which is large and has a p value close to zero.
Thus, we reject the null hypothesis of zero excess kurtosis. That is, the distribution of the log returns has heavy tails.
Problem 5: Summary statistics for daily foreign exchange rates from January 4, 2000 to March 27, 2009. Currency CA EU JP UK
mean Variance Skew Ex. Kurt. Min. Percentage log returns: 2322 data points −0.0068 0.345 −0.239 8.245 −5.072 0.011 0.428 0.126 2.805 −3.003 −0.002 0.440 −0.671 4.581 −5.216 −0.006 0.382 −0.395 7.061 −4.966
Max. 3.807 4.621 2.708 4.435
The sample means of daily log returns of the exchange rates considered are small. The returns all have positive excess kurtosis, indicating heavy tails.
0.4 0.0
0.2
density
0.6
0.8
5
−2
0
2
4
log−return
Figure 1: Empirical density function of the daily log returns of USEU exchange rate: January 4, 2000 to March 27, 2009.
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