The quotient correlation is defined here as an alternative to Pearson’s correlation that is more intuitive and flexible in cases where the tail behavior of data is important. It measures nonlinear dependence where the regular correlation coefficient is generally not applicable. One of its most useful features is a test statistic that has high power when testing nonlinear dependence in cases where the Fisher’s Z-transformation test may fail to reach a right conclusion. Unlike most asymptotic test statistics, which are either normal or χ2, this test statistic has a limiting gamma distribution (henceforth, the gamma test statistic). More than the common usages of correlation, the quotient correlation can easily and intuitively be adjusted to values at tails. This adjustment generates two new concepts—the tail quotient correlation and the tail independence test statistics, which are also gamma statistics. Due to the fact that there is no analogue of the correlation coefficient in extreme value theory, and there does not exist an efficient tail independence test statistic, these two new concepts may open up a new field of study. In addition, an alternative to Spearman’s rank correlation, a rank based quotient correlation, is also defined. The advantages of using these new concepts are illustrated with simulated data and a real data analysis of internet traffic. Show InformationFirst available in Project Euclid: 13 March 2008 Digital Object Identifier: 10.1214/009053607000000866 Primary: 60G70 , 62G10 , 62G20 , 62G32 , 62H15 , 62H20 Keywords: necessary condition of tail independence , Nonlinear dependence , tail dependence , testing (tail) independence The Durbin Watson (DW) statistic is a test for autocorrelation in the residuals from a statistical model or regression analysis. The Durbin-Watson statistic will always have a value ranging between 0 and 4. A value of 2.0 indicates there is no autocorrelation detected in the sample. Values from 0 to less than 2 point to positive autocorrelation and values from 2 to 4 means negative autocorrelation. A stock price displaying positive autocorrelation would indicate that the price yesterday has a positive correlation on the price today—so if the stock fell yesterday, it is also likely that it falls today. A security that has a negative autocorrelation, on the other hand, has a negative influence on itself over time—so that if it fell yesterday, there is a greater likelihood it will rise today. Key Takeaways
The Basics of the Durbin Watson StatisticAutocorrelation, also known as serial correlation, can be a significant problem in analyzing historical data if one does not know to look out for it. For instance, since stock prices tend not to change too radically from one day to another, the prices from one day to the next could potentially be highly correlated, even though there is little useful information in this observation. In order to avoid autocorrelation issues, the easiest solution in finance is to simply convert a series of historical prices into a series of percentage-price changes from day to day. Autocorrelation can be useful for technical analysis, which is most concerned with the trends of, and relationships between, security prices using charting techniques in lieu of a company's financial health or management. Technical analysts can use autocorrelation to see how much of an impact past prices for a security have on its future price. Autocorrelation can show if there is a momentum factor associated with a stock. For example, if you know that a stock historically has a high positive autocorrelation value and you witnessed the stock making solid gains over the past several days, then you might reasonably expect the movements over the upcoming several days (the leading time series) to match those of the lagging time series and to move upward. The Durbin Watson statistic is named after statisticians James Durbin and Geoffrey Watson. Special ConsiderationsA rule of thumb is that DW test statistic values in the range of 1.5 to 2.5 are relatively normal. Values outside this range could, however, be a cause for concern. The Durbin–Watson statistic, while displayed by many regression analysis programs, is not applicable in certain situations. For instance, when lagged dependent variables are included in the explanatory variables, then it is inappropriate to use this test. Example of the Durbin Watson StatisticThe formula for the Durbin Watson statistic is rather complex but involves the residuals from an ordinary least squares (OLS) regression on a set of data. The following example illustrates how to calculate this statistic. Assume the following (x,y) data points: Pair One=(10,1,100)Pair Two=(20,1,200)Pair Three=(35,985)Pair Four=(40,750)Pair Five=(50,1,215)Pair Six=(45,1,000)\begin{aligned} &\text{Pair One}=\left( {10}, {1,100} \right )\\ &\text{Pair Two}=\left( {20}, {1,200} \right )\\ &\text{Pair Three}=\left( {35}, {985} \right )\\ &\text{Pair Four}=\left( {40}, {750} \right )\\ &\text{Pair Five}=\left( {50}, {1,215} \right )\\ &\text{Pair Six}=\left( {45}, {1,000} \right )\\ \end{aligned}Pair One=(10,1,100)Pair Two=(20,1,200)Pair Three=(35,985)Pair Four=(40,750)Pair Five=(50,1,215)Pair Six=(45,1,000) Using the methods of a least squares regression to find the "line of best fit," the equation for the best fit line of this data is: Y=−2.6268x+1,129.2Y={-2.6268}x+{1,129.2}Y=−2.6268x+1,129.2 This first step in calculating the Durbin Watson statistic is to calculate the expected "y" values using the line of best fit equation. For this data set, the expected "y" values are: ExpectedY(1)=(−2.6268×10)+1,129.2=1,102.9ExpectedY(2)=(−2.6268×20)+1,129.2=1,076.7ExpectedY(3)=(−2.6268×35)+1,129.2=1,037.3ExpectedY(4)=(−2.6268×40)+1,129.2=1,024.1ExpectedY(5)=(−2.6268×50)+1,129.2=997.9ExpectedY(6)=(−2.6268×45)+1,129.2=1,011\begin{aligned} &\text{Expected}Y\left({1}\right)=\left( -{2.6268}\times{10} \right )+{1,129.2}={1,102.9}\\ &\text{Expected}Y\left({2}\right)=\left( -{2.6268}\times{20} \right )+{1,129.2}={1,076.7}\\ &\text{Expected}Y\left({3}\right)=\left( -{2.6268}\times{35} \right )+{1,129.2}={1,037.3}\\ &\text{Expected}Y\left({4}\right)=\left( -{2.6268}\times{40} \right )+{1,129.2}={1,024.1}\\ &\text{Expected}Y\left({5}\right)=\left( -{2.6268}\times{50} \right )+{1,129.2}={997.9}\\ &\text{Expected}Y\left({6}\right)=\left( -{2.6268}\times{45} \right )+{1,129.2}={1,011}\\ \end{aligned}ExpectedY(1)=(−2.6268×10)+1,129.2=1,102.9ExpectedY(2)=(−2.6268×20)+1,129.2=1,076.7ExpectedY(3)=(−2.6268×35)+1,129.2=1,037.3ExpectedY(4)=(−2.6268×40)+1,129.2=1,024.1ExpectedY(5)=(−2.6268×50)+1,129.2=997.9ExpectedY(6)=(−2.6268×45)+1,129.2=1,011 Next, the differences of the actual "y" values versus the expected "y" values, the errors, are calculated: Error(1)=(1,100−1,102.9)=−2.9Error(2)=(1,200−1,076.7)=123.3Error(3)=(985−1,037.3)=−52.3Error(4)=(750−1,024.1)=−274.1Error(5)=(1,215−997.9)=217.1Error(6)=(1,000−1,011)=−11\begin{aligned} &\text{Error}\left({1}\right)=\left( {1,100}-{1,102.9} \right )={-2.9}\\ &\text{Error}\left({2}\right)=\left( {1,200}-{1,076.7} \right )={123.3}\\ &\text{Error}\left({3}\right)=\left( {985}-{1,037.3} \right )={-52.3}\\ &\text{Error}\left({4}\right)=\left( {750}-{1,024.1} \right )={-274.1}\\ &\text{Error}\left({5}\right)=\left( {1,215}-{997.9} \right )={217.1}\\ &\text{Error}\left({6}\right)=\left( {1,000}-{1,011} \right )={-11}\\ \end{aligned}Error(1)=(1,100−1,102.9)=−2.9Error(2)=(1,200−1,076.7)=123.3Error(3)=(985−1,037.3)=−52.3Error(4)=(750−1,024.1)=−274.1Error(5)=(1,215−997.9)=217.1Error(6)=(1,000−1,011)=−11 Next these errors must be squared and summed: Sum of Errors Squared =(−2.92+123.32+−52.32+−274.12+217.12+−112)=140,330.81\begin{aligned} &\text{Sum of Errors Squared =}\\ &\left({-2.9}^{2}+{123.3}^{2}+{-52.3}^{2}+{-274.1}^{2}+{217.1}^{2}+{-11}^{2}\right)= \\ &{140,330.81}\\ &\text{}\\ \end{aligned}Sum of Errors Squared =(−2.92+123.32+−52.32+−274.12+217.12+−112)=140,330.81 Next, the value of the error minus the previous error are calculated and squared: Difference(1)=(123.3−(−2.9))=126.2Difference(2)=(−52.3−123.3)=−175.6Difference(3)=(−274.1−(−52.3))=−221.9Difference(4)=(217.1−(−274.1))=491.3Difference(5)=(−11−217.1)=−228.1Sum of Differences Square=389,406.71\begin{aligned} &\text{Difference}\left({1}\right)=\left( {123.3}-\left({-2.9}\right) \right )={126.2}\\ &\text{Difference}\left({2}\right)=\left( {-52.3}-{123.3} \right )={-175.6}\\ &\text{Difference}\left({3}\right)=\left( {-274.1}-\left({-52.3}\right) \right )={-221.9}\\ &\text{Difference}\left({4}\right)=\left( {217.1}-\left({-274.1}\right) \right )={491.3}\\ &\text{Difference}\left({5}\right)=\left( {-11}-{217.1} \right )={-228.1}\\ &\text{Sum of Differences Square}={389,406.71}\\ \end{aligned} |
