ON PREDICTION OF TIME SERIES ON THE BASIS OF RANK CORRELATION COEFFICIENTS

Transcription

1 Janusz L. Wywiał University of Economics in Katowice ON REDICTION OF TIME SERIES ON THE BASIS OF RANK CORRELATION COEFFICIENTS. Basic properties of ran correlation coefficient Let rans of the series of observations of a time series be denoted by: a,t, t,,. Let l,t be number of the implication of the type: a,t > a,h, if t > h, where t,,, h,, t. Let l ',t be number of the following implication: a,t < a,h, if t > h, where t,,, h,, t. The Kendall s (958) ran correlation coefficient is as follows: where: ' ( L L ) Q, () ( ) t t,, ' ' L l, t. t L l Moreover, the statistic Q can be rewritten as follows, see Höffding (97): where Q G () L G, 0 G. ( )

2 0 Janusz L. Wywiał The values of the statistics L, ' L and Q will be denoted by l, ' l and q, respectively. When all permutations (a,, a,,, a, ) are equally probable than E(Q ) ρ 0. In this case the expression (3) show the distribution of the ran correlation coefficient under the assumption that. Now let us assume that the permutation (a,, a,, a,3, a, ) is chosen with probability proportional to the values ( + l ) 3. In this case the probabilities (a,,,a, ) are given in the ninth column of Table. This leads to the distribution of the ran correlation given by the expression () and E(Q ) ρ 0.37 and D(Q ) Let us note that more about the conception of determining distribution of the statistics Q and L can be found in the paper by Höffding (97). Ran permutation and distributions of the ran correlation coefficient Table No. a,,, a, a 3,,, a 3,3 a,, a, l ' ' l l l q (a,,, a,),,3,,,3, ,57,,,3,,3, 5 /3 0,099 3,3,,,3,, 5 /3 0,099,3,,,,3, /3 0,057 5,,,3,3,, /3 0,057 6,,3,,3,, ,09 7,,3,,,3, 5 /3 0,099 8,,,3,,3, /3 0,057 9,3,,,3,, /3 0,057 0,3,,,,3, ,09,,,3,3,, ,09,,3,,3,, /3 0,0 3 3,,, 3,,, /3 0,057 3,,,,,3, ,09 5 3,,, 3,,, ,09 6 3,,,,,3, /3 0,0 7 3,,,,3,, /3 0,0 8 3,,,,3,, /3 0,0 9,,,3 3,,, ,09 0,,3, 3,,, /3 0,0,,,3 3,,, /3 0,0,,3, 3,,, 5 /3 0,0037 3,3,, 3,,, 5 /3 0,0037,3,, 3,,, ,0005 Source: Own calculations.

3 ON REDICTION OF TIME SERIES ( Q q 0) / /8 5/ / 5/ /8 / q q / 3 q / 3 ρ q 0, (3) q / 3 q / 3 q 0,57 q 0,98 q / 3 0,87 q / 3 ( Q q ρ 0.37) 0,76 q 0, () 0,07 q / 3 0,007 q / 3 0,00 q. Ran prediction Let us start consideration with the following example. Example. On the basis of the lines no. 5,,3 and of Table we calculate the following probability: (a 3, 3, a 3,, a 3,3, ρ 0) (a, 3,a,, a,3, a,, ρ 0) + (a,, a,, a,3, a, 3, ρ 0) + (a,, a, 3, a,3, a,, ρ 0) + (a,, a, 3, a,3, a,, ρ 0) / /6. For instance: So, ( a, a3, 3, a3,, a3,3, ρ 0) ( a,, a, 3, a,3, a,, ρ 0) /. a 3, a, a, ρ 0 / 6 ( ) 3, 3, ( a a a 3, a, a, ρ 0) 3,3, 3, 3, 3,3 a,,3,.

4 Janusz L. Wywiał Now on the basis of the line no. of Table we have: (a 3,, a 3, 3, a 3,3, Q 0, ρ 0) (a,, a,, a,3, a, 3, Q 0, ρ 0) /. Hence: (a, 3 a 3,,a 3, 3, a 3,3, Q 0, ρ 0) a,, a,, a,3, a, 3, Q 0, ρ 0 a, a 3, a, Q 0, ρ 0 ( ) ( ) 3, 3, 3,3 /. / So, it means that a, 3 under the condition that a 3,, a 3, 3, a 3,3, ρ 0 and,q 0 with probability one. In this case let us suppose that in the periods t,,3 the observations of the time series are y 0., y 0.8 and y The rans of this values are (,3,). According to the obtained results the rans of all four values of the time series are the elements of the sequence: (a,,a,,a,3, a, ) (,,,3). Hence, in the fourth period the predicted value of the time series has the ran 3. This leads to the conclusion that the predicted value of the time series in the fourth period is between values y 0. and y 0.8. Hence 0. < y < 0.8. Example. Similarly lie in the previous example the following probabilities can be calculated on the basis of the lines no. 5 and of Table : because a a 3, a, a, Q, ρ 3 a, 3, 3, 3, a, a3, 3, a3,, a3,3, Q, ρ 0 3 a, 3, a,, a,3, a,, Q, ρ 0 3 / / a3, 3, a3,, a3,3, Q, ρ 0 3 a, 3 a3, 3, a3,, a3,3, Q, ρ 0 3 a,, a,, a,3,a, 3, Q, ρ 0 3 / / a3, 3, a3,, a3,3, Q, ρ 0 3 a 3 a

5 ON REDICTION OF TIME SERIES 3 The obtained result can be interpreted as follows. Under the assumptions that ρ 0 and Q /3 and a 3, 3, a 3,, a 3,3, the probability that th ran is equal to 3 () is equal to 0.5. The generalization of the derived in Examples and results is as follows. ( b a a, a a,..., a a, Q q, ρ ) a, -, -, -,- - 0 ρ0 ( a ), b, a,- b-, a -,- b-,...,a, b, Q q0, ρ ρ0 ( a a, a a,..., a a, Q q, ρ ρ ) -, -, -,- - 0 (5) Example 3. Now, let us consider the problem of the prediction of two rans. For instance on the basis of the lines no. 6, 0 and of Table we have: /3 a 3, b ( a,3 a,a, b a,, a,, Q 0, ρ 0) /3 a, b /3 a, b 3 because: ( a,3 3,a, a,, a,, Q 0, ρ 0) ( a,, a,,a,3 3,a,, Q 0, ρ 0 ) / ( a,,a,, Q 0, ρ 0) 3/ 3 ( a,3,a, a,, a,, Q 0, ρ 0) ( a,, a, 3,a,3,a,, Q 0, ρ 0 ) / ( a,,a,, Q 0, ρ 0) 3/ 3 ( a,3,a, 3 a,, a,, Q 0, ρ 0) ( a,, a,,a,3,a, 3, Q 0, ρ 0 ) / ( a,,a,, Q 0, ρ 0) /8 3 Similarly, lie in Example let us assume that y 0. and y 0.8. On the basis of the above results we can write that 0. < y < y 3 < 0.8 with probability /3, y 3 < 0. < y < 0.8 with probability /3, y < 0. and y 3 > 0.8 with probability /3. The obtained result can be straightward generalized into the case of the prediction of m-rans provided that the -rans and values of Q +m and ρ +m are fixed as follows: ( a,m+ b m+...;a ;a,m+ a b m+ ; Q ;...; a q; b a ρ a;a ) a m,m m +m +m 0 ( a ), b;...;a,m+ bm+;...;a, b; Q+m q0; ρ +m ρ0 ( a a ;a a ;...;a a ; Q q ; ρ ρ ) m, m, m,m,n m ρ n +m m, 0 +m m, 0 0 ;... (6)

6 Janusz L. Wywiał Example. Similarly, lie in the example the expressions (), (5) and the lines no. 5, of Table lead to the following: ( b a 3,a,a, Q 0., ρ 0.37) a, 3, 3, 3,3 because (a 3, 3, a 3,, a 3,3, Q 0., ρ 0.37) (a, 3, a,, a,3, a,, Q 0., ρ 0.37) + + (a,, a,, a,3, a, 3, Q 0., ρ 0.37) , ( a 3, a, a, Q 0., 0.37) a, 3, 3, 3,3 ρ ( a ), 3, a,, a,3,a,, Q 0., ρ ( a 3, a, a, Q 0., ρ 0.37) 3, 3, 3,3. b 3 b Hence, under the assumptions that (3,,) are the rans of the time series in three periods and ρ 0.37 and Q 0., the probability that ran of the time series in the fourth period is equal to 3 () is equal to (0.97). Let us assume that y 5.8, y 5., y 3.9. The above results let us infer that y > 5.8 with probability else 5. < y < 5.8 with probability Example 5. Let (,) are the rans of the time series in two periods. We are going to predict the rans of the time series in next two periods under the assumptions that Q > 0.5 and ρ Under these assumptions, the expression (6) and the lines no.,, 3 of Table lead to the following: ( a,a b a, a, Q > 0.5, 0. 37) because instance a,3,,, ρ a a 3 a b b b 3 ( a,3,a, a,, a,,q > 0.5, ρ 0. 37) ( a,,a, 3,a,3,a,,Q > 0.5, ρ 0.37) ( a,a,q > 0.5, ρ 0.37),, Hence, if in the first two periods the time series increases then in two next periods the rans of the time series are and with probability 0.79 or and 3 with probability 0.79 or 3 and with probability 0.. Hence, the rans of the time series can be determined by the sequence (,3,,) with probability 0.79 or

7 ON REDICTION OF TIME SERIES 5 the sequence (,,,3) with probability 0.79 or the sequence (,,3,) with probability 0. provided that the distribution of the Kendall s ran coefficient Q is defined by the expression () and Q > 0.5. Hence, when y 5. and y 6. are observed then 5. < y 3 < 6. < y with probability 0.79, y 3 > y > 6. with probability 0.79 and y > y 3 > 6. with probability 0.. The ran coefficient distribution Q can be estimated on the basis of a time series observations. In order to do it the time series representing by the sequence: (y, y,, y t,, y N ) can be divided into segments: (y,, y ), (y +,, y ),, (y h+,, y (h+) ),,(y (H-)+,, y N ) where NH. Next, the each segment is transmed into sequence of rans denoted by () () () () ( h) ( h) ( H ) ( H ) ( a,..., a ), ( a,..., a ),...,( a,..., a ),...,( a,..., a ). Finally, the frequencies of the each permutation of the rans is calculated. This let us evaluate the distribution of the permutation as well as the distribution of the ran coefficient. Let us note that the presented procedure can be useful in the case when the time series is long and segments rather short. The large size of the segment leads very quicly to enormous number of permutations of the rans sequence. Conclusions The proposed method can be useful especially to prediction of a stationary time series treated as the sequence of independent and identically distributed random variables. In this case we can assume that all permutations of the rans are equally probable, so ρ 0. It seems that the same procedure can be useful in the case when random variables is not stationary. In this case the probability distribution of the ran permutations should be estimated. The accuracy of the considered prediction method can be based on the ex-post analysis of frequency of exact prediction of the ran of future observations of the time series. The quality of the proposed prediction procedure can be assessed on the basis of simulation analysis of the actual empirical or artificial time series. But it needs the separate research. Other correlation coefficients lie the well nown Spearman s ran correlation coefficient should be involved in such a research. Acnowledgements The author is grateful to Reviewers valuable comments.

8 6 Janusz L. Wywiał References Höffding (97), On the Distribution of the Ran Correlation Coefficient τ When the Variates Are Not Independent, Biometria, Vol. 3, No. 3/, pp Kendall M.G. (958), Ran Correlation Methods, C. Griffin and Company, London. ON REDICTION OF TIME SERIES ON THE BASIS OF RANK CORRELATION COEFFICIENTS Summary In the paper the problem of prediction of a time series is considered. Time series observations can be measured on order scale. On the basis of observed rans of values of the variables observed in the past periods a ecast of the ran of the observation in the future period is determined. The proposed method results from the derivation of the distribution of the well nown Kendall s ran coefficient. The paper was inspired by a lecture of Jean H.. aelinc who gave it at the University of Economics in Katowice when he received the title of doctor horis causa of the University in 987.