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A Criterion on Practical Use

Suppose a method of technical analysis is known to generate Markov times {t(} as signals of market turning points. a forecaster may, in addition, want to know the size of the probability СЂ(С‚(- < < �), i = I, 2, . . . before investing resources in applying this rule. Indeed, if this probability is less than one, then the rule may never give a signal. This may be uneconomical. Yet, in terms of formal statistical criteria, there is nothing wrong with a Markov time that fails to be finite. In this section, I discuss which categories of technical analysis are likely to yield finite Markov times.

Definition.   We say thai a Markov time С‚ is finite if

/>(С‚<СЃСЃ) = I.

It is clear that a Markov time that is not finite may fail to be a financially rewarding method of forecasting since it may never give a positive or negative signal in spite of being well defined.

It turns out that only in very few cases the {С‚,-} generated by technical analysis will be finite, hence usable, in the sense above. The major exception is the method of moving averages. I show below the condi �tions under which the moving average method generates finite Markov times.

Proposition 1. If the observed process {X,} is stationary and m-de-pendent, all moving average methods characterized by (3)-(4) generate finite Markov times.

Proof. Let Z, be given by (3). If X, is stationary, then Z, and Z,Z,_, are stationary (e.g., Breiman 1968, proposition 6.6). Also note that, due to stationarity, ВЈ[Z(] = 0, hence 0 < P{Z, > 0) < I, unless Z, = 0 almost surely. Let

Clearly, P(Y, a 0) < 1. Now, I apply the theorem provided in the Appendix. Consider

P{ K,<0, at least once for t^n) = 1 - P(Y0>0, Yx > 0,. .. Kn>0).

The theorem in the Appendix requires this probability be one, as n goes to infinity. To show that this is indeed true, note that, if A", is m-dependent, the K/s sufficiently apart will also be independent. Thus, select an integer Рё so that Y, and Y,+u are independent. We utilize such K/a sufficiently apart to write, for large n.

W0>0,   РЈ,>0,... Yn>0)*P{Y0>Q)P(Yu>0)...P(Yka>0)

y, = z,z,

ii -1 *

r = 0, 1

by stationarity and m-dependence. As we let к — � 00,

P(Y0> 0)*-*0,

since P{Y0 > 0) < I, as shown above. Thus,

P(Yns Oat least once) - 1.

Hence, all conditions of the theorem supplied in the Appendix are satisfied and Markov limes т,, т2.....т„ are finite.

Note that assumptions such as stationarity and mixing, a simple form of which is m-dependence, are needed to obtain this result. This might seem unnecessary, but without similar assumptions, one cannot guarantee the finiteness of these Markov times. Indeed, if the process X, is explosive enough, then a moving average method may not gener �ate finite Markov times.4

One implication of this is that trend crossing methods of technical analysis might not always yield finite Markov times—even after a pre-

4. Here is an example provided by the referee. Let Xt be generaled by an explosive AR(I) model:

-Y( = 0*(_,+e(1   /=1,2....

where 0 > 2. and �, are independently and identically distributed random variables with uniform distribution over the interval [ - I, 1), Note that Ihe event E = {X, is always larger than I and tends to �} has positive probability. Now. consider two moving aver �ages with 1 and 2 terms, respcclivcly. Then Z, defined by formula (3) is equivalent to

Z, = (|/2)|<0 - i)X,-t + �,].

With this Z,, we have С‚ = oo on the event E.

cise definition is adopted. To illustrate this, note that a trend line T(s) such as the one shown in figure 5, will admit the representation,

ns) = a, + b,s   s>t, (12)

where a, and b, > 0 are /,-measurable intercept and slope. Now, the difference,

D, = x, - a, - b,s,   s>t,

is clearly not stationary. So the assumptions of proposition 1 are not satisfied for /),, s > t. Hence, even with stationary and m-dependent {x,), as j goes to infinity, P(xt - a, - b,s < 0) may equal one, and xi may never cross the trend line 7"im again. Under these conditions, implied Markov times may be infinite even though they are well de �fined. For example, this may be the case if {x,} is given by

x, = �, + .5e,_,

where the distribution of the independently and identically distributed (i.i.d.) errors {€,} has finite support:

/>(€, sta) = 0   0 < a < �.

Remark. Note that, even if a rule generates signals that are finite with high probability, with, say, /*(т, < со) = .9. this may still create major problems for practical users. In fact, such a probability implies that one out of every 10 signals may be infinite—assuming that the signals are sufficiently apart, and that they arc not correlated. For a forecaster working in real time, a long waiting period then implies either a large (but finite) т, or, with smaller probability, an outcome where no signal will be given. In this latter case, the forecaster should switch to other rules. Since technical analysis never specifies how one rule should be abandoned in favor of others, the requirement that P(j. < со) = I is less trivial than it seems at the outset.