We have shown in principle that Gaussian distributions, which are commonly observed in empirical psychology, result from a combination of a stochastic and a deterministic process. Thus we claim that causation and chance together generate the ubiquitous normal distributions. This composite process can be modeled by the Fokker-Planck approach.
Based on this line of thinking, we need to discuss how the parameters of the Fokker-Planck equation can be estimated from empirical data. These parameters are x0,k, andQ. x0is the deterministic shift of the attractor of the system to a new value.
kstands for the deterministic force that is exerted on any statexof the system, which is proportional to the slope of the attractor at any positionx.Qis the stochastic impact of randomfluctuations on the system, which expands and dissipates all structures of the system. To answer this question, we first recall the definitions of a probability distributionP(x) and of mean values (for more information see Info-Box4.5). We assume that in a series of measurements of the quantityx, wefind the valuesx¼x1, x¼x2,. . .,x¼xn,. When we repeat the measurement very often, we generate a time series where some values ofxappear more often than others. This is quantified by the relative frequency, which is defined by the number of measurementsnjthat yield a specificx¼xjdivided by the number of all measurements,n. Thus,
relative frequency¼nj=n ð4:21Þ We equate this relative frequency with the probability P(xj) to find, in an experiment or observation, the valuex¼xj. In practical applications, we may have to be satisfied with few measurements instead of the very many and use interpola- tions based on the hypothesis of a Gaussian.
Now we may define mean values (averages) that characterize distributions. The mean value ofx, which we denote byx, is given by
x¼x1P xð Þ þ1 x2P xð Þ þ2 . . .þxnP xð Þn ð4:22Þ Similarly,
x2 ¼x21P xð Þ þ1 x22P xð Þ þ2 . . .þx2nP xð Þn ð4:23Þ The variances2is defined by
s2 ¼x2x2 ð4:24Þ We mention the values ofxandx2for the idealized case of a displaced Gaussian, where the displacement isx0(cf. Info-Box4.5).
P xð Þ ¼ðk=ð2πQÞÞ1=2exp k
2Qðxx0Þ2
ð4:25Þ
x¼x0, ð4:26aÞ
x2x2¼Q
k ð4:26bÞ
In this way, by using mean values, we can directly determine the shiftx0of the attractor, which is simply the new mean of observations. This of course is the common approach in psychological methodology. Following (4.26a) and (4.26b), the mean values also contain information on the slope parameterkand the dissipa- tion parameterQ, however in the shape of the ratio ofQandk—we need to separate kfromQ. How can we isolate the parametersQandk, respectively? It is possible to measurekdirectly by means of relaxation experiments (Tschacher et al.,2015b). In such experiments, a state characterized by a value ofx¼xi(istands for“initial”) is observed (or experimentally established) at some initial time. The value ofx¼xi must be different from the equilibrium value x¼ x0. Then the (mean) time τ is measured as the duration until the equilibrium state atx¼x0is reached from the initial state. According to Info-Box4.5,
1
k¼τðrelaxation timeÞ ð4:27Þ By means of this result together with (4.26a) and (4.26b) we can also finally calculateQ. Info-Box4.5provides mathematical details on how we can explicitly calculate (4.26a) and (4.26b) based on (4.25). To do this, we derive the time dependence of the mean, x tð Þ, from the time-dependent Fokker-Planck equation.
We also discuss the time-dependent equation of the variance x2x2. The time- independent solution is shown to yield again (4.26b).
4.7 How Can We Measure the Fokker-Planck Parametersx0,k,Q? 55
Info-Box 4.5: Time-Independent and Time-Dependent Mean Values In the time-independent case, the mean values connected with the time- independent, displaced Gaussian (4.25) can be calculated as follows:
x¼
Z 1
1xP xð Þdx¼ Z 1
1xNexp k
2Qðxx0Þ2
dx ð4:28Þ
Replacexbyξ¼xx0. Then
x¼x0
Z 1
1
Nexp k 2Qξ2
dξþ Z 1
1
Nξexp k 2Qξ2
dξ ð4:29Þ
In this equation, thefirst integral yields 1 because of the definition of the normalization constantN, whereas the second integral yields 0 because the contributions forξ>0 cancel out those forξ<0. Thus we obtain (4.26a).
To derivex2we writeP xð Þin the form withð α¼k=2QÞ ð4:30Þ P xð Þ ¼Nexpðαðxx0ÞÞ2 ð4:31Þ Then
x2 ¼N Z 1
1
x2expαðxx0Þ2
dx ð4:32Þ
The normalization constantNis given by N1¼
Z 1
1
expαðxx0Þ2
dx ð4:33Þ
Replacing x by ξ ¼ x x0 leaves N unchanged, whereas (4.32) is transformed into
x2 ¼N Z 1
1ξ2expαξ2
dξþ2x0N Z 1
1ξexpαξ2 dξ þx20N
Z 1
1expαξ2
dξ ð4:34Þ
In this equation, the second term on the right-hand side vanishes as we have discussed above, while the third term is justx20x2(cf. (4.26a)). Thus we may rewrite (4.34) as
(continued)
Info-Box 4.5 (continued) x2x2¼N
Z 1
1ξ2expαξ2
dξ ð4:35Þ Using calculus or simply a table of integrals we obtain for (4.35)
¼1=ð Þ2α ð4:36Þ or due to (4.30)
x2x2¼Q=k ð4:37Þ Now we turn to the time-dependence of meanxand variances2¼x2x2. To this end we need the time-dependent Fokker-Planck equation (4.16). To derive an equation forx, we multiply both sides by xand integrate overxfrom 1to1, so that
Z 1
1xdP xð Þ;t dt ¼
Z 1
1xd
dxðk xð x0ÞP xð Þ;t Þdxþ Z 1
1xQd2P xð Þ;t dx2 dx
ð4:38Þ In this equation, the left-hand side can be cast into the form
d dt
Z 1
1
xP xð Þ;t dx¼d
dtx tð Þ ð4:39Þ Thefirst term on the right-hand side can be evaluated by“partial integra- tion”to yield
x k xð ð x0ÞP xð Þ;t Þj11 Z 1
1
k xð x0ÞP xð ÞÞ;t dx ð4:40Þ
BecauseP(x;t) must vanish atx¼ 1andx¼ þ1, thefirst term in (4.40) vanishes altogether, while the second term can be written as
k x tð Þ x0
ð4:41Þ The second term on the right-hand side of (4.38) can be evaluated again by partial integration and becomes zero. Thus all in all, we arrive at the following equation for the time-dependent mean valuex tð Þ:
(continued) 4.7 How Can We Measure the Fokker-Planck Parametersx0,k,Q? 57
Info-Box 4.5 (continued) d dtx tð Þ
¼ k x tð Þ x0
ð4:42Þ
which possesses the solution
x tð Þ ¼x0þaekt ð4:43Þ wherea¼xð Þ 0 x0:
Most importantly, (4.43) shows thatx(t) relaxes within a timeτ¼1=kto the central valuex0.
For the sake of completeness, we quote the relaxation equation for x2ð Þt x tð Þ2:
d
dtx2ð Þ t x tð Þ2
¼ 2k x 2ð Þ t x tð Þ2
þ2Q: ð4:44Þ
According to it, the variance relaxes to its equilibrium valuex2x2¼Qkas we have found above (4.26b).
References
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Tschacher, W., Haken, H., & Kyselo, M. (2015b). Empirical and structural approaches to the temporality of alliance in psychotherapy.Chaos and Complexity Letters, 9, 177–191.
Tschacher, W., & Ramseyer, F. (2009). Modeling psychotherapy process by time-series panel analysis (TSPA).Psychotherapy Research, 19, 469–481.