Estimation for quantity of statistics

With only 2000 trajectories and under suitable experimental conditions, such as trajectory length, resolution, modulation depth etc., it seems that IFT works well in principle (deviation < 20%, Figure 2-5). In the simulation, IFT is even confirmed with higher accuracy (deviation < 5%, Figure 2-8) when the number of trajectories is increased to 100,000. Nevertheless, is this number large enough for other conditions?

To show this concern is necessary, we take longer observation time. Figure 2-10 demonstrates that the deviation is increased with the number of period. The example Figure 2-10 (a) has a point separated far from others, which seems to be absurd at first glance. But in fact, this extreme case appears typically.

0 10 20 30 40 50 60 70 80 90 100

0 0.5 1 1.5 2 2.5 3 3.5

(a) 4

Length of trajectories (# of periods)

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Figure 2-10

The mean over 100,000 trajectories for each period with the modulation depth and resolution . Same as the former examples, IFT is calculated 5 times for each period in order to examine the deviation. Notice that there is a point at the upper right corner. Note the red dashed rectangle is just Figure 2-8 and Figure 2-10 (b) is the magnified view of (a) omitting the point at the upper right corner.

The result in Figure 2-10 is due to the structure of the non-uniform average . Because the entropy annihilating trajectories may occur seldom but are exponentially weighted, they contribute substantially to the left hand side of IFT. To keep , each of entropy annihilating trajectories needs a large quantity of entropy producing trajectories to balance. Therefore, the variation on the number of entropy annihilating trajectories would affect the results of IFT enormously, especially when the number of annihilating trajectories is small.

With increased observation time, the mean of total entropy production shifts in a positive direction and spreads outwards in both directions (Figure 2-11 and Figure 2-12). The number of annihilating trajectories also decreases, and it leads to the larger deviation of IFT. In most situations, a large number of entropy producing trajectories

0 10 20 30 40 50 60 70 80 90 100

Length of trajectories (# of periods)

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lacks sufficient number of annihilating ones to balance, which brings about the result of . But sometimes, too many, or even a little more entropy annihilating trajectories are generated, resulting in . This explains the distribution of in Figure 2-10.

Figure 2-11

Histograms of total entropy production with different periods of (a) 20T, (b) 60T, and (c) 100T, respectively. The mean and the width (two standard deviations) of are also shown in each figure.

0 100 200 300

0 100 200 300

-10 -5 0 5 10 15 20 25 30

0 100 200 300

(a)

(b)

(c)

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Figure 2-12

The mean of total entropy production is proportional to the trajectory length. It seems surprising at first glance but in fact can be explained easily; because the total process is Markovian, the change of from periods must be the same as that from and so on.

Except for the rough description from Figure 2-11, the relation between the probability of entropy producing trajectories and entropy annihilating trajectories, in fact, obeys the detailed fluctuation theorem (DFT) [7]

relaxed into the corresponding periodically oscillating distribution. In this case, the

10 20 30 40 50 60 70 80 90 100

Length of trajectories (# of periods)

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trajectory length is very long and thus dominates . Therefore, DFT is valid in principle and suitable for the estimation.

Figure 2-13

The test diagram of DFT for the data set (i). The red asterisk denotes the mean of total entropy production and the points near have more accuracy of the DFT. The blank on the right side represents the missing points due to the lack of realizations; some positive entropy production can not correspond to their negative entropy production .

The dashed line is with slope = 1.

To estimate the trajectory number required to verify IFT, we take an example as follows. Assumed there are two sets of data to verify the IFT of two-state system.

(i) 20 periods and over 100,000 trajectories (j) 40 periods and over 100,000 trajectories.

As shown in Figure 2-10, IFT works very well in the data set (i) but not in the data set (j) and we wonder how large the trajectory number does (j) require to get a

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Whether IFT works well depends on whether DFT works well throughout Figure 2-13. Of course we cannot examine DFT for each , so our method of estimation is to examine DFT for the most frequency value . The number of trajectories with (with error in the simulation) in (i) is approximately 375 and the corresponding number of trajectories with is estimated to be 64.9 according to DFT but is 66 actually.

The little variation doesn’t matter and the number 64.9 of trajectories of is large enough, so that the number of trajectories is also large enough to verify IFT in (i).

Because the trajectory length with decreases exponentially with increased according to DFT, to maintain these rare trajectories to balance those with , the number of total trajectories also has to be increased exponentially, that is

(2-9)

Where is the trajectory number with which IFT can be verified satisfactorily in the data set (i); is the required trajectory number in a certain data set (j) to satisfy IFT with the same accuracy as in (i); and are the means of total entropy production in (i) and (j), respectively.

With this estimation, for each trajectory length can be calculated easily (Figure 2-14). On the other hand, because the total entropy production is linearly weighted in rather than exponentially weighted, would converge to the stable value without large trajectory number. Applying to the calculations of IFT, corrected and satisfactory results are shown in Figure 2-15.

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Figure 2-14

The reference 100000 is in the condition with 20 periods and . With increased length from 20 to 100 periods, is also increased linearly and it results in the exponentially increased .

Figure 2-15

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The results of IFT versus the trajectory length over 100000 trajectories ([blue circles]

taken from Figure 2-10 (b) omitting the separate point) and over [red asterisks]

corrected trajectory number shown in Figure 2-14. Note IFT is calculated five times for each length.

The results of IFT with the corrected numbers of trajectories are obviously more accurate than the original ones. It means this estimation is correct at least in the order of magnitude. Nevertheless, with the longer trajectory length, the results are not as accurate as the result of 20 periods; it means the required number is increased at least exponentially with the mean .