A strange inconsistency between a calculation and a simulation of a TASEP

Question

Consider the following stochastic process, called a totally asymmetric simple exclusion process (TASEP), on the integers $\mathbb{Z}$:

The process evolves over discrete time steps $T = 1, 2, \ldots \infty$. Denote the contents of the integer $n$ as $x(n)$. Initially, at every integer $n$, $x(n)=1$ with probability $0.5$ and otherwise $x(n)=0$.

If for some $n$ we have that $x(n)=1$ and $x(n+1)=0$, then with probability $0.5$, at the next time step we will have $x(n)=0$ and $x(n+1)=1$. (In other words, every $1$ moves right with probability 0.5, assuming there isn't a $1$ blocking it in its new target position).

It's simple to see that the initial distribution (where we have $1$ with probability $0.5$) is stationary. (Edit: Based on page 2 of this paper https://arxiv.org/abs/cond-mat/0101200), this means that in expectation we should expect the number of $1$s passing through $n=0$ to be $T/4$, where $T$ is the number of time steps that have passed.

Now consider the following program, which I simulated on my computer:

Initialize a 0-1 array a[-1000,1000] such that a[n] = 1 with probability 0.5.

Simulate the above Stochastic process for 100 iterations. Count the number of times a[0] goes from 0 to 1.

The result of this program is consistently around $15$, but by the above reasoning we would expect $25$. In fact, it seems it will always be on average a $0.15$ fraction of the number of iterations (even doing 200, or 300 iterations at a time).

So is the math wrong, or is my simulation idea wrong?

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Actual code I used: https://pastebin.com/iPz1S1fK ("count" is the number that comes out as 15; prob(50) means "with probability 50"; Update() performs a single iteration of the TASEP)

Answer 1

The main difference between TASEP in the literature and your model is as follows:

• In your model, we try to move every particle to the right simultaneously each round. So, once a particle makes a move at the given time step, it cannot make another move until other particles are tested.

• On the other hand, in TASEP, each particle tries to move according to its own exponential clock, so it need not wait other particles being tested to make another jump, or if lucky, multiple jumps.

Since a particle must wait to see other particles also attempted move in your dynamics, not only your model is different from TASEP, but it may possible induce a some correlation between particles. As a comparison, here are simulations of both your model and TASEP

TASEP on $\mathbb{Z}/1000\mathbb{Z}$, with the initial distribution as Bernoulli product measure. The following depicts the configurations $(\eta_n : n \in \{1501, \cdots, 2000\})$. The $i$-th row corresponds the configuration $\eta_{i+1500}$ (so that time flows downward), and black dots represent $1$'s and white dots represent $0$'s.

Your model, on $\mathbb{Z}/1000\mathbb{Z}$, with the same initial distribution. It depicts only the configurations between time $1501$ and $2000$, using the same visualization rule as above.

So, even visually we anticipate that your model tends to spread particles more evenly than TASEP, thus showing less granular texture. Plotting the fraction of of particles that moves in each time step shows clear differences:

Answer 2

I ran my own simulation with many more iterations and found the result to be $0.1465\times T$

I also monitored other detail to explain this result. Specifically, I tracked the values of $x[n-1]$, $x[n]$, and $x[n+1]$ for each iteration. There are $8$ possible collective states for these values, and these states are not equally probable: $$\begin{matrix} [0,0,0]\approx0.086\\ [0,0,1]\approx0.121\\ [0,1,0]\approx0.172\\ [0,1,1]\approx0.121\\ [1,0,0]\approx0.121\\ [1,0,1]\approx0.172\\ [1,1,0]\approx0.121\\ [1,1,1]\approx0.086 \end{matrix}$$ Only two of these allow for $x[n]$ to go from $0$ to $1$, those being $[1,0,0]$ and $[1,0,1]$. The probability of being in one of these states is $0.121+0.172=0.293$, and with the state-change probability of $0.5$ this gives the result of $0.1465$ found by the simulation.

Note: No claim is made that the simulation accurately represents the process described in the linked document. In fact, we can say with certainty that it is not an accurate model. The detail above was only intended to explain the unexpected result of the simulation.

I modified the simulation, cycling through the array in reverse to allow for a jump out of $x[n]$ before attempting a jump into $x[n]$. The result was that the $8$ collective states are now equally probable (as expected) and the number of 'particles' moving through a given position is $0.125\times T$, also expected with this model (per Sangchul Lee's comment).