Biased-step entropy production model ($S_n=a(2K-n),a=\ln (p/q)$)

Take $n$ independent steps, each $+1$ with probability $p>1/2$ and $-1$ with probability $q=1-p>0$. Reverse a step sequence by reading it backward and changing each sign; use the same biased law for both experiments. With $K$ positive steps, the forward sequence weight is $p^K{q}^{n-K}$ and its reversed weight is $p^{n-K}{q}^K$. Their logarithm ratio isHere $S_n$ is dimensionless entropy production, not an entropy in joules per kelvin. A Markov chain on a three-state ring gives a state-space realization: start uniformly, take a clockwise step with probability $p$ and a counterclockwise step with probability $q$. The uniform stationary distribution makes endpoint weights cancel. A physical interpretation as exchanged heat requires local detailed balance with bath entropy $\pm k_{B}a$ per directed step; the abstract model alone specifies path probabilities.

For an allowed value $s=a(2k-n)$, the binomial distribution givesThis is an exact same-law finite fluctuation theorem, with no long-time approximation. For $p=3/4$, $n=9$ and $k=6$, $s=3\ln 3$ and $p_n(s)/p_n(-s)=27$. The positive mean is $⟨S_n⟩=na(p-q)$, although the all-negative path has nonzero probability $q^n$. Directly, $⟨e^{-S_n}⟩=(p{e}^{-a}+q{e}^a{)}^n=(q+p{)}^n=1$.

The left panel shows exact probability masses for $p=3/4$, with negative values highlighted; the right panel tests the logarithm ratio against $s$. These are model values, without experimental measurements or simulated measurement noise.

A single step has negative entropy production with probability $q$, independent of how long an observation lasts. A whole window is negative only if negative steps outnumber positive ones:For even $n$, the tie $K=n/2$ has zero entropy production and is excluded. At $p=3/4$, the negative-window probabilities for $n=1,9,21$ are approximately $0.25$, $0.04893$, and $0.00642$. Longer windows still contain negative increments: their expected number is $nq$. Their total is usually positive because the expected number of positive increments is $np$.

Only odd window lengths are plotted to avoid alternating tie effects. The left panel gives the exact negative-window probability and the upper bound $(2\sqrt{pq}{)}^n$; the right gives expected counts $np$ and $nq$. For steps of fixed duration $\delta t$, physical duration is $n\delta t$. These are counts and probabilities predicted by the model, not bead traces. The bound follows from $\mathrm{Pr}(S_n<0)\le ⟨e^{-S_n/2}⟩=(2\sqrt{pq}{)}^n$. Since $2\sqrt{pq}<1$ for $p\ne q$, negative totals are exponentially suppressed even though the negative-step fraction stays $q$.