What does this equation mean?

In equilibrium statistical mechanics, the free-energy difference $\Delta$ F between two control-parameter values is a state function. A rapid drive between them generally performs work W>$\Delta$ F on average because dissipation is positive. It would seem that estimating $\Delta$ F requires an impractically slow, quasistatic protocol. The Jarzynski equality demonstrates otherwise: $\left\langle e^{-\beta W}\right\rangle = e^{-\beta\Delta F}, \qquad \beta = \frac{1}{k_{\mathrm B}T}$. The equality holds for an ensemble of realizations beginning in equilibrium, even when each realization is driven far from equilibrium [ 2 ] . It does not say that the arithmetic mean work equals the free-energy difference. The exponential average disproportionately weights rare, low-work trajectories. That feature makes the theorem exact and can make its estimator statistically difficult: if the trajectories that dominate the exponential average are poorly sampled, finite-data estimates are biased.