How MBAR joins windows into one equilibrium distribution

Evaluate each saved configuration in every window

Pool $N$ saved configurations $\mathbf r_n$. Window $k$ contributes $N_k$ samples with reduced potential $u_k=\beta(U+w_k)$ and dimensionless free energy $f_k=-\ln Z_k$. A saved configuration can be evaluated under every window’s potential.

The mixture density is proportional to $\sum_k N_k e^{f_k-u_k(\mathbf r)}$. Configurations common in many contributing windows need less target weight. The factors $N_k$ correct for unequal sampling amounts.

Source: Michael R. Shirts and John D. Chodera · Giovanni Bussi

Find relative heights by self-consistency

If the constants $f_k$ were known, reweighted sums would estimate each partition function. Requiring those estimates to equal the partition functions already used gives the self-consistent equations. Iterate and subtract the first free energy after each update to fix an arbitrary common zero.

At a shared temperature with known biases, these equations can also be written in the binless WHAM form used by the PLUMED tutorial. Evaluate exponential sums in the logarithmic domain to avoid numerical underflow. Convergence means the equations for these samples were solved. Trapped trajectories can also produce a stable solution.

Source: Michael R. Shirts and John D. Chodera · Giovanni Bussi

Convert constants into target weights

For the unbiased target $u_0=\beta U$, first compute unnormalized weight $v_n$ for every saved configuration. Normalize by their sum. A target expectation is the resulting weighted average. The target itself need not have a directly sampled trajectory.

At a shared temperature, with windows differing only by known biases, the common $e^{-\beta U}$ cancels. Bias energies alone then suffice for reweighting. The hidden coordinates still have to be sampled; algebraic cancellation cannot explore them.

A complete example with sixteen configurations

The target has equal-energy left and right configurations. Window A penalizes the right by reduced energy $\ln 3$; B penalizes the left by the same amount. A contributes twelve observations, nine left and three right. B contributes four, one left and three right. Their equal partition functions allow $\hat f_1=\hat f_2=0$.

The mixture denominator is $12+4/3=40/3$ on the left and $12/3+4=8$ on the right. Unnormalized target weights are $3/40$ and $1/8$. Ten left observations sum to three quarters, as do six right observations. Normalization recovers one half on each side. Pooling raw counts instead gives left probability $10/16=0.625$.

Counts were chosen equal to each window’s expected proportions so the calculation fits on paper. Random data will not generally recover one half exactly. Independent repeats and the next chapter assess that error.

A final histogram still has its own error

MBAR does not bin input energies. Drawing $F(s)$ still requires a density estimate. A bin of width $\Delta s_j$ and total target weight $P_j$ has density approximately $P_j/\Delta s_j$. Wide bins smear barriers; narrow ones can have no observations. An arbitrary pseudocount cannot turn an unvisited region into a measurement.

Inspect overlap too. Floating-point calculations may retain tiny connections across practically disconnected regions. Those connections do not establish a well-measured relative height. The experiment retains unsupported regions and reports usable coverage.

Source: Michael R. Shirts and John D. Chodera · Giovanni Bussi