Taking the log of E [ e − − β β W ] = e − − β β Δ Δ F , and use the cumulant expansion up to the second cumulant, we obtain E [ W ] − − Δ Δ F ≈ ≈ 1 2 β β σ σ W 2 . The left side is the work dissipated into the heat bath, and the right side could be interpreted as the fluctuation in the work due to thermal noise.