Variational Principles in Conformation Dynamics - FU Berlin, FB MI

46 A. AppendixMultiplying the last expression by e −θt gives the desired result.Now we can confirm the claims about the expectation and variance.stochastic integrals we use can be found again in [Evans, ch. 4].The properties ofLemma A.2: The solution X t from Lemma A.1 satisfiesE [X t ]=e −θt E [X 0 ] ,ω 2 [X t ]=e −2θt ω 2 [X 0 ]+(1− e −2θt )α,(A.6)(A.7)with α = k BTb= σ22θ .Proof. Directly using Equation A.1, wefind tE [X t ]=e −θt E [X 0 ]+σE dB s e −θ(t−s) = e −θt E [X 0 ] , (A.8)0since the expectation of stochastic integrals over scalar functions is always zero. Similarly,using that the expectation of the square of a stochastic integral over a scalar function equalsthe normal integral over the square of that function, we computeE Xt2 = e −2θt E t t 2X02 +2σe −θt E [X 0 ] E dB s e −θ(t−s) + σ 2 E dB s e −θ(t−s)0= e −2θt E tX0 2 + σ 2 E ds e −2θ(t−s) (A.9)(A.10)0= e −2θt E X02 σ 2+2θ (1 − e−2θt ).(A.11)Therefore,ω 2 [X t ]=E Xt2 − (E [Xt ]) 2 = e −2θt E X02 + α(1 − e −2θt ) − e −2θt (E [X 0 ]) 2 (A.12)= e −2θt ω 2 [X 0 ]+α(1 − e −2θt ). (A.13)0Next, let us determine the transition kernel p(x, y; τ). Ifthecurrentpositionoftheprocessis known to be x ∈ Ω, itscurrentdistributionisadeltafunctioncentredatx. Consequently,by the preceding Lemma, the distribution at time τ>0 has expectation value e −θτ x andvariance α(1 − e −2θτ ). We can verify that this distribution is a Gaussian function withprecisely these shape parameters: