“…Hence, for ω ∈ C 1 ∩ {τ 3} and t ∈ [τ, 4], we can choose X to be the pathwise unique nonpositive solution todX t = 1 X t dt + dB t , t ∈ [τ, 4]and therefore we can construct X :C 1 → C([0, 4]) fulfilling equation (3.2). Let b(t, x) := b(t, x) − ½ {t 1,x<0}1x+ ½ {t 3,x 0} 1 x − For ω ∈ C 2 and t ∈ [0, 4], let X fulfill Xt (ω) = t 0 b(s, Xs (ω))ds + B t (ω)such that X coincides with the two Bessel bridges on [0, 1] and[1,2] so that X1 = 2 and X2 = 1. This is possible by the same arguments as in the proof of Proposition 3.1.…”