“…which happens with probability 0, since x = (x 1 , x 2 , x 3 ) ⊤ = 0 and, for each s ∈ (0, T ], the distribution of (Y s , X s ) is absolutely continuous, because, as in the proof of part (b) in the proof of Theorem A.1 in Bolyog and Pap [11], the conditional distribution of (Y s , X s ) given (Y 0 , X 0 ) is absolutely continuous.…”