the theorem are satisfied), we have that on the filtration ( F Y t ) t<t$_{1}$ the stochastic differential equation for Y is with where B Y t is Brownian motion on F Y t , and γ$_{s}$ satisfies the following ODE with initial condition γ$_{0}$ = σ $^{2}$. Notice that γ$_{s}$ = Σ$_{z}$( t ) + σ 2 − α 2 $_{1}$t is the unique solution of this ODE and initial condition. Therefore on ( F Y t ) t ≤ $_{t$_{1}$}$, the process Y satisfies The unique strong solution of this stochastic differential equation on [0 , t$_{1}$ ) is Y$_{s}$ = B Y s (see [10], Example 5.2.4). Hence, on the interval [0 , t$_{1}$ ), the process Y is a Brownian motion on its own (completed) filtration. By continuity of Y , this process is a Brownian motion on [0 , t$_{1}$ ]. To prove that α$_{1}$Y$_{t}$$_{1}$ = Z$_{t}$$_{1}$ , notice that where λ ( t ) = exp { − ∫ t 0 α$_{1}$ Σ$_{z}$( t )+ σ $^{2}$− α 2 $_{$_{1}$t}$ds } . t α$_{1}$ Note that a random variable ∫ t 0 α$_{1}$ λ ( s $_{)}$dB 2 s − ∫ t 0 σ$_{z}$ ( s ) λ ( s ) dB 1 s is normally distributed with mean 0 and variance ∫ t 0 α 2 $_{1}$+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds . Therefore, due to condition (4.9), if lim$_{t}$$_{→}$$_{t}$$_{1}$ ∫ t 0 α 2 $_{1}$+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds < ∞ , then ξ$_{t}$$_{1}$ = α$_{1}$Y$_{t}$$_{1}$ = Z$_{t}$$_{1}$ . 26 α$_{1}$Y ∗ t = Z$_{t}$ + λ ( t ) ( − v + ∫ t 0 α$_{1}$ λ ( s ) dB 2 s − ∫ t 0 σ$_{z}$ ( s ) λ ( s ) dB 1 s ) dY$_{s}$ = ( B Y s − Y$_{s}$ ) α 2 1 Σ$_{z}$( t ) + σ 2 − α 2 $_{1}$t ds + dB Y s . ˙ γ$_{s}$ = σ 2 $_{z}$( s ) − γ 2 $_{s}$α 2 1 ( Σ$_{z}$( t ) + σ 2 − α 2 $_{1}$t ) 2 dY$_{s}$ = ( m$_{s}$ − α$_{1}$Y$_{s}$ ) α$_{1}$ Σ$_{z}$( t ) + σ 2 − α 2 $_{1}$t ds + dB Y s , dm$_{s}$ = γ$_{s}$α$_{1}$ Σ$_{z}$( t ) + σ 2 − α 2 $_{1}$t dB Y s ,