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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 $^{2}$. 

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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 

<img src='content_image/326560.jpg'>

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 } . Note that a random variable t α$_{1}$ $_{)}$dB 2 s 

∫ 0 λ ( 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}$ . 

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