content_image/324156.jpg content_image/324157.jpg content_image/324158.jpg content_image/324159.jpg content_image/324160.jpg
Proof. Fix any T ∈ [0 , 1). From Theorem 10.3 of [12] (note that, due to Assumption 3.1, the conditions of the theorem are satisfied), we have that on the filtration ( F Y t ) t ≤ T the stochastic differential equation for Y is 

with 

<img src='content_image/324156.jpg'>

<img src='content_image/324157.jpg'>

where B Y t is Brownian motion on F Y t , and γ$_{s}$ satisfies the following ordinary differential equation 

(ODE) 

with initial condition γ$_{0}$ = σ $^{2}$. 

<img src='content_image/324158.jpg'>

Notice that γ$_{s}$ = Σ$_{z}$( s ) − s + σ 2 is the unique solution of this ODE and initial condition. Therefore on ( F Y t ) t ≤ $_{T}$, the process Y satisfies 

<img src='content_image/324159.jpg'>

The unique strong solution of this stochastic differential equation on [0 , T ] is Y$_{s}$ = B Y s (see [10], Example 5.2.4). Hence, on the interval [0 , 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 , 1]. To prove that Y$_{1}$ = Z$_{1}$ , notice that 

where λ ( t ) = exp { − ∫ t 0 1 Σ$_{z}$( s )+ σ $^{2}$− $_{s}$ds } . Note that a random variable t 0 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 1+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds . Therefore, due to the Assumption 3.1 (and in particular condition (3.3)), if lim$_{t}$$_{→}$$_{1}$ ∫ t 0 1+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds < ∞ , then Y$_{1}$ = Z$_{1}$ . 

<img src='content_image/324160.jpg'>

