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with ˜ Z$_{t}$ = Z$_{t}$ − Z$_{t}$$_{i}$ , thus d ˜ Z$_{t}$ = σ$_{z}$ ( t ) dB 1 t and ˜ Z$_{t}$$_{i}$ = 0. From Theorem 10.3 of [12] (note that, due to (4.10), the conditions of the theorem are satisfied), we have that on the filtration ( F Y t ) t ∈ [ t$_{i}$,t$_{i}$$_{+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$_{t}$$_{i}$ = 0. 

<img src='content_image/325692.jpg'>

<img src='content_image/325691.jpg'>

<img src='content_image/325690.jpg'>

Notice that γ$_{s}$ = Σ$_{z}$( t ) − Σ$_{z}$( t$_{i}$ ) − α 2 i $_{+1}$( t − t$_{i}$ ) is the unique solution of this ODE and initial condition. Therefore on ( F Y t ) t ∈ [ t$_{i}$,t$_{i}$$_{+1}$ $_{)}$, the process Y satisfies 

<img src='content_image/325689.jpg'>

The unique strong solution of this stochastic differential equation on [ t$_{i}$, t$_{i}$$_{+1}$ ) is Y$_{s}$ = B Y s (see [10], Example 5.2.4). Hence, on the interval [ t$_{i}$, t$_{i}$$_{+1}$ ), the process Y is a Brownian motion on its own (completed) filtration. By continuity of Y , this process is a Brownian motion on [ t$_{i}$, t$_{i}$$_{+1}$ ]. To prove that α$_{i}$$_{+1}$ ˜ Y$_{t}$$_{i}$$_{+1}$ = ˜ Z$_{t}$$_{i}$$_{+1}$ , notice that 

where λ ( t ) = exp { − ∫ t t$_{i}$ α$_{i}$$_{+1}$ Σ$_{z}$( t ) − Σ$_{z}$( t$_{i}$ ) − α 2 i $_{+1}$( t − t$_{i}$ $_{)}$ds } . 

Therefore, due to conditions (4.8)-(4.11) we have, exactly as in the previous case, α$_{i}$$_{+1}$ ˜ Y$_{t}$$_{i}$$_{+1}$ = ˜ Z$_{t}$$_{i}$$_{+1}$ . 

<img src='content_image/325688.jpg'>

