On the other hand, if lim$_{t}$$_{→}$$_{t}$$_{1}$ ∫ t 0 α 2 $_{1}$+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds = ∞ , consider the process and a change of time τ ( t ) given by Then, W$_{s}$ = X$_{τ}$$_{(}$$_{s}$$_{)}$ is a Brownian motion. Hence, we can use the law of iterated logarithm to get or, in the original time, where Ξ( t ) = ∫ t 0 α 2 $_{1}$+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds . Due to the conditions (4.8)-(4.11) in this case we have therefore it follows that lim$_{t}$$_{→}$$_{t}$$_{1}$ λ ( t ) X$_{t}$ = 0, thus ξ$_{t}$$_{1}$ = α$_{1}$Y$_{t}$$_{1}$ = Z$_{t}$$_{1}$ . 27 d ˜ Y$_{t}$ = ( ˜ Z$_{t}$ − α$_{i}$$_{+1}$ ˜ Y$_{t}$ ) α$_{i}$$_{+1}$ Σ$_{z}$( t ) − Σ$_{z}$( t$_{i}$ ) − α 2 i $_{+1}$( t − t$_{i}$ ) dt + dB 2 t lim t → t$_{1}$ λ $^{2}$( t )Ξ( t ) log log (Ξ( t )) = 0 , lim sup t → t$_{1}$ X$_{t}$ √ 2Ξ( t ) log log(Ξ( t )) = 1 lim inf t → t$_{1}$ X$_{t}$ 2Ξ( t ) log log(Ξ( t )) = − 1 lim sup s →∞ W$_{s}$ √$_{2}$$_{s}$$_{log log}$$_{s}$ = 1 lim inf s →∞ W$_{s}$ √$_{2}$$_{s}$$_{log log}$$_{s}$ = − 1 ∫ τ ( t ) 0 α 2 1 + σ 2 $_{z}$( s ) λ $^{2}$( s ) ds = t. X$_{t}$ = ∫ t 0 α$_{1}$ λ ( s ) dB 2 s − ∫ t 0 σ$_{z}$ ( s ) λ ( s ) dB 1 $_{s}$, II) Suppose α$_{j}$ ( Y$_{t}$$_{j}$ − Y$_{t}$$_{j}$$_{−}$$_{1}$ ) = Z$_{t}$$_{j}$ − Z$_{t}$$_{j}$$_{−}$$_{1}$ for any j ≤ i . Consider the interval [ t$_{i}$, t$_{i}$$_{+1}$ ]. At t = t$_{i}$ we have ξ$_{t}$$_{i}$ = Z$_{t}$$_{i}$ , and ˜ Y$_{t}$ = Y$_{t}$ − Y$_{t}$$_{i}$ satisfies the following stochastic differential equation on [ t$_{i}$, t$_{i}$$_{+1}$ ]: