On the other hand, if lim$_{t}$$_{→}$$_{1}$ ∫ t 0 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}$$_{)}$ is a Brownian motion. Hence, we can use the law of iterated logarithm to get or, in the original time, where Ξ( t ) = ∫ t 0 1+ σ 2 $_{z}$( s ) λ $^{2}$( s ) ds . Since, due to the Assumptions 2.2, 3.1 and 3.2, we have it follows that lim$_{t}$$_{→}$$_{1}$ λ ( t ) X$_{t}$ = 0, therefore Y$_{1}$ = Z$_{1}$ . With this lemma at hand, establishing that the pair ( H $^{∗}$, θ$^{∗}$) given in the Theorem 3.1 is indeed an equilibrium is straightforward, as the following proposition demonstrates. ■ Proposition 3.2 Suppose that Assumptions 2.1, 2.2, 3.1 and 3.2 are satisfied. Then the pair ( H $^{∗}$, θ$^{∗}$) , where H $^{∗}$( y, t ) satisfies the partial differential equation (PDE) (3.7) with terminal condi- tion (3.8), and the process θ ∗ t is given by (3.9), is an equilibrium. 18 lim t → $_{1}$λ $^{2}$( t )Ξ( t ) log log (Ξ( t )) = 0 , lim sup t → 1 X$_{t}$ √ 2Ξ( t ) log log(Ξ( t )) = 1 lim inf t → 1 X$_{t}$ 2Ξ( t ) log log(Ξ( t )) = − 1 lim sup s →∞ W$_{s}$ √$_{2}$$_{s}$$_{log log}$$_{s}$ = 1 lim inf s →∞ W$_{s}$ √$_{2}$$_{s}$$_{log log}$$_{s}$ = − 1 ∫ τ ( t ) 0 1 + σ 2 $_{z}$( s ) λ $^{2}$( s ) ds = t. X$_{t}$ = ∫ t 0 1 λ ( s ) dB 2 s − ∫ t 0 σ$_{z}$ ( s ) λ ( s ) dB 1 $_{s}$,