content_image/320778.jpg content_image/320777.jpg content_image/320776.jpg content_image/320775.jpg content_image/320774.jpg
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 

<img src='content_image/320778.jpg'>

<img src='content_image/320777.jpg'>

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, 

<img src='content_image/320776.jpg'>

<img src='content_image/320775.jpg'>

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$_{1}$ . 

<img src='content_image/320774.jpg'>

■ 

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. 

