Let $$ E^+ = \{ e \in E : e(t) \ge 0 \text{ for all } t \ge 0 \} $$ be the set of positive excursions of Brownian motion.
For $e \in E^+$ and for $a > 0$, define $$ T_a(e) = \inf\{ t \ge 0 : e(t) = a \} \quad (\text{with the usual convention } \inf \varnothing = +\infty). $$
(1) Let $\varepsilon > 0$ and define $$ g_{T_\varepsilon} = \sup\{ t \in [0, T_\varepsilon] : B_t = 0 \}, \quad d_{T_\varepsilon} = \inf\{ t \ge T_\varepsilon : B_t = 0 \}. $$ Justify the fact that the measure (depending on $\omega$) which to every Borel set $A \subset \mathbb{R}_+$ associates $$ \int_{g_{T_\varepsilon}}^{d_{T_\varepsilon}} 1_A(B_s) \,\mathrm ds $$ has (almost surely) a continuous density with respect to Lebesgue measure on $\mathbb{R}_+$.
My sol: I just used Density of occupation time formula for continuous semi-martingales (here the Brownian motion) to get that the density is $L^{a}_{d_{T_{\epsilon}}}-L^{a}_{g_{T_{\epsilon}}}$. As, Brownian motion is a continuous local martingale so $(L^{a}_{t}(B))_{a \in \mathbb{R}}$ is continuous in $a$ and hence the density function is continuous when seen as a function of $a$.
(2) Show that, $n^+(de)$-almost surely, there exists a continuous function from $\mathbb{R}_+$ into $\mathbb{R}_+$, denoted $$ a \mapsto l^a(e), $$ such that for every nonnegative measurable function $\varphi$ on $\mathbb{R}_+$ one has $$ \int_0^{\sigma(e)} \varphi(e(t)) \,\mathrm dt = \int_{\mathbb{R}_+} \varphi(a) \, l^a(e) \,\mathrm da. $$
My idea: Analogously, $l^{a}(e)$ should be the local time spent by the excursion $e$ at level $a$. However, $e$ is fixed and I don’t know how to translate the density of occupation time for semi martingales to this case. Could you please help me in this case?
(3) Show that $n^+(de)$-almost surely, we have $$ l^0(e) = 0 \quad \text{and} \quad l^a(e) = 0 \text{ for every } a \ge \sup\{ e(t) : t \ge 0 \}. $$
My idea: Again if $l^{a}(e)$ is the local time at a after the excursion is finished, then we get the desired results as $e$ hits 0 at the last time and hence there is no contribution as the support of the local time is a subset of the set of times where the Brownian motion hits that level. How do I make it rigorous again?
(4) Let $a > 0$. Show that under the conditional probability $$ n^+\Bigl( de \mid T_a(e) < \infty \Bigr), $$ the random variable $l^a(e)$ follows an exponential law with mean $2a$, with Laplace transform given by $$ n^+\Bigl( \exp\bigl(-\lambda\, l^a(e)\bigr) \mid T_a(e) < \infty \Bigr) = \frac{1}{1+2\lambda a}. $$
(5) Deduce from the previous question that for every $a > 0$, $s > 0$, and $\lambda > 0$: $$ E\Bigl[\exp\Bigl(-\lambda\, L^a_{\tau(s)}\Bigr)\Bigr] = \exp\Bigl(-\frac{\lambda\, s}{1+2\lambda a}\Bigr), $$ where $L^a_{\tau(s)}$ denotes the local time at level $a$ accumulated up to time $$ \tau(s) = \inf\{ t \ge 0 : l^0(t) > s \}. $$ Relate this formula to Ray--Knight’s theorem.
I guess I can do part (5) once I get accustomed to the concepts used in earlier parts so this part can be left.
