Compound Poisson process: distribution of stopped local time $L_T$

Let $X_t$ be a compound Poisson process with positive drift $$X_t = t + \sum_{i=1}^{N_t}Y_i,$$ where $N_t \sim$Poisson$(\lambda t)$ and the $Y_i$ are i.i.d. with pdf $f$.
With $M_t$ the process $X$ reflected from its maximum, $$M_t = \bar X_t - X_t,$$ where $\bar X_t = \sup_{s\le t}X_s$, let $L_t$ be the local time at zero for $M_t$.
Let $T$ be the hitting time of $M_t$ for $c>0$, $$T = \inf_{t\ge0}\{M_t \ge c\}.$$ Find the distribution of $L_T$.

Since $X_t$ is a compound Poisson process, then $$L_t = \int_0^t 1_{\{M_s=0\}}ds.$$ Whence, by conditioning on $T$, $$E\left( e^{i \theta L_T }\right) =E\left( E \left( \left. e^{i \theta \int_0^T 1_{\{M_s=0\}}ds }\right|T\right). \right)$$ I can't go further.
Is it the right way to approach this problem?
I know a few results and technics on Levy processes, but I don't see which one applies here.

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I think you can solve this problem for a birth and death chain, i.e. occupation time at zero before some passage time, and maybe the answer is opaque enough to discourage you. I think this can be done a couple of ways, but one is using a brownian embedding and quoting the ray-knight theorem. – mike Feb 28 at 21:46
maybe it's not so bad. I suppose $M_t$ is markov, and the occupation time at 0 up to the first etc. must be exponentially distributed. – mike Mar 1 at 13:22

Assume that $T=\inf\{t\geqslant0\mid M_t\geqslant c\}$ (I do not understand the formula in the post). The process $(M_t)_{t\geqslant0}$ is a Markov process on $[0,+\infty)$ starting from $M_0=0$ with the following transition probabilities: if $M_t=x$ with $x\geqslant0$, then $M_{t+\mathrm dt}=0$ with probability $1-\mathbb P(Y\lt x)\lambda\mathrm dt+o(\mathrm dt)$ and $M_{t+\mathrm dt}$ is distributed like $x-Y$ conditionally on $Y\lt x$ with probability $\mathbb P(Y\lt x)\lambda\mathrm dt+o(\mathrm dt)$.
This implies that, once at $0$, $M$ stays at $0$ for an exponential time with parameter $a=\lambda\mathbb P(Y\lt0)$, and, once it leaves $0$, $M$ reaches $c$ before coming back to $0$ with some probability $p$. Thus, $L_T$ is the time spent at $0$ before the first heads in a sequence of heads and tails with probability of heads $p$, each tails counting for an exponential time with parameter $a$. Thus, $L_T$ is exponential with parameter $ap$.
Conditioning on the size of the first jump, one sees that $$p\,\mathbb P(Y\lt0)=\mathbb P(Y\lt-c)+\int_0^cp_x\,\mathbb P(-Y\in\mathrm dx),$$ where, for every $0\lt x\lt c$, $p_x$ is the probability that $M$ starting from $x$ hits $c$ before $0$.
From which one might be able to deduce the value of $p$...