3

i'd like to prove that $$\mathbb{E}\Big[\int_t^T n e^{-n(s-t)} L_s ds | \mathcal{F}_t\Big] \underset{n \rightarrow +\infty}{\longrightarrow} L_t$$ where $L$ is a progressive process such that $$\mathbb{E}[\underset{s \in [0,T]}{\sup} |L_s|^2] < +\infty.$$ (this question comes from a step in Pham book where existence of RBSDE is proven)

My idea is to use conditional dominated convergence theorem: $$\Big| \int_t^T n e^{-n(s-t)} L_s ds \Big| \leq \underset{s \in [0,T]}{\sup} |L_s| (1-e^{-n(T-t)}) \leq \underset{s \in [0,T]}{\sup} |L_s| \in L^1$$

and from integration by part formula (assuming i can use it, see question 1)

$$d(e^{-n(s-t)}L_s) = -ne^{-n(s-t)}L_sds + e^{-n(s-t)}dL_s,$$

so

$$\int_t^T n e^{-n(s-t)} L_s ds = L_t - e^{-n(T-t)}L_T + \int_t^T e^{-n(s-t)}dL_s \underset{n \rightarrow +\infty}{\longrightarrow} L_t.$$

I have two question:

  1. Is it defined the object $\int_t^T e^{-n(s-t)}dL_s$? I'm not sure one could assume L is a semi-martingale.
  2. How to prove that $\int_t^T e^{-n(s-t)}dL_s \rightarrow 0$?
  3. If all of this is wrong, how should i prove the fact?

Thank you for your help!

Snoop
  • 18,347
alexcrespao
  • 53
  • 1
    Does the process $L_t$ have continuous sample paths? Are you talking about convergence almost surely or in probability / in L^1? – Rabbithawke Jan 12 '25 at 18:45
  • Thank you for commenting! I'm talking about almost surely convergence. The only hypotesis over L is the one i wrote. Do you have any idea? – alexcrespao Jan 13 '25 at 08:08

1 Answers1

1

The measures $(\mu_n)_n$ given by $\mu_n(ds)=ne^{-n(s-t)}ds$ are finite sub-probability measures on $[t,T]$ which converge weakly to $\delta_t(ds)$ as $n\to \infty$. If $(L_s(\omega))_{s \in [t,T]}$ is at least rcll (which should be for all $\omega \in \Omega$) then $\lim_{s\downarrow t}L_s(\omega)=L_t(\omega)$ and its path is bounded in $[t,T]$ in the sense that $\sup_{s\in [t,T]}|L_s(\omega)|<\infty$. Let $U_{L(\omega)}$ be the set of points of discontinuity of the path $s\mapsto L_s(\omega)$ over $[t,T]$; we see that $t \notin U_{L(\omega)}$. We also see that $\delta_t(\{t\})=1$ so $\delta_t(U_{L(\omega)})=0$. Then, we have pathwise weak convergence: $\int_t^TL_s(\omega)ne^{-n(s-t)}ds\to \int_t^TL_s(\omega)\delta_t(ds)=L_t(\omega)$. One can then proceed with dominated convergence to prove the claim.

Snoop
  • 18,347
  • Thank you for your help. I'd like to have some clarification: it seems that the idea is to use weak convergence over almost all path $s \mapsto L_s(\omega)$. I'm okay with them being bounded by clearly they are not continuous. It seems you are implying that it suffice to prove that the "limit" measure of discontinuities is zero: is this a theorem? Could you give me a name/reference? Also i have some trouble proving that L is right continuous in t, could you help me? thank you! – alexcrespao Jan 17 '25 at 21:14
  • @alexcrespao here are some facts and references which may clear out your doubts: (i) rcll functions are bounded on compact intervals; (ii) the theory of stochastic processes is usually built over the space of rcll functions; (iii) for (sub)probability measures, $\mu_n\to^\textrm{w}\mu$ iff $\int fd\mu_n\to \int f d\mu$ $\forall f$ bounded measurable with $\mu(U_f)=0$, can be found in (Klenke, Th13.16 p. 254) – Snoop Jan 17 '25 at 22:57