- #1
Paalfaal
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I'm currently working my way through the existence theorem of strong solutions for the stochastic differential equation
## X_t = X_0 + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)Bs ##,
Where ## \int_0^t \sigma(s,X_s)Bs ## is the Ito integral. The assumptions are:
1: ## b,\sigma ## are jointly measurable and adapted to the filtration ## \{ \mathcal{F}_t\}_{t\geq0} ## .
2: ## b,\sigma ## satisfy the Lipschitz- and linear growth bound conditions.
3: ## \left| | X_0 | \right|_{L^2(\Omega)} < \infty ## and ## X_0 ## is ## \mathcal{F}_0##-measurable.
The iterative scheme
##
\begin{align}
\begin{cases}
X_t^0 &= X_0, \\
X_t^{n+1} &= X_0 + \int_0^t b(s,X_s^{n})ds + \int_0^t \sigma(s,X_s^{n}) dB_s,
\end{cases}
\end{align}
##
is introduced. A bit into the proof we need to use Doobs martingale inequality on ##X_t^{n+1} - X_t^n ##. My problem is that I fail to see how the previous expression is a martingale (wrt the filtration ## \{ \mathcal{F}_t\}_{t\geq0} ##). I know it suffices to show that ## X_t^n ## is a martingale. It is true that the Ito integral is a martingale, but what happens with the Lebesgue (or Riemann) integral? I have read the proof in several textbooks (Øksendal, Klöden-Platen, Kuo), none of the argued explicitly why ## X_t^n ## is a martingale, but all og them use Doobs martingale inequality.
By the assumptions it seems clear that ## X_0 ## is a martingale, but the Lebesgue integral seems to get in the way of proving it for general ## n ##.
## X_t = X_0 + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)Bs ##,
Where ## \int_0^t \sigma(s,X_s)Bs ## is the Ito integral. The assumptions are:
1: ## b,\sigma ## are jointly measurable and adapted to the filtration ## \{ \mathcal{F}_t\}_{t\geq0} ## .
2: ## b,\sigma ## satisfy the Lipschitz- and linear growth bound conditions.
3: ## \left| | X_0 | \right|_{L^2(\Omega)} < \infty ## and ## X_0 ## is ## \mathcal{F}_0##-measurable.
The iterative scheme
##
\begin{align}
\begin{cases}
X_t^0 &= X_0, \\
X_t^{n+1} &= X_0 + \int_0^t b(s,X_s^{n})ds + \int_0^t \sigma(s,X_s^{n}) dB_s,
\end{cases}
\end{align}
##
is introduced. A bit into the proof we need to use Doobs martingale inequality on ##X_t^{n+1} - X_t^n ##. My problem is that I fail to see how the previous expression is a martingale (wrt the filtration ## \{ \mathcal{F}_t\}_{t\geq0} ##). I know it suffices to show that ## X_t^n ## is a martingale. It is true that the Ito integral is a martingale, but what happens with the Lebesgue (or Riemann) integral? I have read the proof in several textbooks (Øksendal, Klöden-Platen, Kuo), none of the argued explicitly why ## X_t^n ## is a martingale, but all og them use Doobs martingale inequality.
By the assumptions it seems clear that ## X_0 ## is a martingale, but the Lebesgue integral seems to get in the way of proving it for general ## n ##.
Last edited: