I am reading the proof on the existence of solutions to SDEs from the following post: https://almostsuremath.com/2010/02/10/existence-of-solutions-to-stochastic-differential-equations/
Here, $Z^1, \dots, Z^m$ are semimartingales, $X=(X^1,\dots, X^n)$ is a cadlag adapted process and $X \mapsto a_{ij}(X)$ is a map from the space of cadlag functions to the set of $Z^j$-integrable processes that satisfy some Lipschitz continuity condition.
Namely, we allow $a_{ij}(X)_t$ to be a function of the process $X$ at all times up to $t$. And then assume the two properties on it.
For any cadlag adapted $X$, we define the following $n$-dimensional process, $F(X)^i = N^i + \sum_{j=1}^m \int a_{ij}(X)dZ^j$.
In the proof below, the author constructs inductively a process $X^{(r+1)}$ that approximates $X$ based on the first jump sizes with respect to $F(X)$.
$M$ is defined as $1_{[0,\tau)} (X-F(X))$.
My questions are :
- in the proof below, why do we have the identity $$\lambda_r (X^i)^{\tau_r} = \sum_{j=1}^m \int \lambda_r 1_{(0,\tau_r]} (a_{ij}(X)-a_{ij}(0))dZ^j + \lambda_r (\sum_{j=1}^m \int 1_{[0,\tau_r)} a_{ij}(0)dZ^j + (N^i)^{\tau_r}+(M^i)^{\tau_r}?$$
I cannot figure out how to get this kind of an expansion.
Below this equation in the proof, it says that the final term on the right hand side tends ucp to zero as $r \to \infty$. That $\lambda_r M^{\tau_r}$ tends to zero is clear since $M^{\tau_r}$ is bounded by $\epsilon$ and $\lambda_r \to 0$. But why do $\lambda_r N^{\tau_r}$ and $\lambda_r \sum_j \int_{1_{[0,\tau_r})}a_{ij}(0)dZ^j$ tend to zero? $N$ is just assumed to be some cadlag adapted process so I cannot see why stopping it on $\tau_r$ would make it tend to $0$ when multiplied by $\lambda_r$. Also I don't understand why the stochastic integral with integrand $1_{[0,\tau_r)} a_{ij}(0)$ tend to zero when multiplied by $\lambda_r$. Why do the assumptions on $a_{ij}$ give this?
Moreover, I posted this as another question before. I can't figure out why the sequence $X_{\tau_r \wedge t}^*$ being bounded in probability, where $*$ represents the running supremum of the normed process, imply that $X$ is almost surely bounded on the interval $[0,\tau)$ whenever $\tau<\infty$.
I would greatly appreciate if anyone takes a look at this one as well. Bounded in probability of the running maximum of a stopped process implies almost sure boundedness on the interval $[0,\lim \tau_r)$?
- Finally, it seems like the boundedness of $X$ on the interval $[0,\tau)$ for $\tau<\infty$ is used to show that $1_{(0,\tau)}(a_{ij}(X)-a_{ij}(0))$ is a locally bounded process.
However, from (P2) we always have that $$|1_{(0,\tau)}(a_{ij}(X)-a_{ij}(0))|_t \le KX^*_{t-}$$ and $X^*$ is a cadlag adapted process, hence $X^*_{t-}$ will be left continuous with right handed limits adapted, so locally bounded. Hence, isn't $1_{(0,\tau)}(a_{ij}(X)-a_{ij}(0))$ immediately locally bounded from (P2)? Why do we need $X$ almost surely bounded on $[0,\tau)$ whenever $\tau<\infty$ for this?
Thanks for the additional clarification. With this, I'll be able to write a full detailed proof of lemma $5$, sans the proof of the boundedness assumption which I'll cover in the other post.
Setup
$D^n$ is the space of $n$-dimensional cadlag adapted processes. We have $N \in D^n$, and semi-martingales $Z^1,\ldots ,Z^m$. Let $L^1(Z^j)$ denote the space of predictable and $Z^j$-integrable processes. Let $a_{ij} : D^n \to L^1(Z^j)$ be a map with the following property : for every $X,Y \in D^n$, $|a_{ij}(X) - a_{ij}(Y)| \leq K(X-Y)^*_-$ where $L_t^* = \sup_{s \leq t} \|L_s\|$ is the running maximum of $L$.
With this, define $F(X)$ for $X \in D^n$ component-wise by $$ F(X)_t^i = N_t^i + \sum_{j=1}^m\int_0^t a_{ij}(X)_tdZ^j $$
Reducing the proof to a proposition
The idea is to create $X$ as a jump process which remains constant across certain time intervals, within which it does not diverge from its $F$-value by more than $\epsilon$. We then require to prove that $X$ can be continued to infinity.
For this, we define the processes $X^r$ (which are successive extensions in time, finally giving $X$) using a sequence of stopping times $\tau_r$ (which indicate the time points at which the tolerance $\epsilon$ is breached and $X$ needs to jump). We have $X^0 = N_0, \tau_0=0$, and for $r\geq 0$, $$ X^{r+1}_{t} = \begin{cases} X^r_t & t < \tau_r \\ F(X^r)_{\tau_r} & t \geq \tau_r \end{cases} \\ \tau_{r+1} = \inf\{t \geq \tau_r : \|X_t^{r+1} - F(X^{r+1})_t\| \geq \epsilon\} $$
Note that the $\tau_r$ are increasing stopping times, which increase pointwise to some stopping time $\tau$ as $r \to \infty$. Let $X_t = X^r_t1_{\tau_r>t}$. We will show that $X$ is the desired process.
To begin, we show that on $[0,\tau)$ we have $\|X_t-F(X)_t\|<\epsilon$. Indeed, let $t<\tau$. Then, there exists $r$ such that $\tau_{r+1}>t \geq \tau_{r}$ so that $X_t = X^{r+1}_t$. Since $t<\tau_{r+1}$, by definition we have $\|X^{r+1}_t - F(X^{r+1})_t\| < \epsilon$. Since $X = X^{r+1}$ on $[0,t]$, it follows that $F(X)_t = F(X^{r+1})_t$ and therefore $\|X_t - F(X)_t\|<\epsilon$.
An integral formula
Let $\lambda_r \to 0$ be a sequence of positive reals, and let $M = 1_{[0,\tau)} (X-F(X))$ be the difference process that is bounded by $\epsilon$. For a process $Y$, we define the stopped process (at $\tau$) $(Y^\tau)_t = Y_{\tau \wedge t}$. With this, we now study the sequence of processes $\lambda_r X^{\tau_r}$, and show that it goes ucp to $0$.
We will ignore the $\lambda_r$ for now, and write some equalities that will hold on $[0,\tau)$. For one, note that $X^{\tau_r} = M^{\tau_r} + F(X)^{\tau_r}$. Hence, for every component $i$ we have $$ X^{i,\tau_r} = M^{i,\tau_r} + F(X)^{i,\tau_r} = M^{i,\tau_r} + N^{i,\tau_r} + \sum_{j=1}^m \int 1_{[0,\tau_r]}a_{ij}(X)dZ^j $$
Write $a_{ij}(X) = a_{ij}(X) - a_{ij}(X_0) + a_{ij}(X_0)$ and note that $X_0=0$. This is done to invoke the difference property of the $a_{ij}$. With this, we get $$ X^{i,\tau_r}= \left(\sum_{j=1}^m \int 1_{[0,\tau_r]}(a_{ij}(X) - a_{ij}(0))dZ^j\right)+\left(M^{i,\tau_r} + N^{i,\tau_r} + \sum_{j=1}^m \int 1_{[0,\tau_r]}a_{ij}(0)dZ^j\right) $$
Finally, we multiply by $\lambda_r$ to get $$ \lambda_rX^{i,\tau_r}= \left(\sum_{j=1}^m \lambda_r\int 1_{[0,\tau_r]}(a_{ij}(X) - a_{ij}(0))dZ^j\right)+\lambda_r\left(M^{i,\tau_r} + N^{i,\tau_r} + \sum_{j=1}^m \int 1_{[0,\tau_r]}a_{ij}(0)dZ^j\right) $$
This doesn't perfectly match with Lowther, but it turns out we need not be very accurate.
ucp convergence
We will now show that both terms go to zero in ucp.
For the second, we begin by seeing that $\sup_{s \leq t} \|M_s\| <\epsilon$ for any $t$, therefore $\sup_{s \leq t} \|\lambda_rM_s\| <\lambda_r\epsilon \to 0$ as $r \to \infty$. The other two expressions can be resolved with the same idea.
Indeed, let $Y$ be any cadlag adapted process, and let $\sigma_r$ be a sequence of increasing stopping times with $\mu_r$ a sequence of positive reals converging to $0$. We can show that $\mu_rY^{\sigma_r} \to 0$ in ucp as follows : pick a $t,K$ and let $\epsilon>0$. Note that on $[0,t]$, because $Y$ has cadlag paths, it is a.s. bounded , see e.g. here. That is, we know that $$ P(\sup_{[0,t]} \|Y_s\| = \infty) = 0 \implies \lim_{L \to \infty} P(\sup_{[0,t]} \|Y_s\|> L) = 0 $$
Pick $L$ large enough so that $P(\sup_{[0,t]} \|Y_s\|> L)< \epsilon$. If this is true, then note that $\sup_{[0,t]} \|Y_s\| \geq \sup_{[0,t]}\|Y^{\sigma_r}_s\|$ for all $t$, therefore $P(\sup_{[0,t]} \|Y^{\sigma_r}_s\|> L)< \epsilon$ for all $r$, and multiplying by $\mu_r$ gives $P(\sup_{[0,t]} \|\mu_rY^{\sigma_r}_s\|> \mu_rL)< \epsilon)$ for all $r$. As $\mu_rL \to 0$, we can assume that $r$ is large enough so that $\mu_rL < K$. In that case, $$ P(\sup_{[0,t]} \|\mu_rY^{\sigma_r}_s\|> K)< \epsilon \forall \text{ large $r$} \implies \limsup_{r \to \infty}P(\sup_{[0,t]} \|\mu_rY^{\sigma_r}_s\|> K) < \epsilon $$ for all $\epsilon>0$. Consequently, $\limsup_{r \to \infty}P(\sup_{[0,t]} \|\mu_rY^{\sigma_r}_s\|> K) \leq 0$, which forces $\lim_{r \to \infty}P(\sup_{[0,t]} \|\mu_rY^{\sigma_r}_s\|> K) = 0$, as desired.
Applying this with $\sigma_r = \tau_r$, and with each of $Y_t = N^i_t$ and $Y_t = \sum_{j=1}^m \int_0^t a_{ij}(0)dZ^j$ tells you that the second term goes in ucp to $0$.
For the first term (actually, I wonder if we can use the earlier argument instead of lemma 4, but that's a different topic) , we note that $\lambda_r|(a_{ij}(X) -a{ij}(0))| \leq \lambda_rKX^*_-$ by the difference property. Lemma $4$ applies straightaway since $\lambda_r$ is a convergent hence bounded sequence. It tells you that $\lambda_rX^{\tau_r} \to 0$ in ucp.
The existence of $\lim X_{\tau_r}$
We will now show that , providing $\tau<\infty$ has non-zero probability, $X_{\tau_r}$ has a limit with non-zero probability. We begin by writing the definition of $X_{\tau_r}$. Note that $X_{\tau_r} = X^{r+1}_{\tau_r} = F(X^r)_{\tau_r}$. Writing this down, $$ X_{\tau_r} = N_{\tau_r} + \sum_{j=1}^m \int_0^{\tau_r} a_{ij}(X^r) dZ^j = N_{\tau_r} + \sum_{j=1}^m \int_0^{\tau_-} 1_{[0,\tau_r]}a_{ij}(X^r)dZ^j $$
On $[0,\tau)$, we will now apply dominated convergence in probability. Indeed, $1_{[0,\tau_r]}a_{ij}(X^r) \to 1_{[0,\tau)}a_{ij}(X)$ a.s. by an argument similar to an earlier one in this answer. However, we also have $$ |1_{[0,\tau_r]}a_{ij}(X^r)| \leq |a_{ij}(X^r) - a_{ij}(0)| + |a_{ij}(0)| $$ on $[0,\tau)$. The latter is integrable since $a_{ij}$ maps into $L^1(Z^j)$ for each $j$. The former is integrable since $|a_{ij}(X^r) - a_{ij}(0)| \leq K(X^r)^{*}_-$ which is a locally bounded hence $Z^j$-integrable function. It follows that $$ \lim_{r \to \infty} N_{\tau_r} + \sum_{j=1}^m \int_0^{\tau_-} 1_{[0,\tau_r]}a_{ij}(X^r)dZ^j $$ exists, since $N_{\tau_r} \to N_{\tau^-}$. It follows that $\lim_{r \to \infty} X_{\tau_r}$ exists a.s., provided that $\tau<\infty$ i.e. the limit exists with non-zero probability.
Contradiction
However, $\lim_{r \to \infty} X_{\tau_r}$ doesn't exist a.s., since $$ X^r_{\tau_r} = F(X^{r-1})_{\tau_{r-1}} = X^r_{\tau_{r-1}} = X_{\tau_{r-1}} $$
(the first and second equalities follow from the definition of $X^r$, the second from the definition of $X$) and $$ X_{\tau_{r}} = X^{r+1}_{\tau_{r}} = F(X^{r})_{\tau_r} $$
Therefore $$ \|X_{\tau_r} - X_{\tau_{r-1}}\| = \|F(X^r)_{\tau_r} - X^r_{\tau_{r}}\| \geq \epsilon $$
by the definition of $\tau_r$. Thus, $X_{\tau_r}$ is almost nowhere convergent, contradicting the conclusion of the previous section. The only possibility is that $\tau=\infty$, as desired.