On the proof of the existence of solutions to SDE via step function approximation

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, Z1,,Zm are semimartingales, X=(X1,,Xn) is a cadlag adapted process and Xaij(X) is a map from the space of cadlag functions to the set of Zj-integrable processes that satisfy some Lipschitz continuity condition.

Namely, we allow aij(X)t to be a function of the process X at all times up to t. And then assume the two properties on it.

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For any cadlag adapted X, we define the following n-dimensional process, F(X)i=Ni+mj=1aij(X)dZj.

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,τ)(XF(X)).

My questions are :

  1. in the proof below, why do we have the identity
    λr(Xi)τr=mj=1λr1(0,τr](aij(X)aij(0))dZj+λr(mj=11[0,τr)aij(0)dZj+(Ni)τr+(Mi)τr?

I cannot figure out how to get this kind of an expansion.

  1. Below this equation in the proof, it says that the final term on the right hand side tends ucp to zero as r. That λrMτr tends to zero is clear since Mτr is bounded by ϵ and λr0. But why do λrNτr and λrj1[0,τr)aij(0)dZj tend to zero? N is just assumed to be some cadlag adapted process so I cannot see why stopping it on τr would make it tend to 0 when multiplied by λr. Also I don’t understand why the stochastic integral with integrand 1[0,τr)aij(0) tend to zero when multiplied by λr. Why do the assumptions on aij give this?

Moreover, I posted this as another question before. I can’t figure out why the sequence Xτrt being bounded in probability, where represents the running supremum of the normed process, imply that X is almost surely bounded on the interval [0,τ) whenever τ<.

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?

  1. 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?

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Answer

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

Lemma 5(Lowther) : For all \epsilon>0 there is an X\in D^n with \|X-F(X)\|< \epsilon i.e. \|X_t - F(X)_t\|<\epsilon for all t\geq 0.

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.

Attribution
Source : Link , Question Author : nomadicmathematician , Answer Author : Sarvesh Ravichandran Iyer

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