# Estimation of integral of stochastic process(Krylov estimation)

Let $$X_n$$ be a sequence of Ito diffusions $$dX_n(t)=b_n(t) \, dt+\sigma_n(t) \, dW(t), \qquad 0\leq t\leq T$$ with $$b_n$$ uniformly bounded and $$\sigma_n$$ uniformly elliptic. Then Krylov's estimation gives the following: for any bounded measurable function $$f$$ with compact support $$\mathcal K$$, $$E\int_0^T|f(t,X_n(t)|dt\leq \left(\int_0^T\int_\mathcal K|f(t,x|^2 \, dt \, dx \right)^\frac{1}{2}.$$

Problem: If there exists $$X$$ such that $$\lim_{n\rightarrow\infty}E \left[\sup_{0\leq t\leq T}|X_n(t)-X(t)|^2 \right]=0,$$ then $$E\int_0^T|f(t,X(t)|dt\leq \left(\int_0^T\int_\mathcal K|f(t,x|^2 \, dt \, dx \right)^\frac{1}{2}.$$

The above is what I read in a book. I tried to prove it with Lusin Theorem. For any $$\varepsilon >0$$, it follows from Lusin Theorem that there exists a closed set $$A$$ with $$Leb(A)<\varepsilon$$ and $$f$$ is continuous on $$A$$. \begin{align*} &E\int_0^T|f(t,X(t)| \, dt \\&=E\int_0^T|f(t,X(t)| I\{X(t)\in A\} \, dt+E\int_0^T|f(t,X(t)|I\{X(t)\in A^c\} \, dt. \end{align*} It is easy to get the upper bound for the first part. But I don't think the second part is negligible.

Another method may be to get the representation of $$X$$ and then apply the Krylov estimation. However, I am not sure how to get the form of $$X$$.

Any thoughts on how to prove this? Thanks!

Since

$$\lim_{n \to \infty} \mathbb{E} \left( \sup_{0 \leq t \leq T} |X_n(t)-X(t)|^2 \right) = 0$$

there exists a subsequence $$(X_{n(k)})_{k \geq 1}$$ such that

$$\sup_{0 \leq t \leq T} |X_{n(k)}(t)-X(t)| \xrightarrow[]{k \to \infty} 0 \tag{1}$$

almost surely. To prove the inequality $$\mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) \leq \left( \int_0^T \int |f(t,x)|^2 \, dx \, dt \right)^{1/2} \tag{2}$$ we first prove this inequality for a nice class of functions and then use a density argument:

If $$f: [0,T] \times \mathbb{R}^d \to \mathbb{R}$$ is a function which is continuous with respect to the 2nd variable and which satisfies $$f(t,x)=0$$ for all $$t \in [0,T]$$, $$|x| \geq R$$ for some $$R>0$$, then it follows from the dominated convergence theorem and the Krylov estimate for $$X_n$$ that

\begin{align*} \mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) &= \lim_{k \to \infty} \mathbb{E} \left( \int_0^T |f(t,X_{n(k)}(t)| \, dt \right) \\ &\leq \left( \int_0^T \int |f(t,x)|^2 \, dx \, dt \right)^{1/2}, \end{align*}

i.e. $$(2)$$ holds for any such $$f$$. Now if $$f$$ is a function of the form $$f(t,x) = 1_A(t) 1_C(x) \tag{3}$$ for some Borel set $$A$$ and a closed set $$C$$ with $$C \subseteq B(0,R)$$ for some $$R>0$$, then we can find a sequence $$(g_n)_{n \in \mathbb{N}}$$ of continuous functions supported in $$B(0,R+1)$$ such that $$1_C(x) = \inf_{n \in \mathbb{N}} g_n(x), \qquad x \in \mathbb{R}^d, \tag{4}$$ (see Urysohn's lemma for details). If we define $$f_n(t,x) := 1_A(t) g_n(x)$$ then it follows from the first part of this proof and $$(4)$$ that

\begin{align*} \mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) &\stackrel{(4)}{\leq} \mathbb{E} \left( \int_0^T |f_n(t,X(t))| \, dt \right) \\ &\leq \sqrt{\int_0^T \int |f_n(t,x)|^2 \, dx \, dt} \end{align*}

for all $$n \in \mathbb{N}$$. Taking the infimum over all $$n \geq 1$$ yields, by (4) and the monotone convergence theorem, that

$$\mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) \leq \sqrt{\int_0^T \int |f(t,x)|^2 \, dx \, dt},$$

i.e. $$(2)$$ holds for functions of the form $$(3)$$. Now an application of the (functional) monotone class theorem (see e.g. Theorem 1 here) gives that $$(2)$$ holds for any bounded measurable function $$f: [0,T] \times B(0,R) \to \mathbb{R}$$. As $$R>0$$ is arbitrary, this proves the assertion.

• In order to use Theorem 1, we still need to verify that if $f$ is bounded and there is a sequence of nonnegative functions $f_n$ increasing pointwise to $f$ with $f_n$ satisfying (2), then $f$ satisfies (2). To prove this, note that $f_n$ increasing pointwise to $f$, then $f_n(t,X)$ converges to $f(t,X)$ almost surely. Therefore, $$\mathbb E\left(\int_0^T|f(t,X(t))|dt\right)\leq \mathbb E\left(\int_0^T|f(t,X(t))-f_n(t,X_n(t))|dt\right)+\sqrt{\int_0^T\int|f_n(t,x)|^2dtdx}.$$ Letting $n\rightarrow\infty$, we get that $f$ satisfies (2). Is my computation right? – SHAN Apr 7 at 12:50
• @SHAN Well, yes; alternatively you can just apply the monotone convergence theorem, i.e. use $$\mathbb{E} \int_0^T |f(t,X(t))| \, dt = \sup_n \mathbb{E} \int_0^t |f_n(t,X(t))| \, dt$$ – saz Apr 7 at 13:13
• Yes. Thank you for your help. – SHAN Apr 8 at 2:08
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 51, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998915195465088, "perplexity": 109.25012141657061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00426.warc.gz"}