Numerical solution of BSDEs

Question

In this paper by Samu Alanko and Marco Avellaneda, the authors made the following statement: Let $W_t$ be a standard Brownian motion and $f$ a suciently smooth function satisfying e.g. a polynomial growth condition, then integration by parts with dominated convergence theorem then shows that $$f'(x) = \lim_{\Delta t \rightarrow 0} \mathbb{E}[f(x + W_{\Delta t}) \frac{W_{\Delta t}}{\Delta t}]$$ $$f"(x) = \lim_{\Delta t \rightarrow 0} \mathbb{E}[f(x + W_{\Delta t}) \frac{W^2_{\Delta t}}{\Delta t^2}]$$ Although I understand intuitively why that is true. I failed to see how it is achieved by integration by parts and dominated convergence theorem. Can someone please explain this in detail a bit? Thanks in advance.

Answer 1

There is a basic equality for a Gaussian random variable, which follows from integration by parts. Assuming that $X\sim \cal{N}(0,\sigma^2)$. Assuming that $f$ is differentiable with polynomial growth, then

$$\mathbb{E}[f(X)X]=\sigma^2\mathbb{E}[f'(X)].$$

Applying this to Brownian motion (taking $x=0$ without loss of generality, we get: $$\mathbb{E}[f(W_{\Delta t})W_{\Delta t}]=\Delta t\mathbb{E}[f'(W_{\Delta t})].$$

Thus we need to show that

$$\lim_{\Delta t\to 0}\mathbb{E}[f(W_{\Delta t})\frac{W_{\Delta t}}{\Delta t}]=\lim_{\Delta t\to 0}\mathbb{E}[f'(W_{\Delta t})]=f'(0),$$

Now there are many ways to show the latter, but for example, we may write $$\mathbb{E}[f'(W_{\Delta t})]=\mathbb{E}[f'(\sqrt{\Delta t}W_1)].$$

Assuming that $|f'(x)|\le C|x|^n$ for some $C,n>0$ and all $x\in \mathbb{R}$, we see that for $\Delta t<1$, $$|f'(\sqrt{\Delta t}x)|\le C|\sqrt{\Delta t}x|^{n}\le C|x|^{n}.$$

Thus the functions $f'(\sqrt{\Delta t}x)$ are dominated by the integrable function (with respect to the Gaussian measure) $C|x|^n$. As pointwise $\lim_{\Delta t\to 0} f'(\sqrt{\Delta t}x)=f'(0)$, we thus conclude from the Dominated convergence theorem that $$\lim_{\Delta t\to 0}\mathbb{E}[f'(W_{\Delta t})]=f'(0).$$

The second statement follows from the same aurgument with two applications of the first identity.

Edit:: To prove the first equality. $$\mathbb{E}[f(X)X]=\frac{1}{\sqrt{2\pi \sigma^2}}\int dx e^{-x^2/2\sigma^2}xf(x)dx=-\frac{\sigma^2}{\sqrt{2\pi \sigma^2}}\int dx(\frac{d}{dx}e^{-x^2/2\sigma^2})f(x)dx=$$$$\frac{\sigma^2}{\sqrt{2\pi \sigma^2}} \int dx e^{-x^2/2\sigma^2}f'(x)dx=\sigma^2\mathbb{E}[f'(X)]$$

Answer 2

As a guess: As a first step in the first formula one could replace the function with $$f(x+W_{Δt})=f(x)+\int_0^1f'(x+sW_{Δt})\,ds\,W_{Δt}.$$ There is some shifting the limit inside expectation and integral contained, dominated convergence is one big theorem allowing that. For the second use partial integration to get $$f(x+v)=f(x)+f'(x)v+\int_0^1(1-s)f''(x+sv)\,ds\,v^2.$$ But that seems to imply that the second formula produces a singular term $\frac{f(x)}{Δt}$? Is $f(x)=0$ assumed somewhere? And $$ d(W_t^4)=4W_t^3\,dW_t+6W_t^2\,dt\implies \Bbb E[d(W_t^4)]=6t\,dt\implies \Bbb E[W_t^4]=3t^2, $$ which is also not compatible with the second formula.

Answer 3

In addition to Pax's answer, one might use Ito's lemma for $f(x+W_{\Delta t})W_{\Delta t}$ to get the first equation, which gives $$ \begin{align} f(x+W_{\Delta t})W_{\Delta t} & = \int_{0}^{\Delta t}f{'}(x)W_{t} d W_{t} + \int_{0}^{\Delta t}f(x + W_{t}) d W_{t} + \frac{1}{2}\int_{0}^{\Delta t}f{''}(x)W_{t} dt \\ & + \int_{0}^{\Delta t}f{'}(x) dt, \end{align} $$ divided by $\Delta t$ and taking expectation on both sides yields $$ \lim_{\Delta t\rightarrow 0}\mathbb{E}[f(x+W_{\Delta t})\frac{W_{\Delta t}}{{\Delta t}}] = \lim_{\Delta t\rightarrow 0}\int_{0}^{\Delta t}\frac{f{'}(x)}{\Delta t} dt = f'(x). $$