Numerical solution of BSDEs 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.
 A: 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.
$$\mathbb{E}[f(X)X]=\frac{1}{\sqrt{2\pi \sigma^2}}\int dx e^{-x^2/2\sigma^2}xf(x)dx=-\frac{\sigma^2}{\sqrt{2\pi \sigma^2}}\int dx(\frac{d}{dx}e^{-x^2/2\sigma^2})f(x)dx=$$$$\frac{\sigma^2}{\sqrt{2\pi \sigma^2}} \int dx e^{-x^2/2\sigma^2}f'(x)dx=\sigma^2\mathbb{E}[f'(X)]$$
A: 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.
A: 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
$$
\lim_{\Delta t\rightarrow 0}\mathbb{E}[f(x+W_{\Delta t})\frac{W_{\Delta t}}{{\Delta t}}] = \lim_{\Delta t\rightarrow 0}\int_{0}^{\Delta t}\frac{f{'}(x)}{\Delta t} dt = f'(x).
$$
