Why is $\int_{\mathbb{R}^d}\varphi(y)\mathrm{d}B_y=\int_{\mathbb{R}^d}(B_x-B_y)\varphi'(y)\mathrm{d}y$?

Question

I have encountered this equation

$$\int_{\mathbb{R}}\varphi(y)\mathrm{d}B_y=-\int_{\mathbb{R}}(B_y-B_x)\varphi'(y)\mathrm{d}y$$

here, on page 18. $\varphi$ is a compactly supported smooth function, and $B_x$ is $1$-$D$, two-sided Brownian motion.

Just to check my understanding: does this simply follow from integration by parts, i.e.

$$\int_{\mathbb{R}}\varphi(y)\mathrm{d}B_y=[\varphi(y)B_y]\restriction_{\mathbb{R}}-\int_{\mathbb{R}}(B_y)\varphi'(y)\mathrm{d}y = -\int_{\mathbb{R}}(B_y)\varphi'(y)\mathrm{d}y ,$$

(where the last equality follows from the fact that $\varphi$ has compact support and hence vanishes at the boundary)

?

Because I'm not then sure how the $B_x$ part comes into play.

Answer 1

The integration by parts formula shows

$$\varphi(R) B_R - \varphi(-R) B_{-R} = \int_{-R}^R \varphi(y) \, dB_y + \int_{-R}^R B_y \varphi'(y) \, dy$$

for all $R>0$. Since $\varphi$ has compact support we can let $R \to \infty$ and obtain

$$\int_{\mathbb{R}} \varphi(y) \, dB_y = - \int_{\mathbb{R}} B_y \varphi'(y) \, dy. \tag{1}$$

Using again that $\varphi$ has compact support, we find

$$\int_{\mathbb{R}} B_x \varphi'(y) \, dy = B_x \int_{\mathbb{R}} \varphi'(y) \, dy = B_x \lim_{R \to \infty} (\varphi(R)-\varphi(-R)) = 0$$

for all $x \in \mathbb{R}$, and therefore, by $(1)$,

$$\int_{\mathbb{R}} \varphi(y) \, dB_y = - \int_{\mathbb{R}} (B_y-B_x) \varphi'(y) \, dy.$$