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Oct 24, 2023 at 23:17 history edited Thomas Kojar CC BY-SA 4.0
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Oct 24, 2023 at 23:12 comment added user479223 Sure. It is unclear what he means.
Oct 24, 2023 at 23:06 comment added Thomas Kojar He said $W(r)$ is white noise and so I took that to mean $W(r)=\frac{dB_{r}}{dr}$. If he meant something different, like you said, it is again just integration.
Oct 24, 2023 at 23:03 comment added user479223 Where is the $dB$ coming from? It seems like OPs question there's no actual differential.
Oct 24, 2023 at 16:48 history answered Thomas Kojar CC BY-SA 4.0