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Suppose have a constant stochastic process $X(s)=k$. Suppose we have some other function $B_{s}$ and we are interested in computing

$$\int_{0}^{t} X(s) dB_{s} =\int_{0}^{t} k dB_{s}=k\int_{0}^{t} dB_{s} = kB_{s}|_{0}^{t}=k(B_{t}-B_{0})$$

Now, suppose we have a step function $\phi(s)$ and wish to integrate it. Suppose we have $N$ rectangles underneath the interval $[a,b]$. Then, according to this video, one should write

$$\int_{a}^{b} \phi(s) dB_{s} = \sum_{i=1}^N\int_{a}^{b} k_{i} dB_{s} = \sum_{i=1}^Nk_{i}\int_{a}^{b} dB_{s} = \sum_{i=1}^Nk_{i}(B_b-B_a)$$

However, to me this doesn't make sense because, given my prior result, if I wish to look at each rectangle under the curve, then it would only be on a small interval since the prior result only applies to a single interval. So to me, it should look like

$$\int_{a}^{b} \phi(s) dB_{s} = \sum_{i=1}^N\int_{a_i}^{b_i } k_{i} dB_{s} = \sum_{i=1}^Nk_{i}(B_{b_{i}}-B_{a_{i}})$$

where $\bigcup\limits_{i=1}^{n} [a_{i},b_{i}]=[a,b]$. Do either of these formulations make sense? If so, which one is correct (if either)?

  • Hi: If you can get your hands on Roy Howard's "A Signal Theoretic Introduction to Random Processes", I'm pretty sure that book would clear it up. I was similarly confused for a long time and then I happened to be reading through that book and he explained it way more clearly ( with pictures ) than any other probability or real analysis text that I own and I own a lot of them. The irony is that his book doesn't really fit into either of those categories. It's more related to noise. Anyway, IMHO, without diagrams, it's very difficult to see what's happening and his has beautiful diagrams. – mark leeds Jul 08 '22 at 22:24
  • @markleeds omg great advice. Thanks Mark! will check it out :) – Stan Shunpike Jul 09 '22 at 01:22
  • I hope that you find the book and that it helps. I was unexpectedly surprised because I bought the book for a different reason ( a better understanding of DSP viewpoint of random processes ) and never expected it to cover the lebesgue integral. Good luck. – mark leeds Jul 09 '22 at 03:25
  • Yes, the video is actually wrong, he is missing the indicator function of the $i$-th interval. Your formulation is the right one. – Chaos Jul 09 '22 at 14:34
  • @Chaos perfect. thanks for confirming – Stan Shunpike Jul 09 '22 at 20:49
  • Stan: I didn't look at the video yet but, based on Chaos' reply, there's clearly no need to get the book. Of course, if you want to get it anyway, maybe it's available in the library. Definitely it opened my eyes but yours are already opened :). Chaos, Thanks. I will check out the video. – mark leeds Jul 10 '22 at 02:30
  • @markleeds definitely getting the book :) I just love learning – Stan Shunpike Jul 10 '22 at 06:31
  • okay. great. if you're interested in the dsp viewpoint of stochastic processes, I think you will like it. And the diagrams in the lebesgue integral section are tremendous. !!!! It's interesting for me because I have the statistical viewpoint of stochastic processes so it's almost like a new field even though it's the same topic. – mark leeds Jul 11 '22 at 13:03
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    I’m voting to close this question because the video OP was looking at is wrong. – Kurt G. Dec 26 '23 at 19:40

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