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My knowledge on stochastic calculus tells me that the notation $\text{d} X_t = \mu(X_t) \text{d} t + \sigma(X_t)\text{d} B_t$ is just an abbreviation of $X_t = X_0 + \int_0^t \mu(X_s) \text{d} s + \int_0^t \sigma(X_s)\text{d} B_s$.

Consider for instance $\text{d}{s_t} = c\text{d} t + \text{d}Z_t$, where here $Z_t$ is a Brownian motion and $c\sim \mathcal{N}\left(-\frac{1}{2}\sigma^2,\sigma^2\right)$ just one time (i.e., you are not drawing a $c$ each $t$ but only once). Following my background I guess that's equivalent to saying that $$ s_t = s_0 + ct + Z_t. $$

Now, I've been told that the posterior of $c$ given $s_t$ ($s_t$ is interpreted as a signal of $c$ at time $t$, that's why there's a noisy term $Z_t$ in the equation) is given by $$ c\sim \mathcal{N}\left(\hat{c}_t, \hat{\sigma}_t^2\right) $$ where $\text{d}\hat{c}_t = \hat{\sigma}_t^2\text{d}\hat{Z}_t$ and $\hat{\sigma}_t^2 = \frac{1}{\frac{1}{\hat{\sigma_t}^2}+t}$, and $\text{d}\hat{Z}_t = \text{d}s_t - \mathbb{E}_t\left[\text{d}s_t\right]$.

I'm struggling with two things:

  1. How are posteriors computed in continuous time? Do you know any reference where I can study this?
  2. What's the formal definition of $\mathbb{E}_t\left[\text{d}s_t\right]$? I'm guessing something like $\mathbb{E}_t\left[s_{t+u}\right]$ for any $u>0$... Hence, in this case, we would have that $$ \mathbb{E}_t[s_{t+u}] = s_0 + \mathbb{E}[c](t+u) + \mathbb{E}_t\left[Z_{t+u}\right] = s_0 - \frac{1}{2}\sigma^2(t+u) + Z_t $$ for all $u>0$, thanks to the fact that $\mathbb{E}_t\left[Z_{t+u}\right] = Z_t$ for all $u>0$. Using this, $$ s_{t+u} - \mathbb{E}_t[s_{t+u}] = \left(c-\frac{1}{2}\sigma^2\right)(t+u) $$ for all $u>0$. I expected some Brownian motion term in $\hat{Z}_t$, that's why I don't think I'm interpreting appropriately...

I think I provided all the context. Let me know if there's something missing. thanks!

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To answer your second question, We have

$$ \lim_{u \rightarrow 0} \frac{\mathbb{E}[s_{t+u} - s_{t}]}{u} = \lim_{u \rightarrow 0}\frac{-\frac{1}{2}\sigma^2 (t+u) - \frac{1}{2} \sigma^2 t}{u} = -\frac{1}{2}\sigma^2. $$

Or in other words, $\mathbb{E}[ds_t] = -\frac{1}{2}\sigma^2 dt$. So

$$d\hat{Z}_t = ds_t + \frac{1}{2}\sigma^2t = (c+\frac{1}{2}\sigma^2) dt + dZ_t$$

It is best not to think too rigorously about the notation $\mathbb{E}[ds_t]$. As you point out, it is meant as somewhat of a shorthand, and also to aid with intuition.

There also seems to be an error in your calculation when you write $\mathbb{E}[Z_{t+u}] = Z_t$. It is certainly true that $\mathbb{E}[Z_{t+u} | Z_t] = Z_t$, due to the martingale property. However, $\mathbb{E}[Z_t]= 0$ for all $t$.

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  • $\begingroup$ Ok, maybe my notation was not clear enough. When I wrote $\mathbb{E}_t$ I meant the conditional expectation... (in this finance papers they never explicit what $\sigma$-algebra are using but I guess is the generated by all the variables until time $t$). So when I wrote $\mathbb{E}_t\left[\text{d}s_t\right]$ I was meaning ``conditional expectation''... Or maybe it's a typo in the paper? But for sure they write the subscript $t$ with traditionally means conditional... Thanks! $\endgroup$ Commented Sep 22, 2023 at 11:58
  • $\begingroup$ Yes, I now notice the subscript. The calculation for $\mathbb{E}_t ds_t$ is essentially the same, so I hope the above can still answer your question. $\endgroup$ Commented Sep 22, 2023 at 12:09
  • $\begingroup$ ah, I see: the term $Z_t$ cancels with the term $\mathbb{E}_t\left[Z_{t+u}\right] = Z_t $ by the martingale property so it's still $-\frac{1}{2}\sigma^2$. Amazing. Thanks! By any chance do you have an answer for the first point? $\endgroup$ Commented Sep 22, 2023 at 14:28
  • $\begingroup$ I haven't come across continuous time Bayesian inference, but I don't see why the computation would be different than in the discrete time case. The posterior does not need to be computed continuously, it can be computed at some given time $t^\prime$ and then later at $t^{\prime\prime}$ and so on . $\endgroup$ Commented Sep 23, 2023 at 19:36
  • $\begingroup$ Amazing thanks. I accepted your answer. You have been really helpful. Thank you so much. $\endgroup$ Commented Sep 25, 2023 at 20:58
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The formal "defintion" of stochastic differential is what you wrote at the beginning: $dX_t=...$ is a short form of $X_t=X_0+...\,.$ In particular $ds_t$ has no formal definition other than the equation wrote: $s_t=s_0+ct+Z_t\,.$ The rest follows from $\mathbb E_t[Z_t]=Z_t$ and $\mathbb E_t[Z_{t+u}]=Z_t$ and $$\textstyle\mathbb E_t[ct]=-\frac{\sigma^2t}{2}\,,\quad \mathbb E_t[c(t+u)]=-\frac{\sigma^2(t+u)}{2}$$ assuming that $c\sim N(-\frac{\sigma^2}{2},\sigma^2)$ is independent of the Brownian motion $Z\,.$

These formulas could be used to calculate the derivative $$\tag{1} \frac{d}{dT}\Bigg|_{T=t}\mathbb E_\color{red}{t}\big[s_T\big]=-\frac{\sigma^2}{2}\,. $$ Imho this is no reason to conclude that we can formally define the conditional expectation $$\tag{2} \mathbb E_\color{red}{t}\big[ds_t\big] $$ of $ds_t\,.$ What should the random variable $ds_t$ be? If it is ${\cal F}_t$-measurable then $\mathbb E_\color{red}{t}\big[ds_t\big]=ds_t$ - in contrast to the other answer.

The minimal compromise I accept is that $$ \mathbb E_\color{red}{t}\big[ds_t\big]=-\frac{\sigma^2}{2}\,dt $$ is an "abbreviation" of (1), albeit a very confusing one.

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  • $\begingroup$ I'm not sure you are answering the question(s)... What about $\mathbb{E}_t[\text{d}s_t]$? $\endgroup$ Commented Sep 22, 2023 at 14:29
  • $\begingroup$ I said: $ds_t$ has no formal definition other than being a short form of $s_t=s_0+ct+Z_t\,.$ In particular I am deliberately not answering questions about $E_t[ds_t]\,.$ $\endgroup$ Commented Sep 22, 2023 at 16:48
  • $\begingroup$ Not sure then why you are answering. Even more: not sure then why you are answering something that I already wrote in my question. I guess it's not difficult to understand that when I say "$\text{d}X_t$ is an abbreviation for..." and then I ask for a "formal definition" of $\mathbb{E}_t\left[\text{d}s_t\right]$, what I'm actually asking for is for the thing that the latter is an abbreviation of. Besides, not difficult to be more kind. $\endgroup$ Commented Sep 23, 2023 at 18:30
  • $\begingroup$ Besides: we both wrote that $ds_t=cdt+dZ_t$ is an abbreviation of $s_t=s_0+ct+Z_t\,.$ None of us seems to know of what $\color{red}{E_t}[ds_t]$ should be an abbreviation. I have read OP carefully. Please let me know if I missed something. $\endgroup$ Commented Sep 24, 2023 at 9:00
  • $\begingroup$ In the other answer of this post you can find a suggestion of interpretation. Do we agree that $s_t = s_0 + ct + Z_t$ is formal? Do we agree that $ds_t = c dt + dZ_t$ is not but still it represents what the former says? So my question is: I have $E_t[ds_t]$ and I don't know how to clearly interpret it, do you have an interpretation of what it's suppose to represent? $\endgroup$ Commented Sep 24, 2023 at 14:46

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