I am confused about the different options for modeling white noise in a continuous-time system.
Setup:
I have a signal $x(t) + \epsilon(t)$, where $\epsilon(t)$ is white noise. However, I also want to add a (known) noise to $x(t)$ and model it in the form $$ x(t) = \bar{x}(t) + v(t), $$ where $v(t)$ is white noise, and for simplicity, let's assume $\bar{x}(t) = \bar{x}$ is constant.
Now suppose I want to evaluate $E[\int_0^T x(t)^T x(t) dt]$. This would yield:
\begin{align*} E\left[\int_0^T x(t)^T x(t) dt\right] &= E\left[\int_0^T \bar{x}^T \bar{x} dt\right] + 2E\left[\int_0^T \bar{x}^T v(t) dt\right] + E\left[\int_0^T v(t)^T v(t) dt\right] \\ &= (\bar{x}^T \bar{x}) \cdot T + 2\cdot \int_0^T \bar{x}^T E\left[v(t)\right] dt + \int_0^T E\left[v(t)^T v(t)\right] dt\\ &= (\bar{x}^T \bar{x}) \cdot T + \int_0^T E\left[v(t)^T v(t)\right] dt \\ &= (\bar{x}^T \bar{x}) \cdot T + \mathrm{tr}\left(\int_0^T E\left[v(t) v(t)^T\right] dt\right). \end{align*}
As far as I understand it is allowed to swap the expectations and integrals, and also the trace since they are linear operators. So essentially the problem is reduced to finding the second term in this expression.
Problem:
My issue is, that I have seen different ways of modeling Gaussian white noise, i.e., $v(t)$. What would seem intuitive to me is to simply say that $v(t) \sim \mathcal{N}(0, \Sigma_v)$, and therefore we have: $$ \mathrm{tr} \left(\int_0^T E\left[v(t) v(t)^T\right] dt \right)= \mathrm{tr}\left(\int_0^T \Sigma_v dt \right)= \mathrm{tr}(\Sigma_v)\cdot T. $$
However, I have seen cases (such as how $\epsilon(t)$ is usually modeled) which essentially state that the expectation when modeling white noise is given as: $E[v(t)v(\tau)^T] = \Sigma_v \delta(t - \tau)$ where $\delta(\cdot)$ is the Dirac delta. This way, the expectation obeys the limit: $$ \lim_{dt \rightarrow 0} E[v(t)v(t)^T] dt = \Sigma_v. $$ Writing the integral as a sum of $N$ terms, i.e. the function value at time $t$ times the length $dt = T/N$, we now evaluate the second term as: \begin{equation} \mathrm{tr} \left(\int_0^T E\left[v(t) v(t)^T\right] dt \right)= \mathrm{tr} \left(\lim_{N \rightarrow \infty} \sum_{i=1}^{N} E\left[v(t_i) v(t_i)^T\right] T/N \right) = \mathrm{tr} \left(\lim_{N \rightarrow \infty} \sum_{i=1}^{N} \Sigma_v \right) = \infty. \end{equation} I am not really sure if I am making any mistakes here, but I cannot make this second form to work.
Unfortunately for me, this second form of white noise is what is commonly used for e.g. modeling $\epsilon(t)$. For example, its definition is important in the derivation of the stochastic HJB equations, where terms involving $E[\epsilon(t)\epsilon(t)^T]dt^2$ become $\Sigma_\epsilon dt$, and are thus important in the linear approximation and cannot be thrown away. Clearly, with the 'intuitive' $E[\epsilon(t)\epsilon(t)^T] = \Sigma_\epsilon$, this would not be the case, leading to fundamentally different equations.
Question:
To wrap up, I do not see a consistent way of modeling the known noise $v(t)$ as white noise, while both being able to evaluate the expectation of its norm, and keeping the effects which would appear in the derivation of the stochastic HJB equation. Am I making a mistake, is there a way to resolve this issue? Or, is there any physical reason why I am allowed to model the two types of noise, $v(t)$ and $\epsilon(t)$, differently? In this latter case, the effects of $v(t)$ will not appear in $dt^2$ terms when deriving the HJB, while $\epsilon(t)$ will, even though they are both additive white noise on my signal $\bar{x}(t)$, which seems off.
Thanks for anyone reading this through, and any clarifications would be very much appreciated!
-Tusike
I believe I have figured out the answer to this question.
The first white noise formulation is clear, at any time the signal $v(t)$ is drawn from a distribution $\mathcal{N}(0, \Sigma_v)$. It is well defined at any time $t$.
The second interpretation of white noise, on the other hand, is explicitly defined as to examine how it influences the time derivative of some term (such as $\epsilon(t)$ in dynamic equations $\dot{x} = ...$). In order to cause an abrupt change $X$ at an instantaneous time $t'$, the derivative should be $X \delta(t-t')$. As this is essentially infinite at $t = t'$, its norm cannot be interpreted as a finite value.
Thus, with the second interpretation, the model should actually take the form: $$ x(t) = \bar{x}(t) + w(t), \qquad \dot{w}(t) = v(t) $$ where $v(t)$ is white noise according to the second interpretation. The math for evaluating expectations etc... should be carried out accordingly.