From the material/books/etc. that I have read - an Ito diffusion always has a fixed initial condition $$ dX(t)= \mu dt + \sigma dW(t) \hspace{10mm} (1) \\ X(0)=x_0 $$
Does it make sense to model a diffusion with a normally distributed initial condition as an example $$ dX(t)= \mu dt + \sigma dW(t) \hspace{10mm} (2)\\ X(0)= N(0,1) $$
Additional information 1:
According to this Wiki article Kolmogorov backward equations (diffusion)
Informally, the Kolmogorov forward equation addresses the following problem. We have information about the state $x$ of the system at time $t$ (namely a probability distribution $p_{t}(x))$; we want to know the probability distribution of the state at a later time $s>t$. The adjective 'forward' refers to the fact that $p_{t}(x)$ serves as the initial condition and the PDE is integrated forward in time. (In the common case where the initial state is known exactly $p_{t}(x)$ is a Dirac delta function centered on the known initial state).
So, when the initial state of the diffusion is not known (FK initial condition is not delta)- then what is the representation of the initial state diffusion (equation 1)?
Additional information 2:
Monte Carlo in probability theory:
Let $g(x)$ be a function of a random variable $X$, then $$ E[g(x)]= \intop\nolimits_{-\infty }^{\infty} g(x)f(x)dx $$
Where $f(x)$ is the probability distribution of $X$.
Now, assume $g(x,t)$ is the solution of the Feynman-Kac formula
$$
\frac{\partial g(x,t)}{\partial t} = -v(x)g(x,t) + \mu \frac{\partial g(x,t)}{\partial t} + \frac{ \sigma^2 }{2} \frac{\partial^2 g(x,t) }{\partial x^2} \hspace{10mm} (3)
$$
Example graph:
The blue line (or any parallel line to it) in the graph gives us the conditional solution $[g(x,t)|X(0)=x]$ - specifically $X(0)=0$ in this case.
Ok, great now imagine that I am modeling a system of (100 particles or items).
- Each particle follows the diffusion model defined for $X(t)$ in (1).
- Also, suppose that I have solved the Feynman-Kac for the function $g(x,t)$ but I do not know exactly the initial condition of each particle.
- But, I do know that their initial condition follows a distribution i.e.: $f(x_0) \approx N(0,1)$.
So, if I want to find an average solution $E[g(x,t)]$, I suppose i can write the following:
$$ E[g(x,t)]= \intop\nolimits_{-\infty }^{\infty} g(x,t|X(0)=x_0)f(x_0)dx \approx \frac{1}{N} \sum_{i=i}^{N}g(x^i,t) \hspace{10mm} (4) $$
Thus, if I were to simulate $10,000$ diffusion paths $(N=10000)$ with $X(0)=N(0,1)$ - would approximate the integral.
So, in this context how wrong am I?