Consider the interval $[0,1]$ and a partition $\mathscr{P}_n = \{ [t_i,t_{i+1}), \, i=1,\ldots,N_n \, : \, 0=t_0 < \ldots < t_{N_n} = 1\}$. Suppose that for all $i$ and $t \in [t_i,t_{i+1})$, we define a process $X(t)$ such that $X(t)$ is the same on each sub interval $[t_i,t_{i+1})$, thus $X(t) \neq X(s)$ only if $t \in [t_i,t_{i+1})$, $s \in [s_j,s_{j+1})$ and $i \neq j$. Suppose also that $X(t) \sim N(\mu_i, \sigma^2_i)$ for $t \in [t_i,t_{i+1})$ and $X(s) \perp X(t)$ if $t \in [t_i,t_{i+1})$, $s \in [s_j,s_{j+1})$ and $i \neq j$. Now let the mesh grid of the partition go to zero as $n \rightarrow \infty$. What is the limiting process?
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$\begingroup$ Should there be any limiting process? I suppose $\mu_1$ might be $\ne\mu_2$. So I don't see any meaningful limit. And did you mean $0=t_1<\dots$? And is $N_n$ increasing as you increase $n$? $\endgroup$– Claude ChaunierCommented Jul 25 at 14:16
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$\begingroup$ $\mu_i, \sigma_i^2$ implicitly depend on $n$, and you haven't said how. Everything depends on what they are. $\endgroup$– Nate EldredgeCommented Jul 25 at 17:44
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$\begingroup$ Generally, though, when you try to make a process take independent values on disjoint intervals, like a white noise, you won't get convergence, unless you shrink the variances so much that the limit is deterministic. $\endgroup$– Nate EldredgeCommented Jul 25 at 17:47
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$\begingroup$ Are you trying to construct a Gaussian process $X_t$ such that $E[X_t X_s]=0$ iff $s\neq t$? You can simply use the Kolmogorov existence theorem. math.stackexchange.com/questions/3511263/… $\endgroup$– Thomas KojarCommented Jul 25 at 18:40
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