Three errors. When you computed the partial derivatives, you forgot the sums over $i$. I do not see what $x_{ni},y_{ni},z_{ni}$ means. Last, $\tan \phi = y/x$ does not imply $\phi = \arctan(y/x)$, since $\phi$ is not necessarily in $]-\pi/2,\pi/2[$.

Since the vector $x$ is assumed to be Gaussian and centered, the components $x_i$ must be orthonormal therefore i.i.d. if you want the map $f \mapsto \sum_i f_ix_i$ from $\ell^2$ to $L^2(P)$ to be an isometry.

What happens if $\Omega$ is a disk? Do you have an idea of the answer in this case?

Let us continue this discussion in chat.

@TEX The limit points are almost surely the same and form and interval (because the difference of two consecutive terms goes to 0), possibly random, it depends on the conditional distributions. $M_0(b)=0$ by convention.

@TEX I added explanations. I do not use the L^2 convergence of the martingale here, although I mentioned it because it holds and is easy to prove.

@TEX I corrected a few typos. If $A$ is en event and $\mathcal{B}$ a sigma-field, then $\mathrm{Var}(1_A|\mathcal{B}) = E[(1_A-P(A|\mathcal{B}))^2|\mathcal{B}] \le 1/4$. Taking expectations $E[(1_A-P(A|\mathcal{B}))^2] \le 1/4$, so. $E[(1_A/k-P(A|\mathcal{B})/k)^2] \le 1/(4k^2)$. Actually bounding above by $1$ instead of $1/4$ is trivial and sufficient for the proof. We do not know the limit of $M_n(b)$. The almost sure convergence follows from the martignale convergence theorem.

The statement is not clear at all. I do not see the role of index $k$, as if the entries were constant on each row. Is the matrix considered $(f(a_j-b_k)]_{j,k}$? But in this case, I am also surprised to find no assumptions on $a_1,\ldots,a_N,b_1,\ldots,b_N$, since exchanging $a_1$ and $a_2$ would have the effect to exchange rows $1$ and $2$ and to change the detetminant into its opposite.

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Posting this kind of question on math.stackexchange.com/ would be more appropriate.