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You misunderstood how the arrangement should look. For $n=2$, you have the lines (hyperplanes) $x_1=0$, $x_2=0$, and $x_1=x_2$. So your hyperplane arrangement looks like Remember, the rank of the central arrangement $\mathcal{B}$ is the dimension of the space spanned by the normal vectors of the hyperplanes in $\mathcal{B}$. Now it should make sense why ...


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I'm going to assume that $A = [a_{ij}], \;\; 1 \le i, j \le n, \tag{1}$ where the $a_{ij} \in \Bbb F$, $\Bbb F$ some field; then a moment's reflection reveals that $W$ may be expressed as the space of all polynomials in $A$, that is $W = \Bbb F[A]. \tag{2}$ Suppose $A \ne 0$; then characteristic polynomial of $A$, $p(x) \ne 0$. Now for $f(x) \in ...


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For concreteness, let $\pi$ be the random permutation such that $X_t = R_{\pi(t)}$. Then we have $$ S = \sum_{t=1}^N a_t X_t = \sum_{t=1}^N a_t R_{\pi(t)} = \sum_{t=1}^N a_{\pi^{-1}(t)} R_t. $$ It remains to show that for each $\pi$, the vector $(a_{\pi^{-1}(1)},\ldots,a_{\pi^{-1}(n)})$ has the same distribution as $(a_1,\ldots,a_n)$. One way to show this is ...


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Hint: The Cayley-Hamilton theorem implies that $A^n$ is a linear combination of $I,A,A^2,\ldots,A^{n-1}$.


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The $X_u$ should not be independent: if they were, I think that $X_Z$ would not be measurable.


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Claim: for any measurable function $f : \mathbb{R}^d \to \mathbb{R}$, we have $\int f\,d\nu_A = \frac{1}{P(A)} E[f(X) 1_A].$ You have already proved this for $f$ of the form $f = 1_B$. Then prove it for simple functions $f$ using linearity, for nonnegative $f$ using monotone convergence, and then for general $f$. Finally apply it with $f(x) = e^{i ...


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Bochner's Theorem: A function $F:\mathbb R\to \mathbb C$ is the characteristic function for some random variable if, and only if, $F$ is positive-definite, continuous at the origin and $F(0)=1$. Positive definiteness: The function $f$, being a characteristic function, is positive-definite. Consider, for $\ r,s=1,\ldots,n,\ $ the arbitrary numbers $\ ...


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Looks like it does. Hint: learn about Iverson brackets. They're handy for solving these problems.



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