Proposition. Let $latex mathbf{X} in mathbb{R}^{n times n}$ be a matrix that is symmetric ($latex mathbf{X}^top = mathbf{X}$) and idempotent ($latex mathbf{X}^2 = mathbf{X}$). Then the rank of $latex mathbf{X}$ is equal to the trace of $latex mathbf{X}$. In fact, they are both equal to the sum of the eigenvalues of $latex mathbf{X}$.
The proof is relatively straightforward. Since $latex mathbf{X}$ is real and symmetric, it is orthogonally diagonalizable, i.e. there is an orthogonal matrix $latex mathbf{U}$ ($latex mathbf{U}^top mathbf{U} = mathbf{I}$) and a diagonal matrix $latex mathbf{D}$ such that $latex mathbf{D} = mathbf{UXU}^top$ (see here for proof).
Since $latex mathbf{D}$ is a diagonal matrix, it implies that the entries on the diagonal must be zeros or ones. Thus, the number of ones on the diagonal (which is $latex text{rank}(mathbf{D})…
I recently learned of a fairly succinct proof for the asymptotic distribution of the Pearson chi-square statistic (from Chapter 9 of Reference 1), which I share below.
First, the set-up: Assume that we have $latex n$ independent trials, and each trial ends in one of $latex J$ possible outcomes, which we label (without loss of generality) as $latex 1, 2, dots, J$. Assume that for each trial, the probability of the outcome being $latex j$ is $latex p_j > 0$. Let $latex n_j$ denote that number of trials that result in outcome $latex j$, so that $latex sum_{j=1}^J n_j = n$. Pearson’s $latex chi^2$-statistic is defined as
Theorem. As $latex n rightarrow infty$, $latex chi^2 stackrel{d}{rightarrow} chi_{J-1}^2$, where $latex stackrel{d}{rightarrow}$ denotes convergence in distribution.
Before proving the theorem, we prove a lemma that we will…
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