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Let $X \in \mathbb{R}^{n \times p}$ consist of iid $\mathcal{N}(0,1)$. Assume that $n/p$ converges to a positive constant. Denote by $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_{\min(n,p)} \ge 0$ the singular values. What's the joint distribution of $$\sigma_1 - \sigma_2, \sigma_2 - \sigma_3, \sigma_3 - \sigma_4, \ldots?$$

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The joint distribution of the spacings $x_i=\sigma_i-\sigma_{i+1}$, $i=1,2,\ldots m-1$, $m=\min(n,p)$, follows from the joint distribution of the singular values $\sigma_i$ ($i=1,2,\ldots m$, ordered from large to small):

$$P(\sigma_1,\sigma_2,\ldots\sigma_m)\propto \prod_{i=1}^m e^{-\sigma_i/2}\sigma_i^{\alpha/2}\prod_{i<j}^m (\sigma_i-\sigma_j)$$

with the definition $\alpha=\max(n,p)-\min(n,p)-1$. To obtain the distribution of the spacings you substitute

$$\sigma_i=\sigma_m+\sum_{k=i}^{m-1}x_i,\;\;i=1,2,\ldots m-1$$

The determinant of this transformation is unity, so you immediately arrive at the joint distribution $P(x_1,x_2,x_{m-1},\sigma_m)$ of the $m-1$ spacings and the smallest eigenvalue $\sigma_m$. You may or may not want to integrate out the remaining $\sigma_m$. (I would think that setting $\sigma_m\mapsto 0$ should be quite accurate an approximation.)

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