A natural hypothesis for the success of reservoir computing in generic tasks is the ability of the untrained reservoir to map different input time series to separable reservoir states - a property we term separation capacity. We provide a rigorous mathematical framework to quantify this capacity for random linear reservoirs, showing that it is fully characterised by the spectral properties of the generalised matrix of moments of the random reservoir connectivity matrix. Our analysis focuses on reservoirs with Gaussian connectivity matrices, both symmetric and i.i.d., although the techniques extend naturally to broader classes of random matrices. In the symmetric case, the generalised matrix of moments is a Hankel matrix. Using classical estimates from random matrix theory, we establish that separation capacity deteriorates over time and that, for short inputs, optimal separation in large reservoirs is achieved when the matrix entries are scaled with a factor $\rho_T/\sqrt{N}$, where $N$ is the reservoir dimension and $\rho_T$ depends on the maximum input length. In the i.i.d.\ case, we establish that optimal separation with large reservoirs is consistently achieved when the entries of the reservoir matrix are scaled with the exact factor $1/\sqrt{N}$, which aligns with common implementations of reservoir computing. We further give upper bounds on the quality of separation as a function of the length of the time series. We complement this analysis with an investigation of the likelihood of this separation and its consistency under different architectural choices.

翻译：储层计算在通用任务中取得成功的一个自然假设是：未经训练的储层能够将不同的输入时间序列映射到可分离的储层状态——这一特性我们称之为分离能力。我们提出了一个严格的数学框架来量化随机线性储层的这种能力，证明其完全由随机储层连接矩阵的广义矩矩阵的谱特性所表征。我们的分析主要针对具有高斯连接矩阵的储层（包括对称矩阵和独立同分布矩阵），尽管相关技术可自然推广至更广泛的随机矩阵类别。在对称情形下，广义矩矩阵为汉克尔矩阵。利用随机矩阵理论中的经典估计，我们证明了分离能力随时间推移而衰减，并且对于短时输入，当矩阵元素按因子$\rho_T/\sqrt{N}$缩放时（其中$N$为储层维度，$\rho_T$取决于最大输入长度），大型储层可实现最优分离。在独立同分布情形下，我们证明当储层矩阵元素按精确因子$1/\sqrt{N}$缩放时，大型储层始终能实现最优分离，这与储层计算的常见实现方式一致。我们进一步给出了分离质量随时间序列长度变化的上界估计。此外，我们通过研究这种分离的可能性及其在不同架构选择下的一致性，对上述分析进行了补充。