\sum_{s}p^{(m)}(s)\bra{s} = \sum_{\alpha,s} A_{\alpha_1}^{s_1}A_{\alpha_1,\alpha_2}^{s_2}\cdots
  A^{s_m}_{\alpha_{m}}\ket{s_1,s_2,\ldots,s_m}.\label{eq:mps-representation-2}\end{aligned}$$ In this representation, each probability state is a linear form that maps observables to expected values, or vice versa $$\bar{f} = \sum_s f(s)p(s) = \braket{p|f}\simeq \int\!f(x)p(x)\mathrm{d}x.
  \label{eq:integration-mps}$$ We will call this strategy the *linear MPS representation* because $\bar{f}$ is a linear function with respect to any of the tensors in $\ket{p}$ or $\ket{f}.$

Memory cost {#sec:memory}
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Let us denote $\chi_f,$ $\chi_p$ or simply $\chi$ the largest bond dimensions to encode those functions as MPS. Assuming an $N$-dimensional volume, discretized with $m$ qubits per dimension, the space required by an MPS and an MPO scales as $\mathcal{O}(Nm\chi^2).$ Since $\chi$ dominates this scaling, we need to understand how this dimension behaves in typical problems.

The answer to this question is connected to the bipartite entanglement that is stored in the MPS representations. Since MPS’s are obtained through a recursive Schmidt decomposition [@vidal2003], the entanglement entropy over any sequential bipartition of the state is bounded by $S\leq \log_2(\chi).$ Conversely, if the maximum entanglement is $S,$ we will expect that the bond dimension scales as $\chi\sim 2^S.$ Thus, for the smooth distributions from Sect. \[sec:numerics-1d\], where $S\sim N,$ we expect a scaling of resources of the form $$\mbox{memory} 