Is there a classical analogue to purification of quantum states?

The answer is no -- tracing out part of a system always reduces the entropy, so the marginal distribution will always be less mixed.

This is different for quantum state -- a part of a system can have more randomness than the system as a whole.

Just to add a couple more equations to the other answer, consider a (classical, discrete) joint probability distribution $(p_{ij})_{ij}$. Let the corresponding marginal distribution be $(q_i)_i$ with $q_i=\sum_j p_{ij}$. Then $$S(q)=-\sum_i q_i \log q_i=-\sum_{ij}p_{ij}\log\left(\sum_k p_{ik}\right) \le -\sum_{ij}p_{ij}\log p_{ij}=S(p).$$

On the other hand, in the quantum case, consider the maximally entangled two-qubit state $$\newcommand{\ket}[1]{\lvert #1\rangle}\ket\Psi\equiv\frac{1}{\sqrt2}(\ket{0,0}+\ket{1,1}).$$ This, when measured in the computational basis, corresponds to the probability distribution $p_{ij}=\frac12(\delta_{i,0}\delta_{j,0}+\delta_{i,1}\delta_{j,1})$, which has entropy $S(p)= \log2 .$ However, and this is a crucial point, in the quantum case the entropy depends on the measurement basis, and if a state is pure we can always find a measurement basis with respect to which the entropy is zero.

The corresponding marginal state is totally mixed state: $\operatorname{Tr}_B(\ket\Psi\langle\Psi\rvert)=I/2$. This is not a pure state anymore, and in any measurement basis it corresponds to the outcome probabilities $q_1=q_2 = 1/2$, and thus to the entropy $S(q)=\log2$.

Therefore in the quantum case the marginal distribution can have larger entropy than the joint distribution, which as we showed above is not possible for classical distributions.