Resolve `hidden variables` section confusion in `docs/theory/continuous-treatments.md`

[samjmolyneux]

I came into some confusion while fixing docs/theory/continuous-treatments.md in #178.

The hidden variables section could use some clarity and clearing up some of the ambiguity, but currently I'm not sure what the intended description should be, so I left it largely untouched.

The example of confusion I gave in the #178 pr is copy and pasted below:

From

Given $(C = c, Z = z, L = l)$, the conditional mean of $(U_X, U_Y)$ is defined to be zero for all $(c, z, l)$. We define the conditional covariance matrix of $U_X$ to be the $n_Z \times n_Z$ identity matrix, and the variance of $U_Y$ to be 1. What is left to be modelled is the conditional cross-covariance between $U_X$ and $U_Y$,

$$ \sigma_{czl} \equiv \mathbb E[U_X U_Y \ \vert \ C = c, Z = z, L = l], $$

We have

$$\left. \begin{pmatrix} U_X \\\ U_Y \end{pmatrix} \,\right|\, C=c,\ Z=z,\ L=l \sim \mathcal{N}\left( \begin{pmatrix} \mathbf{0}_{d_X} \\\ 0 \end{pmatrix}, \begin{pmatrix} I_{d_X} & \sigma_{czl} \\\ \sigma_{czl}^{\top} & 1 \end{pmatrix}. \right).$$

which implies

$$C \perp U_X \mid Z,L.$$

Now for the regression equation, originally we had

$$\begin{align} r(x,z,l) &:= \mathbb{E}\!\left[ Y \mid X=x, Z=z, L=l \right] \notag \\\ &= \int f_Y(u_y,x,l)\, p(u_y\mid u,l)\, \mathrm{d}u_y \notag \\\ &= \int f_Y(u_y,x,l)\, \sum_c \pi_{ul}(c)\, p_{\mathcal N}(u_y;m_c,v_c)\, \mathrm{d}u_y. \end{align}$$

But the probability on the second line, $p(u_y \vert u, l)$, should be $p(u_y, \vert u, l, z)$. And when we do that surely we get the following:

$$\begin{align} r(x,z,l) &:= \mathbb{E}\!\left[ Y \mid X=x, Z=z, L=l \right] \notag \\\ &= \int f_Y(u_y,x,l)\, p_{U_Y\mid U_X,Z,L}(u_y\mid u,z,l)\, \mathrm{d}u_y \notag \\\ &= \int f_Y(u_y,x,l)\, \sum_{c=1}^{K} \mathbb{P}(C=c\mid U_X=u, Z=z,L=l)\, p_{\mathcal N}(u_y;m_c,v_c)\, \mathrm{d}u_y, \\\ &= \int f_Y(u_y,x,l)\, \sum_{c=1}^{K} \mathbb{P}(C=c\mid Z=z,L=l)\, p_{\mathcal N}(u_y;m_c,v_c)\, \mathrm{d}u_y, \end{align}$$

Unless we are assuming

$$\mathbb{p}(C=c \mid U_X=u, Z=z, L=l)\\\ = \pi_{ul}(c) = \mathbb{p}(C=c \mid U_X=u, L=l),$$

or equivalently,

$$C \perp Z \mid U_X,L.$$

Combined with $C \perp U_X \mid Z, L$ that would give us

$$\mathbb{P}(C=c \mid U_X=u, Z=z, L=l) = \mathbb{P}(C=c \mid L=l)$$

But surely that contradicts

$$\pi_{ul} \equiv \mathrm{softmax}(f_\pi(u, l; \theta_\pi))$$