Multivariate marginal distribution I am trying to deduce the following formula, from "Information-Theoretically Secure Secret-Key Agreement by NOT Authenticated Public Discussion" (Definition 4, page 9): $P_{\tilde{X}Y}(x,y) = \sum_{x'}\sum_{z}P_{XYZ}(x',y,z) \cdot P_{\tilde{X}|Z}(x,z)$ but can't figure it out. Probably (among other things), what is puzzling me is the $\tilde{X}|Z$ inside the summatory since, on the one hand, the summatory does not iterate over $\tilde{X}$, yet it deals with the conditioned probability of $\tilde{X}$ given $Z$ (and something similar for $Y$).

The notation is a little strange, let me use $\tilde{x}$ for the values corresponding to variable $\tilde{X}$.

In general, for any four random variables, we have:

$$P_{\tilde{X},Y}( \tilde{x},y) = \sum_{x} \sum_{z} P_{X,\tilde X,Y,X}(x,\tilde{x},y,z) = \sum_{x} \sum_{z} P_{X,Y,Z}(x,y,z) P_{\tilde{X} \mid X,Y,Z}(\tilde{x} | x,y,z)$$

Now, in the paper it's asserted that $P_{X,\tilde{X},Y,Z} = P_{X,Y,Z} P_{\tilde{X} \mid Z}$.

That is not true in general. It's true for our variables because (def. 4) **$\tilde {X}$ is generated from $Z$** , so it's reasonable to assume $ P_{\tilde{X} \mid X,Y,Z}= P_{\tilde{X}\mid Z}$

Then

$$P_{\tilde{X},Y}( \tilde{x},y) = \sum_{x} \sum_{z} P_{X,Y,Z}(x,y,z) P_{\tilde{X} \mid Z}(\tilde{x} | z)$$