What can be said about Precision-Recall when combining binary classifiers using unanimous voting? Suppose I combine binary classifiers $A$ and $B$, having precisions $Pr_A$ and $Pr_B$ respectively, where $Pr = \frac{True Positives}{True Positives + False Positives}$, and recall $Re_A$ and $Re_B$ respectively, where $Re = \frac{True Positives}{True Positives + False Negatives}$, yielding classifier $C$ with precision $Pr_C$ and recall $Re_C$.
Each training instance $i$ consists of an input vector $v_i$ with $k$ dimensions and an output label $y_i$.    
Both classifiers are trained using the same training instances, and tested using the same test instances.
We combine both classifiers using unanimous voting, where $C$ predicts the label $0$ if either $A$ or $B$ predict $0$, else it predicts $1$. $0$ = negative example, $1$ = positive example.
What can be said about the Precision-Recall of $C$ when compared to $A$ and $B$? 
 A: Since $C$ only classifies instances as positive when $A,B$ agree, $TP_C \le TP_A$ and $TP_C \le TP_B$. ($TP$ meaning the count of true positives for a classifier.) An analogous pair of inequalities holds for $FP_C$, e.g $FP_C \le FP_A$.
Applying that fact:
$$TP_C \le TP_A \\ \frac{TP_C}{TP_C + FN_C} \le \frac{TP_A}{TP_C + FN_C} = \frac{TP_A}{TP_A + FN_A}$$
(The equality at right is true because the sum represents the total positive instances, and this sum is the same across classifiers.) 
Since a similar inequality holds for $B$, we've shown that the recall is upper-bounded by the lesser of $Re_A, Re_B$. This should make intuitive sense: Recall captures how many positives we "caught," and a conjunction of classifiers can only catch, at most, as many as the lower-recall of the two classifiers.
Building the same equations for precision gives you much less to say. We know
$$
TP_C \le TP_A \\ TP_C \le TP_B \\ FP_C \le FP_A \\FP_C \le FP_B
$$
and want to show something about $\frac{TP_C}{TP_C + FP_C}$. $FP_C = 0$ and $TP_C=0$ will both satisfy these equations—regardless of the metrics for the base classifiers—and yield respective precisions of $1,0$.
This too makes intuitive sense: If the second classifier classifies none of the first classifiers false positives, the conjunction will have no false positives. If it classifies none of the first classifier's true positives, it will have no true positives. In sum, without more information or assumptions, it doesn't seem to me that much can be said about precision of the conjunction.
Conditional probabilities
If you'd like to interpret precision as an estimate of $p(x_+|a_+, b_+)$, let's start by decomposing that probability, per Bayes' rule:
$$
p(x_+|a_+, b_+) = \frac{p(a_+, b_+ | x_+)p(x_+)}{p(a_+, b_+)} = \frac{p(a_+, b_+ | x_+)p(x_+)}{p(a_+, b_+|x_+)p(x_+) + p(a_+, b_+|x_-)p(x_-)}
$$
This alone should lend a few helpful intuitions:


*

*Precision grows with the prior probability of $x_+$. Sensible, since it's easier to be right when more instances are positive.

*It grows as $p(a_+,b_+|x_-)$ declines, confirming the intuition from the equalities above.


In other words, precision of the conjunction is best when the classifiers are differently incorrect, i.e. their incorrect classifications are different instances. (This should sound like a violation of conditional independence.)
But, let's assume conditional independence:
$$
p(x_+|a_+, b_+) = \frac{p(a_+|x_+)p(b_+ | x_+)p(x_+)}{p(a_+|x_+)p( b_+|x_+)p(x_+) + p(a_+|x_-)p( b_+|x_-)p(x_-)}
$$
Then compare to either of the classifiers:
$$\frac{p(a_+|x_+)p(b_+ | x_+)}{p(a_+|x_+)p( b_+|x_+)p(x_+) + p(a_+|x_-)p( b_+|x_-)p(x_-)} \stackrel{?}{\ge} \frac{p(a_+|x_+)}{p(a_+|x_+)p(x_+) + p(a_+|x_-)p(x_-)} \\ p(b_+|x_+) \stackrel{?}{\ge} p(b_+|x_-)$$
Naturally, to outperform both classifiers, this would have to be true of both. (By the way, the false discovery rate would estimate the quantity at right.)
But remember that this inequality applies to the unobservable distributions: Applying it to the analogous estimates of any given pair of classifiers doesn't guarantee improved precision over a given dataset.
