Distribution of *conditional* frequencies when frequencies follow a Dirichlet distribution Context: we have a large number of individuals characterized by two binary traits; call these $T$ with values $\{0,1\}$, and $T'$ with values $\{0',1'\}$. So there are four types of individuals: $00'$, $01'$, $10'$, $11'$, which appear in the population with unknown relative frequencies $f_{00'}$, $f_{01'}$, $f_{10'}$, $f_{11'}$ summing up to one.
Suppose that our degree of belief about these frequencies (assumed continuous) is expressed by a Dirichlet distribution with parameters $(Aa_{00'}, Aa_{01'}, Aa_{10'}, Aa_{11'})$, the $a$s summing up to one:
$$\mathrm{p}[f_{00'}, f_{01'}, f_{10'}, f_{11'} \mid
A, (a_{00'}, a_{01'}, a_{10'}, a_{11'})] \propto
\prod_{i=0}^1\prod_{j'=0'}^{1'} f_{ij'}^{A a_{ij'}-1}\;\delta\bigl({\textstyle\sum_{ij'}}f_{ij'}-1\bigr).$$
We can also consider the marginal frequencies of individuals having trait $T'$ only, for example: $f_{0'} \equiv f_{00'} + f_{10'}$ and $f_{1'} \equiv f_{01'} + f_{11'}$. Owing to the "aggregation" property of the Dirichlet distribution (Kotz & al 2000, also Basu & al 1982), these marginal frequencies also have a Dirichlet distribution with parameters $\bigl(A(a_{00'} + a_{10'}), A(a_{01'} + a_{11'})\bigr)$ (a Beta distribution).
Question: Consider now the conditional frequencies of trait $T$ given $T'$, for example
$$f_{1\mid 0'}  \equiv \frac{f_{10'}}{f_{00'}+f_{10'}}.$$
What distribution expresses our degree of belief about such a conditional frequency, given the context above?
While I sit down and calculate (or sample), I'd be grateful for any literature or calculation hints on this. Thank you!
Additional motivation: For inference about sequential data, like for example text, speech, genes, some literature express the degree of belief about conditional frequencies $f_{i\mid j}$ (of, say, one word given the previous one) with a Dirichlet distribution (e.g. MacKay & al 1995):
$$\mathrm{p}[f_{i \mid j} \mid
A, (a_{i\mid j})] \propto
\prod_{i} f_{i\mid j}^{A a_{i\mid j}-1}\;\delta\bigl({\textstyle\sum_{i}}f_{i\mid j}-1\bigr), \qquad\text{for every }j.$$
This approach is different from using a Dirichlet distribution for the joint frequencies $f_{ij}$, and I wonder how different is the distribution for the conditional frequencies that we obtain by assuming Dirichlet for the joint frequencies instead, as in my question above.
References:
– Basu, de Bragança Pereira: On the Bayesian analysis of categorical data: the problem of nonresponse (1982) https://doi.org/10.1016/0378-3758(82)90004-0, §§ 3–4.
– Kotz, Balakrishnan, Johnson: Continuous Multivariate Distributions. Vol. 1 (2nd ed. Wiley 2000), §49.1.
– MacKay, Peto: A hierarchical Dirichlet language model (1995) https://doi.org/10.1017/S1351324900000218, https://pdfs.semanticscholar.org/01fa/57bd91f731522c861404d29e4604ba6ac6d3.pdf.
 A: Edit:
(I decided not to delete this answer, as it contains a proof of the distributive property of the Dirichlet Distribution. I have however now managed to answer the original post, which I put in a separate answer)
I generally think about these problems using the fundamental theorem of calculus. You make a new variable (let's call it a), write down an integral for the probability that $a<A$, $P(a<A)$, differentiate wrt A (this will usually use the fundamental theorem of calculus, or in more complex cases, Leibniz rule for differentiating under the integral sign, a generalisation of the former), and the result gives you the pdf for a evaluated at A, or $p(A)$. Let's see how this works, to derive an expression for $a=q_{1}+q_{2}$, or what you refer to as the associative property of the Dirichlet Distribution.
$P(a<A)=\int_{0}^{A}dq_{1}\int_{0}^{A-q_{1}}dq_{2}\int_{0}^{1-q_{1}-q_{2}}dq_{3}D(q_{1}, q_{2}, q_{3}, 1 - q_{1} - q_{2} - q_{3};\alpha)$
Writing this as $\int_{0}^{A}f(A,q_{1})dq_{1}$ and differentiating, one obtains:
$f(A,A) + \int_{0}^{A}\frac{\partial}{\partial A}f(A,q_{1})dq_{1}$
Because 
$f(A,q_{1})=\int_{0}^{A-q_{1}}dq_{2}\int_{0}^{1-q_{1}-q_{2}}dq_{3}D(q_{1},q_{2},q_{3}, 1-q_{1}-q_{2}-q_{3})$
we can see that $f(A,A)=0$, so we only have to worry about
$\int_{0}^{A}dq_{1}\frac{\partial}{\partial A}\int_{0}^{A-q_{1}}dq_{2}\int_{0}^{1-q_{1}-q_{2}}dq_{3}D(q_{1},q_{2},q_{3}, 1-q_{1}-q_{2}-q_{3})$
which is a relatively simple case of the fundamental theorem of calculus, basically replace all instances of $q_{2}$ with $(A-q_{1})$
$\int_{0}^{A}dq_{1}\int_{0}^{1-q_{1}-(A-q_{1})}dq_{3}D(q_{1}, a-q_{1}, q_{3}, 1 - q_{1} - (A-q_{1})-q_{3};\alpha)$
which, explicitly now subbing in the function form of the Dirichlet Distirbution, is given by
$\frac{1}{B(\alpha)}\int_{0}^{A}dq_{1}q_{1}^{\alpha_{1}-1}(A-q_{1})^{\alpha_{2}-1}\int_{0}^{1-A}dq_{3}q_{3}^{\alpha_{3}-1}(1-A -q_{3})^{\alpha_{4}-1}$ 
This is now a product of integrals rather than a double integral. The first is solved with the substitution $v=\frac{q_{1}}{A}$ and the second with $u=\frac{q_{3}}{1-A}$. The first gives $A^{\alpha_{1}+\alpha_{2}-1}B(\alpha_{1}, \alpha_{2})$ and the second gives $(1-A)^{\alpha_{3}+\alpha_{4}-1}B(\alpha_{3},\alpha_{4})$
Putting these together with the original normalisation constant, you get the desired result.
However for $a=\frac{q_{1}}{q_{1}+q_{2}}$, I've hit a stumbling block which might turn out to be very hard to resolve and mean this doesn't have a closed form solution, or perhaps just that this method doesn't work. The sticking point, is that in order to define the region in 2-d space such that $0< \frac{q_{1}}{q_{1}+q_{2}}<1$, is actually quite hard. For example, if $q_{1}=1$ and $A=\frac{1}{4}$, no (permitted) $q_{2}$ can satisfy $\frac{1}{1+q_{2}}<\frac{1}{4}$. 
In fact, if $A>\frac{1}{2}$, this is always possible but if $A<\frac{1}{2}$, then $q_{1}<\frac{A}{1-A}$ is needed. But the condition $q_{1}<\frac{A}{1-A}$ isn't enough, because when $A>\frac{1}{2}$, this constraint is too weak, $q_{1}$ needs to be smaller than a number which is greater than 1, so actually the constraint is $q_{1}<min(1, \frac{A}{1-A})$, and this cannot be differentiated, so I don't see how to differentiate under the integral sign.
A: Let $\gamma_{ij} \sim \texttt{gamma}(A a_{ij})$ independently. Recall that construction of the Dirichlet distribution as a normalization of gamma random variables. Then the Dirichlet distribution we start with is equal in distribution to 
$$
\left(\frac{\gamma_{00}}{\sum \gamma_{ij}}, \ldots, \frac{\gamma_{11}}{\sum \gamma_{ij}}\right) \sim \texttt{dirichlet}(A a_{00}, \ldots, Aa_{11}). 
$$
But now we have, for example, 
$$
f_{1 | 0} = \frac{f_{10}}{f_{00} + f_{10}} \stackrel{d}{=} \frac{\gamma_{10}}{\gamma_{00} + \gamma_{10}}. 
$$
And we can get a similar expersion for $f_{0|0} = \gamma_{00}/(\gamma_{00} + \gamma_{10})$, whence it follows that 
$$
(f_{0|0}, f_{1|0}) \sim \texttt{dirichlet}(A a_{00}, A a_{10}). 
$$
Easy. 
To answer your question about how the conditional model varies from the joint model: if you specify Dirichlet priors on the conditionals of the first component given the second, and a very particular Dirichlet prior on the marginals of the second, then you get a Dirichlet prior for the joint. So you get a little bit more flexibility by using Dirichlet priors on the marginals and conditionals separately; the most general sort of prior you get this way is a special case of what has been called a hyper-Dirichlet, or Dirichlet-tree prior. The hyper-Dirichlet is also conjugate to multinomial sampling; in addition to the Dirichlet, it also contains all stick-breaking constructions you can get from the Beta distribution, and many other possibilities as well. 
