In Convex Optimization by Boyd there is a section on optimal detectors and hypothesis testing where the following equivalence for conditional probabilities are asserted.
P is an n x m matrix where:
$$ p_{kj} = prob( X = k | Q = j ).$$
Where X is a random variable with a distribution that depends on parameter Q.
(See Boyd §7.3, p. 364 bottom, here the greek letter theta has been swapped with Q to avoid potential formatting errors)
T is an m x n matrix where:
$$t_{ik} = prob( G = i | X = k )$$
Where G is the guess for the value of parameter Q given observation X = k. (See Boyd §7.3.1, p. 365)
They then define a detection probability matrix as:
$$D_{ij} =( TP )_{ij} = prob( G = i | Q = j )$$
(See Boyd §7.3.2, p. 366)
In other words the detector is making the calculation: $$ P( G = i|Q = j) = \sum_{k} P(G_i|X_k)*P(X_k|Q_j)$$
I may be making a very elementary error but my understanding is: $$ P( G = i|Q = j) = \sum_{k} P(G_i\bigcap X_k|Q_j)$$
And as $$ \sum_{k} P(G_i|X_k)*P(X_k|Q_j) \neq \sum_{k} P(G_i\bigcap X_k|Q_j)$$
I am unclear as to how this conclusion was reached. Is this just an estimate given a lack of knowledge? Boyd addresses it tangentially in a lecture as a distinction between a bayesian vs frequentist approach. However the book then goes on to state that in a Bayesian detector design we would minimize the probability of the error stated as: $$q^TP^e$$ where the probability of error = 1 - Probability that we guess correctly. i.e.
$$ P^e = 1 - D_{ii}$$ and $$q_i = P(Q =i)$$
This does not seem that this fully addresses the discrepancy as $$ P( G = i|Q = i) \neq \sum_{k} P(G_i|X_k)*P(X_k|Q_i)* q_i$$
I'm sure I must be doing something incorrectly in terms of calculating these conditional probabilities so any thoughts or insights would be appreciated thanks!
The Law of Total Probability does state:
$\qquad\def\={{\,=\,}}\mathsf P(G\=i\mid Q\=j)=\sum_k\mathsf P(G\=i\mid X\=k, Q\=j)\,\mathsf P(X\=k\mid Q\=j)$
However, do note that $\mathsf P(G\=i\mid X\=k, Q\=j)=\mathsf P(G\=i\mid X\=k)$ (for all $i,j,k$) exactly when: $\boldsymbol G$ and $\boldsymbol Q$ are conditionally independent when given $\boldsymbol X$.
Your text should have made that statement clear, although it may be hidden in a Bayesian Directed Acyclic Graph or something.
Eg: The factorisation for $Q\to X\to G$ is:
$\qquad\mathsf P(G\=i, X\=k, Q\=j)=\mathsf P(G\=i\mid X\=k)\,\mathsf P(X\=k\mid Q\=j)\,\mathsf P(Q\=j)$
Anyway, when $G\perp Q\mid X$ we do have:
$\qquad\mathsf P(G\=i\mid Q\=j)=\sum_k \mathsf P(G\=i\mid X\=k)\,\mathsf P(X\=k\mid Q\=j)$
So, then $D_{ij} = \sum_k t_{ik}\,p_{kj}$