I have two matrices: $A = {a_1,...,a_n}$ and $B = b_1,...,b_k$. They are the input and output alphabets of a discrete memoryless channel, with transition probabilities $q_{ij}$. Let $\tilde B = {\beta}_1,...,{\beta}_m$, with m ≥ 2, be a new output alphabet, and let $u_{jt}$, $j = 1,...,k$ and $t = 1,...,m$, be probabilities satisfying $$\sum_{t=1}^m u_{jt} = 1$$ for each $j = 1,...,k$. We make $\tilde B$ into the new output alphabet of the channel by declaring that $b_j$ will be read as $\beta_t$ with probability $u_{jt}$.
That is, whenever $b_j$ is the output letter, a probabilistic experiment is performed with outcomes $\beta_1$,...,$\beta_m$ and corresponding probabilities $u_{j1},...,u_{jm}$ to determine which of the new output letters will be the output.
Question 1: Supposing $B$ has been replaced by $\tilde B$ as described, how to express the new matriculates of transition probabilities $\tilde Q$ = $[\tilde q_{it}]$ for the new channel with input alphabet $A$ and output alphabet $\tilde B$ in terms of the old matrix of transition probabilities $Q$ and the matriculates of probabilities $U = [u_{jt}]$?
Attempt: I am not sure where to start. Question seems simple - are we just rewriting the matrix, replacing the $q$ for the new alphabet? This is what I have as an answer: $\tilde Q=$ \begin{pmatrix} q_{11} & q_{12}....q_{1m}\\\ q_{21} & q_{22} .... q_{2m} \\\ q_{n1} & q_{n2} .... q_{nm}\end{pmatrix}
Question 2: Suppose that $A = \{0,1\}, B = \{0,1,*\}$ and $Q=\begin{pmatrix}q_{00} & q_{01} & q_{0*}\\\ q_{10} & q_{11} & q_{1*}\end{pmatrix}$$ = $\begin{pmatrix}.9 & .02 & .08\\\ .05 & .88 & .07\end{pmatrix} Find a $3 \times 2$ matrix $U = [u_{jt}]$ that will amalgamate $B$ to $\tilde B = A$ so that the resulting channel is binary symmetric, and $u_{00} = u_{11} = 1$.
Attempt: all I have is a $3 x 2$ matrix U = $\begin{pmatrix}1 & 0 \\\ 0 & 1 \\\ 0.5 & 0.5\end{pmatrix}$. When $a_j = 0$ then 0 is received. When $a_j = 1$ then 1 is received. I don't think this is an answer to the question, though. I am confused.