Let $n\in\mathbb Z_{\ge1}$ be a strictly positive integer, let $T=\mathbb R_{\ge0}$ be the nonnegative real numbers, let $S=\cup_{m=0}^n\{0,1\}^m,$ let $\mu_1,\dots,\mu_n\in\mathbb R_{\ge0}$ be nonnegative real numbers, let $\nu\in\mathbb R_{>0}$ be a strictly positive real number and let $X=(X_t)_{t\in T}$ be a Markov chain on $S$ with rates $Q(x,y)$ defined as follows: For all $m\in[n]$ and all $x\in S$ such that $|x|=m$ (here, $|x|$ denotes the "length" of $x$), $$Q(x,y)=\sum_{j=1}^{m-k+1}\left[x_{-kj}=y\right]\mu_k\enspace\left(=\left|\left\{j\in[m-k+1]:x_{-kj}=y\right\}\right|\mu_k\right)\quad(k\in[m],|y|=k),$$ $$Q(x,x_{+j})=\nu\quad(j\in[m],x_j=0),$$ $$Q(x,x)=-\sum_{y\in S:y\ne x}Q(x,y)\enspace\left(=-m\nu-\sum_{k=1}^m(m+1-k)\mu_k\right),$$ and everything else is $0.$ Here, $$x_{-kj}=x_{-k,j}=x_{-,k,j}=(x_1,\dots,x_{j-1},x_{j-1+k},\dots,x_m)$$ denotes the element of $S$ obtained by deleting consecutive $k$ digits from $x,$ starting from the $j^\text{th}$ digit, and $$x_{+j}=x_{+,j}=(x_1,\dots,x_{j-1},1,x_{j+1},\dots,x_m)$$ denotes the element of $S$ obtained by flipping the $j^\text{th}$ digit of $x$ (assuming $x_j=0$). Now, let $s\in S$ such that $|s|=n$ and $s\ne(1)_{k\in[n]}$ ($s$ is not the all-$1$ vector of length $n$), and assume $X_0\sim\delta_s$ ($X_0$ starts at $s$). Next, let $A=\{x\in S:\forall k\in[|x|]\enspace x_k=1\}$ be the set of elements of $S$ with every digit equal to $1.$ Then, let $\tau=\inf\{t\in T:X_t\in A\}$ be the hitting time of $A.$ What is the distribution of $\tau$? (E.g., the CDF.)
(This problem came up when I was studying mutations of DNA sequences.)
If $\forall k\in[n]\!\setminus\!\{1\}\enspace\mu_k=0,$ then we can let $\tilde X=(\tilde X_t)_{t\in T}$ be a Markov chain on $\tilde S:=\{0,1,2\}^n$ (think of $2$ as "deletion") with the following rates $Q(x,y)$: For all $m\in[n]$ and all $x\in\tilde S$ such that $|x|=m,$ $$Q(x,x_{+1j})=\nu\quad(j\in[n],x_j=0),\qquad Q(x,x_{+2j})=\mu_1\quad(j\in[n],x_j\in\{0,1\}),$$ $$Q(x,x)=-\sum_{y\in\tilde S:y\ne x}Q(x,y)\enspace(=-m(\nu+\mu_1)),$$ and everything else is $0.$ Here, $$x_{+1j}=x_{+1,j}=(x_1,\dots,x_{j-1},1,x_{j+1},\dots,x_m)$$ is the element of $\tilde S$ obtained by replacing the $j^\text{th}$ digit of $x$ with the digit $1$ and $$x_{+2j}=x_{+2,j}=(x_1,\dots,x_{j-1},2,x_{j+1},\dots,x_m)$$ is the element of $\tilde S$ obtained by replacing the $j^\text{th}$ digit of $x$ with the digit $2.$ Then, let $\tilde s=s,$ assume $\tilde X_0\sim\delta_{\tilde s}$ ($X_0$ starts at $\tilde s$), let $D=\{i\in[n]:s_i\not\in\{1,2\}\}$ be the indices at which $s_i\not\in\{1,2\}$ and for each $i\in D,$ let $T_i=\inf\{t\in T:(\tilde X_t)_i\in\{1,2\}\}$ be the time it takes for the $i^\text{th}$ digit to reach $\{1,2\}.$ Next, let $\sigma=\max_{i\in D}T_i$ be the time it takes for all the digits to become $1$ or $2.$ Then $\sigma\overset d=\tau,$ by construction. In addition, since $T_i~(i\in D)$ are independent and are each exponentially distributed with parameter $\nu+\mu_1$ (exponential race), we have that for all $t\in T,$ $\mathbb P(\tau\le t)=(1-e^{-(\nu+\mu_1)t})^d,$ where $d=|D|$ is the number of indices $i$ for which $s_i\not\in\{1,2\}.$
However, when I try to generalize this argument to the case $\exists k\in[n]\!\setminus\!\{1\}\enspace\mu_k\ne0,$ it doesn't seem to work -- the $T_i$ are no longer independent.