I have been working on an example provided in the answers on Math Stackexchange involving transition Kernels for discrete time continuous state space Markov Chains.
Let $\{X_n\}$ be a sequence given by $X_{n+1}=(X_n+U_n)\mod 1$, where $\{U_n\}_{n=0}^\infty$ are iid uniform variables on $[0,\frac{1}{3}]$. For example, given $X_n=\frac{1}{3}$, the kernel defines $X_{n+1}$ to be $U(\frac{1}{3},\frac{2}{3})$ distributed. Then the probability that $X_{n+1}$ falls into $A=[\frac{2}{5},\frac{4}{5}]$ is given by $P(\frac{1}{3},A)=\displaystyle\int_\frac{2}{5}^\frac{2}{3}3dx+\int_\frac{2}{3}^\frac{4}{5}0dx=3\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{4}{5}$.
Now my question involves two sided conditional probabilities. For a discrete state space Markov chain these probabilities can be written as one sided probabilities as follows:
\begin{align*} \mathbb{P}[x_n|x_{n-k}^{n-1},x_{n+1}^{n+k}] &=\frac{\mathbb{P}[x_{n-k}^{n-1},x_n,x_{n+1}^{n+k}]} {\displaystyle\sum_{x\in\mathcal{A}}\mathbb{P}[x_{n-k}^{n-1},x,x_{n+1}^{n+k}]}\\ &=\frac{\mathbb{P}[x_{n-k}]\left(\displaystyle\prod_{i=n-k+1}^{n-1}\mathbb{P}[x_i|x_{n-k}^{i-1}]\right)\mathbb{P}[x_n|x_{n-k}^{n-1}]\mathbb{P}[x_{n+1}|x_{n-k}^{n}]\left(\displaystyle\prod_{i=n+2}^{n+k}\mathbb{P}[x_i|x_{n-k}^{i-1}]\right)} {\displaystyle\sum_{x\in\mathcal{A}}\mathbb{P}[x_{n-k}]\left(\displaystyle\prod_{i=n-k+1}^{n-1}\mathbb{P}[x_i|x_{n-k}^{i-1}]\right)\mathbb{P}[x|x_{n-k}^{n-1}]\mathbb{P}[x_{n+1}|x_{n-k}^{n-1}x]\left(\displaystyle\prod_{n+1}^{n+k}\mathbb{P}[x_i|x_{n-k}^{n-1}xx_{n+1}^{i-1}]\right)}\\ &=\frac{\mathbb{P}[x_n|x_{n-k}^{n-1}]\mathbb{P}[x_{n+1}|x_{n-k}^{n}]\displaystyle\prod_{i=n+2}^{n+k}\mathbb{P}[x_i|x_{n-k}^{i-1}]} {\displaystyle\sum_{x\in\mathcal{A}}\mathbb{P}[x|x_{n-k}^{n-1}]\mathbb{P}[x_{n+1}|x_{n-k}^{n-1}x]\prod_{i=n+2}^{n+k}\mathbb{P}[x_i|x_{n-k}^{n-1}xx_{n+1}^{i-1}]}\\ &\overset{Markov}{=}\frac{\mathbb{P}[x_n|x_{n-1}]\mathbb{P}[x_{n+1}|x_n]\displaystyle\prod_{i=n+2}^{n+k}\mathbb{P}(x_i|x_{i-1})}{\displaystyle\sum_{x\in\mathcal{A}}\mathbb{P}[x|x_{n-1}]\mathbb{P}[x_{n+1}|x]\displaystyle\prod_{i=n+2}^{n+k}\mathbb{P}(x_i|x_{i-1})} =\frac{P_{x_{n-1},x_n}P_{x_n,x_{n+1}}}{\displaystyle\sum_{x\in\mathcal{A}}P_{x_{n-1},x}P_{x,x_{n+1}}} \end{align*} where $P_{a,b}$ defines the transition probability $\mathbb{P}(X_n=b|X_{n-1}=a)$.
Now my question is how to calculate such a probability for the continuous case. I need to formulate a proposition and proof it for my master thesis and in order to do that I would like to understand it for the example above.
For this let $X_{n-1}=\frac{15}{30},X_n=\frac{24}{30},X_{n+1}=\frac{27}{30}$. I would like to know how to write "$P(X_{n-1},X_{n+1},A)$" (new notation) or $\mathbb{P}(X_n\in A|X_{n-1},X_{n+1})$. I was thinking of a joint conditional distribution. We know that $X_n\in[\frac{17}{30},\frac{25}{30}]$, since $[\frac{15}{30},\frac{15}{30}+\frac{1}{3}]=[\frac{15}{30},\frac{25}{30}]$ can be reached from $X_{n-1}$, but $X_{n+1}$ can be reached from $[\frac{27}{30}-\frac{1}{3},\frac{27}{30}]=[\frac{17}{30},\frac{27}{30}]$. Then for each $A$ for which $[\frac{17}{30},\frac{25}{30}]\subset A$ we have $P(X_{n-1},X_{n+1},A)=1$ and for all other $A$ one can compute the probability by using the function $$f_{X_{n-1},X_{n+1}}(x)=\begin{cases}\frac{30}{8},&x\in[\frac{17}{30},\frac{25}{30}]\\0,&\text{otherwise}\end{cases}$$
Now my final question is: how is this related to the discrete case and how can a formula be derived that is only given by one sided conditional probabilities and only using the functions $f_{X_i}(x)$? I have been puzzling for a long time, with intersections, unions, inverse functions, other distributions (normal) and can't come up with a nice formula which gives the right answers. Can anyone give me a hint?
For every $(x,y)$ on the unit circle $\mathbb R/\mathbb Z$ and every measurable subset $A$ of $\mathbb R/\mathbb Z$, $$P(X_n\in A\mid X_{n-1}=x,X_{n+1}=y)=P(x+U_{n-1}\in A\mid x+U_{n-1}+U_n=y)$$ By hypothesis, $(U_{n-1},U_n)$ is i.i.d. uniform on $[0,c]$ where, from now on, $c=\frac13$. Thus, the conditional distribution of $X_n$ conditionally on $[X_{n-1}=x,X_{n+1}=y]$ has PDF $f_{x,y}$ defined by $$f_{x,y}(z)=\frac{f(z-x)f(y-z)}{(f\ast f)(y-x)}$$ where $f$ denotes the common PDF of $U_{n-1}$ and $U_n$, and $f\ast f$ is the PDF of $U_{n-1}+U_n$, defined by $$(f\ast f)(w)=\int_\mathbb R f(u)f(w-u)\,du$$ To summarize, $$f_{X_n}(z\mid X_{n-1},X_{n+1})=\frac{f(z-X_{n-1})f(X_{n+1}-z)}{(f\ast f)(X_{n+1}-X_{n-1})}$$ hence, conditionally on $(X_{n-1},X_{n+1})$, $X_n$ is uniformly distributed on the interval $$[X_{n+1}-c,X_{n-1}+c]$$ and the formula makes sense only when $X_{n-1}<X_{n+1}<X_{n-1}+2c$, which happens with full probability.