reverse of Transition Probability Density function

Question

Given 2 distributions with the probability density functions $p(x)$ and $q(y)$, and their transition probability density function $T(y,x)$, we have

$$q(y) = \int p(x)T(y,x) \mathrm dx$$

In which situation, there would exist a "reverse of transition probability density function" $R(y,x)$ such that

$$p(x) = \int q(y)R(y,x) \mathrm dy$$

and how to compute it?

I suppose in general $R(y,x)$ might not exist. For example, if $\forall x$, it maps to a particular $y^*$.

But in some situation, $R(y,x)$ exists and can be computed. For example, for a diffusion process $\mathrm dX = \mu \mathrm dX + \sigma \mathrm dW$, the (forward) transition probability density function $T(y,x)$ can be derived from Kolmogorov forward equation (ie Fokker–Planck equation); while the backward transition probability density function $R(y,x)$ can be derived from Kolmogorov backward equation.

So, in general, when would such $R(y,x)$ exist and be computable? if so, how to compute it?

Answer 1

Some observations, done in the discrete setting:

• Given a probability vector $p$ and another probability vector $q$ there is always a stochastic matrix $T$ such that $Tp=q$ (I am considering the matrices acting on column vectors on the right rather than on row vectors from the left).

For the moment I have just a visual justification. For example let $n=3$. We can consider a stochastic matrix as a transition probability $T_{ji}=P(X_{n+1}=j|X_{n}=i)$ of a Markov chain, where $i,j=1..3$ are the states. We can than plot next to the other the probability distributions $p,q$ in the following visual way:

, where the total length of the big rectangles is $1$ and the length of the subrectangles are proportional to $p_i$ or $q_i$.

Using such plot we can build a Markov chain that evolves $p$ to $q$. For this purpose to evolve state $i$ we can check the projection of the rectangle associated with $p_i$ onto $q$. For example state 2 will be evolved into state 1 or 2 with a ratio of probabilities proportional to the length of the indicated arrows.

This justifies the fact that we can find stochastic $T$, $R$ such that $Tp=q$ and $Rq=p$ (actually we can find infinite many...). Of course $T,R$ need not be the inverse of the other here we are considering $p,q$ fixed.

• Stochastic matrices may not be invertible, e.g.:

$$\frac{1}{2}\begin{pmatrix}1 & 1\\\ 1 & 1\end{pmatrix}$$ is not invertible. If the determinant is not zero, than it is invertible, but the inverse may not be a stochastic matrix (columns sum to 1 but the values may become negative, see for example here Inverse of a regular stochastic matrix ).

Related:

https://www.tandfonline.com/doi/abs/10.4169/college.math.j.44.2.108?journalCode=ucmj20

This should also very much be connected to time reversibility and convergence to equilibrium, but I guess it would take much longer to discuss this...