Reversing the probability distribution for falling ball

Solution 1:

Some Bayesian analysis might be needed in an extension of this study, but not here, not yet.

This is just a common coordinate transformation in science and engineering.

Denote the starting position as $Y_1$ and the position afterwards as $Y_2$, measured with respect to a certain external reference frame (the two positions are measure in the same reference frame). This external frame doesn't even need to be an absolute (fixed) one, presumably the possible movement of the frame (seemingly just translation and no rotation, no scaling, and no skewing) is already taken care of during the modeling of the displacement distribution $f_X(x)$ that is given.

When one takes the "before" position $Y_1$ as the temporary origin, the displacement for any $\pmb Y$ with respect to this origin is $X\equiv Y - Y_1$, including $Y_2\,.~$ For this point $Y_2$ that is of interest, denote $X \equiv Y_2 - Y_1$.
^{usually the word "displacement is used to emphasize that it can be negative, while sometimes people think "difference" is always non-negative.}

Similarly, When one takes the $Y_2$ as the temporary origin, then the displacement with respect to this origin is $X' \equiv Y - Y_2$ for any $Y$. Note the symmetry of this definition: the displacement is always defined with respect to a specific origin in the same way.

The point $Y_1$ now is of interest, and it's displacement with respect to $Y_2$ is $X' \equiv Y_1 - Y_2 = -X$.

That's it. The quantities being deterministic or stochastic doesn't change how a displacement (coordinate) should be defined. When an (temporary) origin is chosen, it is the "object" $Y_1$ that is chosen, not a value (fixed point) $Y_1 = y_1$ that is chosen. Of course, the chosen object can be degenerate thus fixed, or the object could be auxiliary and doesn't correspond to an actual physical particle.