I am currently working on a probabilistic model of a known linear system. It is desired to estimate the value of a system process at discrete time steps.
Each time step could be described as follows:
y[k] = y[k-1] + V/L*dT
Parameter dT is fixed.
Both "V" and "L" parameters are constant but do have a bounded tolerance limit - say +/- dV and +/-dL.
Assume continuous uniform distribution - that is, the probability of "V" in range of "V0-dV" to "V0+dV" is 1/(2 dV).
If y[k-1] is 0 (known value), what is the range of y[k] and y[k+1]?
This is not a problem belonging to the tag "probability" because you did not define the probabilities belonging to the different cases.
For example, let the probabilities belonging to $\pm dV$ and $\pm dL$ be all $\frac12$ and let these quantities be independent.
Now, to answer the question: There are four possibilities for the sign of $dV $ and $dL $: $++,+-,-+,++$ all these occur with probability $\frac 14$. The possible outcomes for $$y (k)=\begin{cases} \frac {dV}{dL}\text { with probability }\frac12\\ -\frac {dV}{dL}\text { with probability }\frac12 \end {cases}.$$
The range is given then.
For $k+1$ we have
$$y (k+1)=\begin{cases} \frac {dV}{dL}+\frac {dV}{dL}\text { with probability }\frac14\\ \frac {dV}{dL}-\frac {dV}{dL}\text { with probability }\frac14 \end {cases}$$
and
$$y (k+1)=\begin{cases} -\frac {dV}{dL}+\frac {dV}{dL}\text { with probability }\frac14\\ -\frac {dV}{dL}-\frac {dV}{dL}\text { with probability }\frac14 \end {cases}.$$
The range can be seen now.