# Intuition for Markov process

Why is the process $$Q(t)$$ defined below a Markov process ? $$Q(t)=Q(0)+A(t)- S\left(\int_0^t Q(s)\,\mathrm ds\right)$$

where $$A$$ and $$S$$ are unit rate Poisson process.
Since the integral depends on the past, shouldn't this be non-Markov ? What is the intuition for this ?

• Why do you think it is a Markov process? The pathwise definition of $Q$ is by means of an integral equation. Jan 10, 2023 at 19:43
• The paper I am reading says it's Markov
– abc
Jan 10, 2023 at 20:01

In classical calculus, it would be translated by the fact that the derivative of the considered quantity at time $$t$$ is independent of past values; however, as stochastic processes are not differentiable, we are constrained to study their increments instead.
In the present case, the variation of the stochastic process $$Q$$ over a time interval $$[s,t]$$ is given by $$Q(t) - Q(s) = A(t) - A(s) - \left(S\left(\int_0^tQ(\tau)\mathrm{d}\tau\right) - S\left(\int_0^sQ(\tau)\mathrm{d}\tau\right)\right),$$ with $$A(t) \sim S(t) \sim \mathcal{Poisson}(t)$$, hence $$A(t)-A(s) \sim \mathcal{Poisson}(t-s)$$ and $$S\left(\int_0^tQ(\tau)\mathrm{d}\tau\right) - S\left(\int_0^sQ(\tau)\mathrm{d}\tau\right) \sim \mathcal{Poisson}\left(\int_s^tQ(\tau)\mathrm{d}\tau\right)$$, such that in fine $$Q(t)-Q(s) \sim \mathcal{Poisson}\left(t-s - \int_s^tQ(\tau)\mathrm{d}\tau\right).$$ This expression doesn't contain times prior to $$s$$, that is why the increments of $$Q$$ only depend on the "present" values over the interval $$[s,t]$$, hence memorylessness / Markov property.