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 ?
 A: The Markov property means that the evolution of a stochastic process doesn't depend on the past, but the current state can be a function of past values (at least of the initial state).
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.
