Is the product of a random variable and a Markov process still a Markov process? Let $N_t$ be a Poisson process, Y^i independent, identical Markov processes, say Brownian motions $+ 1$. Let $J$ be independent, non-trivial random variables iid (not a constant), almost surely non-zero. The question is:
Is $\sum_{i=1}^{N_t}J_i \cdot Y^i_{t-T_i}$ still a Markov process - kind of a modified compound Poisson process? $T_i$ are the jump times of $N_t$. My guess is no; I'll try to come up with a counterexample. It shouldn't be too difficult. Is the guess correct? The process should depend on the jumps, so maybe that will exclude Markov property, because you "cannot" know the jump size from the last value of the process.
 A: For every $t\geqslant0$, let $X_t=\sum\limits_{i=1}^{N_t}J_i\cdot Y^i_{t-T_i}$. A simple example is when the sequence $(J_i)_{i\geqslant i}$ is i.i.d. Bernoulli $\pm1$ centered and, as suggested in the question, $(Y^i)_{i\geqslant i}$ are i.i.d. standard Brownian motions. Then the process $(X_t)_{t\geqslant0}$ is distributed like $(Z_t)_{t\geqslant0}$ defined by
$$
Z_t=\sum_{i=1}^{N_t}J_i+\int_0^t\sqrt{N_t}\mathrm dY_s,
$$
where $Y=(Y_t)_{t\geqslant0}$ is a standard Brownian motion. In words, conditionally on $Z_t=z$ and $N_t=n$, when $s\to0^+$ and neglecting $o(s)$ terms, $Z_{t+s}=z+\varepsilon H+\sqrt{ns}G$ and $N_{t+s}=n+H$ for some independent $(H,\varepsilon,G)$, where $H=1$ and $H=0$ with probability $s$ and $1-s$ respectively, $\varepsilon=+1$ and $\varepsilon=-1$ with probability $\frac12$ and $\frac12$ respectively, and $G$ is a standard centered normal random variable. 
Thus, $(Z_t,N_t)_{t\geqslant0}$ is a Markov process on the state space $\mathbb R\times\mathbb N_0$.
For every function $u$, let $Du:z\mapsto \frac12(u(z+1)+u(z-1)-2u(z))$ denote the discrete Laplacian of $u$. The dynamics described above yields the differential equation
$$
\frac{\mathrm d}{\mathrm dt}\mathrm E(u(Z_t))=\mathrm E(Du(Z_t))+\frac12\mathrm E(N_tu''(Z_t)).
$$
In particular, 
$$
\left.\frac{\mathrm d}{\mathrm ds}\mathrm E(Z_{t+s}^2\mid Z_t,N_t)\right|_{s=0}=1+N_t.
$$
The proof is complete if $N_t$ is not measurable for $Z_t$.
But the conditional distributions of $Z_t$ conditionally on $N_t=n$, for every $n\geqslant1$, are all absolutely continuous with respect to each other (the case $n=0$ being different since, conditionally on $N_t=0$, $Z_t=0$ almost surely), hence one can reconstruct $N_t$ from $(Z_s)_{0\leqslant s\leqslant t}$ (count the jumps) but not from $Z_t$ alone. 
This proves that $\mathrm E(Z_{t+s}^2\mid Z_t,N_t=n)$ does depend on $n$, at least when $s\to0^+$, hence that $(Z_t)_{t\geqslant0}$ is not a Markov process with respect to its own filtration.
A: Let us put $Y_{1,t} = t$ and $Y_{k,t} = B_t$ for $k\geq 1$. Then if we denote
$$
X_t  =\sum\limits_{k=1}^{N_t}J_k Y_{k,t}
$$
then $X_t$ is not a Markov process. Indeed, $\mathsf P\{X_{t+1}\leq x|X_s,s\leq t\}$ and $\mathsf P\{X_{t+1}\leq x|X_t = 1\}$ will be different in the case when $X_s = s$ for all $s\leq t$. This is the main idea - but you may want me to write here these distributions explicitly if this idea is not enough for the proof.
A: OK, this should be a counterexample - let's simply ignore the random jumps in my question.
Consider the sum of two independent geometric Brownian motions $S^1_t$ and $S^2_t$ (with $\sigma_1$ and $\sigma_2$) with a drift, s.t. they are martingales. So they read as:
$S^{1/2}_t = S^{1/2}_0 exp[-\frac{\sigma_{1/2}}{2} + \sigma_{1/2}dW^{1/2}_t ]$
Let 
$X_t = Y_1 S^1_t + Y_2 S^2_t$
Then we have
$dX_t = Y_1 \sigma_1 dW^1_t + Y_2 \sigma_2 dW^2_t$. Then it can be seen, that the distribution of $X_t$ depends on $Y_1 \sigma_1 S^1_t$ and $Y_2 \sigma_2 S^2_t$, i.e. the variance of the normal distribution. Now unless the $\sigma$s do not equal one (in that case the variance becomes just $X_t$), this means that:
$\mathbb{P}(X_t<x|X_u,S^1_u, S^2_u, u\leq s) \neq \mathbb{P}(X_t<x|X_s)$
Nevertheless, $X_t$ is of course a (local) martingale. Objections?
