Let
$dX_{t} = m_1(l_1-X_{t})dt+\sigma_1 dW_{t}$
and
$dY_{t} = m_2(l_2-Y_{t})dt+\sigma_2(\rho dW_{t}+\sqrt{1-\rho^2}dW_{t}^{1})$
where the $m_i$'s, $l_i$'s and $\sigma_i$'s are constants, $\rho \in [-1,1]$, and $W_{t}$ and $W_{t}^{1}$ are independent Brownian motions.
I know that $X_t$ and $Y_t$ are Markov processes, is also their sum $X_t+Y_t$ another Markov process?
Best Answer
It this case, if for example $m_1\neq m_2, \sigma_1\sigma_2\rho\neq 0$, no, because the drift is not a function of $X_t + Y_t$.
In a general situation, when $X,Y$ are two non independant Markov chains, $X+Y$ is not.
To understand this with a more simple example, in a discret time setting, take for example $(X_n)$ be a simple random walk on $\mathbb Z$, and take $Y_n = n1_{X_1 = 1}$ (is is a Markov process, because is $Y_{n+1} = Y_n + 1_{X_1 = 1}$).
Then, $S = X+Y$ is not a Markov process: the law of $S_n|S_{n-1}$ depends on $X_1 = S_1$.