The sum of two random processes with independent increment also has independent increment?

Question

$X_n$ and $Y_n$ are random processes. Both of them have independent increment. Does $X_n + Y_n$ has that property?

Answer 1

No, in general this is not the case. As the example below shows, the processes may depend on each other and therefore adding them up this destroys the independence of the increments. It is not difficult to show that $(X_n+Y_n)_n$ has independent increments if the processes $(X_n)_n$ and $(Y_n)_n$ are independent.

Example Let $(B_t)_{t \geq 0}$ a Brownian motion and set $$X_n := B_n \qquad \qquad Y_n :=X_{n-1} = B_{n-1}$$ and $Y_0 := 0$. For any $n \geq 3$, we have

$$Z_n :=X_n+Y_n = B_n+B_{n-1}$$

Then,

$$\begin{align*} & \mathbb{E}((Z_n-Z_{n-1}) \cdot (Z_{n-1}-Z_{n-2})) \\ &= \mathbb{E}((B_n-B_{n-2}) \cdot (B_{n-1}-B_{n-3})) \\ &= \mathbb{E}(B_n \cdot B_{n-1}) - \mathbb{E}(B_n \cdot B_{n-3}) - \mathbb{E}(B_{n-2} \cdot B_{n-1}) + \mathbb{E}(B_{n-2} \cdot B_{n-3}) \\ &= (n-1)-(n-3)-(n-2)+(n-3) = 1 \end{align*}$$

using that $\mathbb{E}(B_s \cdot B_t) = \min\{s,t\}$ for any $s,t \geq 0$. On the other hand, $\mathbb{E}B_t = 0$ implies

$$\mathbb{E}(Z_n-Z_{n-1}) = \mathbb{E}(Z_{n-1}-Z_{n-2})=0$$

This shows that $Z_n-Z_{n-1}$ and $Z_{n-1}-Z_{n-2}$ are not independent.