# How to prove that the increments are independent?

I shall prove that if two independent processes have each independent increments, then the sum have independent increments.

What I have tried is this:

Assume we have a probability space $(\Omega, \mathcal{F},P)$, and two stochastic processes $X_t$, $Y_t$. Let $\sigma(X_{t^*})=\{X_{t^*}^{-1}(B)\}$, the sigma-algebra generated by the variable. And let $\sigma(X_{t_1},X_{t_2})=\sigma(\sigma(X_{t_1}),\sigma(X_{t_2}))$ the sigma algebra generated by the two sigma algebras etc.

Let us assume that $s _1\le s_2 \le t_1 \le t_2$. What I need to show is that : $\sigma(X_{t_2}+Y_{t_2}-X_{t_1}-Y_{t_1})$ is independent of $\sigma(X_{s_2}+Y_{s_2}-X_{s_1}-Y_{s_1})$. What I know already is that $\sigma(X_{t_2}-X_{t_1})$ is independent of $\sigma(X_{s_2}-X_{s_1})$ and $\sigma(Y_{t_2}-Y_{t_1})$ is indepent of $\sigma(Y_{s_2}-Y_{s_1})$. And lastly since the two processes are independent i know that $\sigma(X_{t_2},X_{t_1},X_{s_2},X_{s_1})$ is independent of $\sigma(Y_{t_2},Y_{t_1},Y_{s_2},Y_{s_1})$.

What I need to end up with is that:

$P(X_{t_2}+Y_{t_2}-X_{t_1}-Y_{t_1}\in A, X_{s_2}+Y_{s_2}-X_{s_1}-Y_{s_1} \in B)=P(X_{t_2}+Y_{t_2}-X_{t_1}-Y_{t_1}\in A)\cdot P(X_{s_2}+Y_{s_2}-X_{s_1}-Y_{s_1} \in B)$.

But I do get stuck here. My only idea is to work with some kind of reduction, where I look at the variables $X_{t_2}-X_{t_1}=Z_1,Y_{t_2}-Y_{t_1}=Z_2,X_{s_2}-X_{s_1}=Z_3,Y_{s_2}-Y_{s_1}=Z_4$. The the LHS of what is above is $P(Z_1+Z_2 \in A, Z_3+Z_4 \in B)$. Uintil this point I haven't really gotten anywhere it seems I just have rewritten the problem, maybe the next step is showing that $Z_1, Z_2$, but I am not sure how that follows either(if variables are independent, are their differences also independent.?).

Any hints or help on how to solve this?

• By following your approach up to the point "What I need to end up...", you can conclude using the fact that if $\mathcal{G}$ and $\mathcal{H}$ are both independent of $\mathcal{C},$ then also $\sigma(\mathcal{G}, \mathcal{H})$ is independent of $\mathcal{C}.$ This you can prove, by checking the independence on the intersection stable generator $\mathcal{P}:= \{ F | F = G \cap H, G \in \mathcal{G}, H \in \mathcal{H} \}$ – Kore-N Feb 15 '16 at 20:17

It is much easier to consider the random vector $$Z := (X_{s_2}-X_{s_1},Y_{s_2}-Y_{s_1}, X_{t_2}-X_{t_1},Y_{t_2}-Y_{t_1})$$ and show that the components are independent. In fact, we have

\begin{align*} &\quad \mathbb{P}( X_{s_2}-X_{s_1} \in A_1, X_{t_2}-X_{t_1} \in A_2, Y_{s_2}-Y_{s_1} \in A_3, Y_{t_2}-Y_{t_1} \in A_4) \\ &= \mathbb{P}(X_{s_2}-X_{s_1} \in A_1, X_{t_2}-X_{t_1} \in A_2) \cdot \mathbb{P}( Y_{s_2}-Y_{s_1} \in A_3, Y_{t_2}-Y_{t_1} \in A_4) \\ &= \mathbb{P}(X_{s_2}-X_{s_1} \in A_1) \mathbb{P}(X_{t_2}-X_{t_1} \in A_2) \mathbb{P}(Y_{s_2}-Y_{s_1} \in A_3) \mathbb{P}(Y_{t_2}-Y_{t_1} \in A_4) \end{align*}

where we have used first the independence of $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ and then the independence of the increments. Now we can use a general result from probability theory.

Lemma: Let $Z=(Z_1,\ldots,Z_n)$ be a vector of independent (real-valued) random variables and $g: \mathbb{R}^k \to \mathbb{R}$ a measurable function for some $k \leq n$. Then $g((Z_1,\ldots,Z_k)),Z_{k+1},\ldots,Z_n$ are independent.

See below for a proof. Applying this statement twice, we find that $(X_{t_2}-X_{t_1})+(Y_{t_2}-Y_{t_1})$ and $(X_{s_2}-X_{s_1})+(Y_{s_2}-Y_{s_1})$ are independent and this finishes the proof.

Remark: It is not too difficult to show that a stochastic process $(X_t)_{t \geq 0}$ has independent increments if, and only if, $X_t-X_s$ is independent of $\mathcal{F}_s$ for any $s \leq t$ (here $\mathcal{F}_t$ denotes the canonical filtration). Using this characterization, it is a little bit easier to prove the assertion.

Proof of the Lemma:

\begin{align*} \mathbb{P}(g(Z_1,\ldots,Z_k) \in A, Z_{k+1} \in B_1,\ldots,Z_{n} \in B_{n-k}) &= \mathbb{P}((Z_1,\ldots,Z_n) \in g^{-1}(B) \times B_1 \times \ldots \times B_{n-k}) \\ &= \mathbb{P}((Z_1,\ldots,Z_k) \in g^{-1}(B)) \prod_{j=k}^n \mathbb{P}(Z_j \in B_j) \\ &= \mathbb{P}(g(Z_1,\ldots,Z_k) \in B) \prod_{j=k}^n \mathbb{P}(Z_j \in B_j) \end{align*}

for any measurable sets $B$, $B_j$.