# Showing independence of increments of a stochastic process

The textbook on stochastic calculus I am now reading says that if $$X\colon [0,\infty)\times\Omega\rightarrow\mathbb R$$ is a stochastic process such that

1. $$X(t)-X(s)\sim N(0,t-s)$$ for all $$t \geq s \geq 0$$,

2. $$E[X(t)X(s)]=\min\{s,t\}$$ for all $$s,t \geq 0$$,

then, $$X$$ exhibits independence increment, i.e. for every $$0 \leq t_1<..., $$X(t_1)$$, $$X(t_2)-X(t_1)$$, …, $$X(t_n)-X(t_{n-1})$$ are independent.

Here $$X(t)$$ denotes a random variable $$X(t):\Omega\rightarrow \mathbb R$$ such that $$X(t)(\omega)=X(t,\omega)$$.

But I guess this is not true. I suspect that we need an additional condition that $$X$$ is a Gaussian process. (Then, it is easy to show the independence)

Am I on the right track? If so, can you give me some counterexamples?

Or can it be shown without assuming Gaussian process?

Any hint would be appreciated! Thanks and regards.

• Seems this will return to oblivion. @Mhr, could you let me know which textbook you're referring to? Maybe there's a hint in it. – AddSup Dec 25 '18 at 19:31
• @AddSup yes also interested to get the original source to check for myself as well! – Ezy Dec 29 '18 at 4:34

Let $$t_0=0$$ and look at the sequence $$Y_i=X(t_i)-X(t_{i-1})$$ for $$i\in[1:n]$$. These are Gaussian of mean $$0$$ and respective variance $$t_i-t_{i-1}$$. Let's look at all covariances, for $$i< j$$ (without loss of generality) \begin{align*} \mathbb{E}[Y_i Y_j] &= \mathbb{E}[(X(t_i)-X(t_{i-1})(X(t_j)-X(t_{j-1}))]\\ &=\mathbb{E}[X(t_i)X(t_j)]-\mathbb{E}[X(t_{i-1})X(t_j)]-\mathbb{E}[X(t_i)X(t_{j-1})]+\mathbb{E}[X(t_{i-1})X(t_{j-1})]\\ &=t_i-t_{i-1}-t_i+t_{i-1}\\ &=0 \end{align*} So the covariance matrix of $$(Y_1,\dots,Y_n)$$ is a diagonal matrix and so they are all independents. You should be able to conclude that you don't need $$X$$ to be a Gaussian process from there, the only difference is that you add $$X(0)$$ to the first element and that $$X(0)$$ is uncorrelated with all $$Y_i$$ for $$i\in[2:n]$$.

The only problem I can see there is if $$t_1=0$$ but in this case you can reduce $$n$$ by one and remove the first element to prove your result.

All this is true only if $$Y=(Y_1,\dots,Y_n)$$ is jointly Gaussian. A sufficient condition of $$Y$$ to be joint Gaussian is that for any vector $$\mathbf a$$ of size $$n$$, $$\mathbf a^T Y$$ is a Gaussian random variable.

I am not sure on how to prove that but some thing of use may be that $$Y_i+Y_{i+1}+\dots + Y_{j-1}+Y_{j} = X(t_{j})-X(t_{i-1})$$ is Gaussian for any $$i.

• That's just uncorrelatedness, not independence, I guess? How do you know that $(Y_1,...,Y_n)$ is jointly normal? – Mhr Nov 4 '18 at 14:13

I believe you can use a similar trick that is used for the Levi characterisation of BM namely applying ito on the characteristic function of $$X_t$$ and then using the iterated expectations to show the independence of the various $$X_{t_i}-X_{t_j}$$ again by showing the characteristic function expectation factorizes.

• Ezy, thank you, that seems promising. But it also seems that to apply Levy's characterization, we need $X$ to be a square-integrable continuous martingale (or to be a continuous local martingale and $[X]_t=t$). Could you show me how we can get this condition from the two assumptions in the original post? – AddSup Dec 27 '18 at 8:19
• To be specific, in page 3 of the note you linked, how do we know $f(M_t,t)$ is a martingale when $M$ is only known to satisfy the two conditions in the OP? – AddSup Dec 27 '18 at 8:55
• @AddSup for the QV part i dont think you need independence, just that up to moment 4 of increments factorize which i believe you can do similarly to the derivation made in the other answer above. See p4 of ocw.mit.edu/courses/sloan-school-of-management/… . For continuity i agree i went perhaps too quickly but i believe jump processes would have quadratic variation always strictly higher than $t$ due to the jump contribution. But i agree i need to find a more convincing proof for that (if that’s correct) – Ezy Dec 27 '18 at 12:09