If $I_1, I_2, I_3, ...$ is a iid process. Then can I say that
$$M_n = \frac{1}{n}\sum_{i=1}^n I_i$$
Is a independent stationary increments?
I think it shouldn't, but I am not sure if my answer is right, but essentially I know that
$$M_2 - M_1 = \frac{1}{2}I_1 + \frac{1}{2}I_2 - I_1 = \frac{1}{2}I_2-\frac{1}{2}I_1$$
$$M_1-M_0 = I_1$$
And hence
$$M_2 - M_1 = \frac{1}{2}I_2 - \frac{1}{2}(M_1-M_0)$$
And hence $M_n$ does not have independent increments.
Does that work?
Yes that works. To give slightly more detail, suppose that $M_2-M_1$ were independent of $M_1-M_0$. Since $I_2$ is independent of $M_1-M_0$, this would imply that $-I_2+2(M_2-M_1)=M_1-M_0$ is also independent of $M_1-M_0$. But the only way a random variable can be independent of itself is if it is constant a.s. So the claim holds except in the degenerate case where $I_1$ is a.s. constant.