Great hints. I have just gone through it and have shown everything and shown that $M_t$ is a Martingale. Now I am not 100% sure how this relates to the quadratic variation expression I am proving though.

Sorry. I just thought about it longer and I think I got it. Thank you very much

@Jim Great. You are welcome. If you find the answer helpful, you can upvote/accept it by clicking on the up arrow/tick next to it.

To clarify, step 1 is because it is a square integrable martingale right?

@Jim We don't need square-integrability for this. Note that any martingale has constant expectation, i.e. $$\mathbb{E}(M_t) = \mathbb{E}(M_0)$$ for any martingale $(M_t)_{t \geq 0}$.

Sorry I should have clarified my question. Why did you assume that $X_0=0$

@Jim Well, because that's a standing assumption if you talk about Lévy processes and use the probability measure $\mathbb{E}$. If the process is started at $x \in \mathbb{R}$, then the probability measure $\mathbb{E}^x$ is used. (However, for $x \neq 0$ the claim does not hold.)

Thanks for the help. Could you give me one more line of hint when using the stationary and independent increment properties at the end of step 3. I thought I did it but it turns out I was wrong about it and am now stuck.

@Jim Where exactly are you stuck?

I know I am supposed to end up with $X_s^2 - E(X_s^2)$. The first term I think simplifies to $E(X_t^2-X_s^2|F_s)$ but even that I am doubting myself. I think all this increment and filtration stuff is confusing me

@Jim Let $X$ be a random variable and $\mathcal{F}$ a filtration such that $X$ and $\mathcal{F}$ and independent. How to calculate $$\mathbb{E}(X \mid \mathcal{F})$$...?

If X is independent of the filtration F, then $E(X|F)=E(X)=0 in this case$

@Jim Note that in general $\mathbb{E}(X) \neq 0$. So, what happens if you applies this with $X := (X_t-X_s)^2$ and $\mathcal{F} := \mathcal{F}_s$?

$E((X_t-X_s)^2|F_s)=E((X_t-X_s)^2)=0$ I'm pretty sure this doesnt seem right

@Jim The first equality is correct; the second not. Why should $\mathbb{E}((X_t-X_s)^2)$ equal $0$? Use instead the stationarity of the increments and (afterwards!) step 2.

Oh right. I get it now. Ok. For the second term, does this make sense: $2E(X_s(X_t-X_s)|F_s)=2E(X_s|F_s)E(X_t-X_s|F_s)=2X_sE(X_t-E_s)=0$

@Jim It's correct what you have written. Note however that we can only pull $X_s$ outside because it is $\mathcal{F}_s$-measurable. (It is (in general) not true that $$\mathbb{E}(X \cdot Y \mid \mathcal{F}) = \mathbb{E}(X \mid \mathcal{F}) \cdot \mathbb{E}(Y \mid \mathcal{F}).$$

Perfect. I am so sorry for making you go through all this. I just started this course and facts are still new to me. For the third term, I am pretty sure $E(X_s^2|F_s)=X_s^2$. Why is $X_s^2\ F_s$ measurable? I thought only $X_s is F_s$ measurable

@Jim Let $X$ be (again) a random variable and $f: \mathbb{R} \to \mathbb{R}$ a (Borel-)measurable function. Is $f(X)$ again a random variable (i.e. measurable)?

Right. I understand. f(X) will also be measurable and in this case $f=x^2$. So after all the algebra, I obtain $X_s^2-sE(X_1^2)$ and so $X_t^2-tE(X_1^2)$ is a martingale. Thus by definition, $<X>_t=tE(X_1^2)$

@Jim Yes, exactly.

@Jim So why did you un-accept my answer? (Don't get me wrong, I don't care about the reputation. I'm just wondering...)

Sorry. I was just clicking random stuff