Let $W$ be the standard Brownian motion with $$ \Delta W_{t_{i+1}^{n}} := W_{t_{i+1}^{n}}-W_{t_{i}^{n}}\; \; \;\operatorname{and}\; \; \; \Delta t_{i+1}^{n}:=t_{i+1}^{n}-t_{i}^{n}.$$ In lecture notes, we have written in one line that
$\mathbb E[(\Delta W_{t_{i+1}^{n}}^{2}-\Delta t_{i+1}^{n})^{4}]=M\Delta t_{i+1}^{n}$ where $M$ is some constant. I do not believe that this statement is trivial. I have proven that
$\mathbb E[\Delta W_{t_{i+1}^{n}}^{2m}]=\frac{(m-1)!!}{(\Delta t_{i+1}^{n})^{2m}}(\Delta t_{i+1}^{n})^{2m}=L(\Delta t_{i+1}^{n})^{2m}$ for any $m \in \mathbb N$ and some constant $L$.
With this in mind, I attempt to compute $\mathbb E[(\Delta W_{t_{i+1}^{n}}^{2}-\Delta t_{i+1}^{n})^{4}]$.
My attempt:
I want to evaluate the random variable $\Delta W_{t_{i+1}^{n}}^{2}-\Delta t_{i+1}^{n}$. Clearly, by definition of $W$ being a standard Brownian motion, we have $\mathbb E[\Delta W_{t_{i+1}^{n}}^{2}-\Delta t_{i+1}^{n}]=\Delta t_{i+1}^{n}-\Delta t_{i+1}^{n}=0.$
However, I am quite certain that $W^{2}$ and $W^{4}$, do not have increments that are normally distributed. Do I really need to expand the expression $(\Delta W_{t_{i+1}^{n}}^{2}-\Delta t_{i+1}^{n})^{4}$ in order to prove the statement, is there no other way?
For fixed $s \leq t$ set $X:=W_t-W_s$ and $u:= t-s$. Claim: It holds that $$\mathbb{E}[(X^2- u)^4] = M \cdot u^4 \tag{1}$$ for some constant $M$ not depending on $s$ and $t$.
Proof: Since $(W_t)_{t \geq 0}$ is a Brownian motion, the random variable $X$ is Gaussian with mean zero and variance $t-s$. In particular, $X = \sqrt{t-s} W_1$ in distribution. Thus,
$$\mathbb{E}[(X^2-u)^4] = \mathbb{E}[((t-s) W_1^2- (t-s))^4] = (t-s)^4 \underbrace{\mathbb{E}[(W_1^2-1)^4]}_{=:M<\infty}.$$
Applying $(1)$ it follows that
$$\mathbb{E}[(\Delta W_{t_{i+1}^n}^2-\Delta t_{i+1}^n)^4] = M (\Delta t_{i+1}^n)^{\color{red}{4}}.$$