# Variance of the square variation process

Let $X = (X_n)_{n\in\mathbb{N}_0}$ a square integrable $(\mathcal{F_n})_{n\in\mathbb{N}_0}$-martingale. The predictable process $\langle X \rangle_n = \sum_{i=1}^n \Bigl(\mathbf{E}\bigl[X_i^2\vert \mathcal{F}_{i-1}\bigr] - X_{i-1}^2\Bigr)$ is called the square variation process of $X$.

My question is why the following holds: $$\mathbf{E}\bigl[\langle X \rangle_n\bigr] = \mathbf{Var}[X_n - X_0]$$

I guess I know the answer:

We use the telescoping property of the sum:

$$\mathbf{E}\bigl[\langle X \rangle_n \bigr]= \mathbf{E}\biggl[\sum_{i=1}^n \Bigl(\mathbf{E}\bigl[X_i^2\vert \mathcal{F}_{i-1}\bigr] - X_{i-1}^2\Bigr)\biggr] = \sum_{i=1}^n \Bigl(\mathbf{E}\bigl[\mathbf{E}\bigl[X_i^2\vert \mathcal{F}_{i-1}\bigr]\bigr] - \mathbf{E}\bigl[X_{i-1}^2\bigr]\Bigr) = \\ \sum_{i=1}^n \Bigl(\mathbf{E}\bigl[X_i^2\bigr] - \mathbf{E}\bigl[X_{i-1}^2\bigr]\Bigr)= \mathbf{E}\bigl[X_n^2\bigr] - \mathbf{E}\bigl[X_0^2\bigr]\, .$$

Using the tower property and the martingale property we obtain that $$\mathbf{E}\bigl[X_n X_0\bigr] =\mathbf{E}\Bigl[\mathbf{E}\bigl[X_n X_0|\mathcal{F}_0\bigr]\Bigr]=\mathbf{E}\Bigl[X_0\,\mathbf{E}\bigl[X_n |\mathcal{F}_0\bigr]\Bigr] = \mathbf{E}\bigl[X_0^2\bigr]\, .$$ We also know that since $X_n$ is a martingale its expectation is constant, i. e. $\mathbf{E}\bigl[X_n\bigr] = \mathbf{E}\bigl[X_0\bigr]$, so \begin{align} \mathbf{Var}\bigl[X_n - X_0\bigr] &= \mathbf{E}\bigl[(X_n - X_0)^2\bigr] - \mathbf{E}\bigl[X_n - X_0\bigr]^2 \\ & = \mathbf{E}\bigl[(X_n - X_0)^2\bigr]\\ &=\mathbf{E}\bigl[X_n^2\bigr] - 2\,\mathbf{E}\bigl[X_n X_0\bigr] + \mathbf{E}\bigl[X_0^2\bigr] \\ &= \mathbf{E}\bigl[X_n^2\bigr] - \mathbf{E}\bigl[X_0^2\bigr] \,.\; \end{align} $\square$

Is that right?

Please answer if you know it, this is very important for me! Thank you!

The argument is correct. Maybe in the third line of the computation of $\mathbf{Var}\bigl[X_n - X_0\bigr]$, I would refer (and give a name to) the displayed equation where $\mathbf{E}\bigl[X_n X_0\bigr]$ is computed.