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.

  • $\begingroup$ Thank you again. If you could also take a look at the linked post? That would be so nice! $\endgroup$ – Qyburn Apr 3 '15 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.