$\mathbb{E}((M(\tau)-M(\rho))^2|F_{\rho})=\mathbb{E}(M^2(\tau)-M^2(\rho)|F_{\rho})$

Question

$\rho,\tau$ - stopping times such that $0\le \rho\le\tau\le T$ and $M(t),F_t)$ square integrable martingale. Prove that $\mathbb{E}((M(\tau)-M(\rho))^2|F_{\rho})=\mathbb{E}(M^2(\tau)-M^2(\rho)|F_{\rho})$ Can anyone prove it?

Answer 1

This is a very common identiy in the formulation of Ito calculus.

\begin{align} \mathbb{E}[(M(\tau) - M(\rho))^2|F_\rho] &= \mathbb{E}[M(\tau)^2 - 2M(\rho)M(\tau)+M(\rho)^2|F_\rho]\\ \tag{1} &= \mathbb{E}[M(\tau)^2|F_\rho] - 2M(\rho)\mathbb{E}[M(\tau)|F_\rho]+\mathbb{E}[M(\rho)^2|F_\rho]\\ \tag{2} &= \mathbb{E}[M(\tau)^2|F_\rho] - 2M(\rho)^2+\mathbb{E}[M(\rho)^2|F_\rho]\\ \tag{3}&= \mathbb{E}[M(\tau)^2|F_\rho] - 2\mathbb{E}[M(\rho)^2|F_\rho]+\mathbb{E}[M(\rho)^2|F_\rho]\\ &= \mathbb{E}[M(\tau)^2|F_\rho] - \mathbb{E}[M(\rho)^2|F_\rho]\\ &= \mathbb{E}[M(\tau)^2 - M(\rho)^2|F_\rho] \end{align} So, (1) is by the linearity of conditional expectations, where things $F_\rho$-measurable are treated as scalars (they can just come out of the expectation). (2) is by definition of a Martingale. (3) is another example of the property of conditional expectation that if $X \in \mathcal{F}$, $X = \mathbb{E}[X|\mathcal{F}]$. Anyway, hope this helps. A good reference is Durret, Probability: Theory and Example Theorem 4.1.14 p. 212, Theorem 4.4.8 Conditional Variance Formula p. 237.

Answer 2

Well $M_{\rho}$ is certainly $\mathscr{F}_{\rho}$ measurable. So upon expanding the square on the left hand side, notice that $2 \mathbb{E}[M_\rho M_ \tau | \mathscr{F}_{\rho}] = 2 M_\rho \mathbb{E}[M_\tau | \mathscr{F}_{\rho}] = 2 M_{\rho}^{2}$ by the optional stopping theorem, which is applicable here since the stopping times are bounded (assuming your $T$ is a constant). This solves the problem.