Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale

Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.

Let $\sequence {X_t}_{t \ge 0}$ be a continuous-time martingale such that $\size {X_t}^2$ is integrable for each $t \in \hointr 0 \infty$.

Let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$.

Let:


 * $s = t_0 < t_1 < \ldots < t_n = t$

be a finite subdivision of $\closedint s t$.

Then:


 * $\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_s} = \expect {X_t^2 - X_s^2 \mid \FF_s} = \expect {\paren {X_t - X_s}^2 \mid \FF_s}$ almost surely.

Proof
We have, for $i < j$:

Since $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale, we have:


 * $X_{t_i}$ is $\FF_{t_i}$-measurable

and so:


 * $2 \expect {X_{t_j} X_{t_i} \mid \FF_{t_i} } = 2 X_{t_i} \expect {X_{t_j} \mid \FF_{t_i} }$

from Rule for Extracting Random Variable from Conditional Expectation of Product.

Using the martingale property, we have:


 * $\expect {X_{t_j} \mid \FF_{t_i} } = X_{t_i}$

From Conditional Expectation of Measurable Random Variable, we have:


 * $\expect {X_{t_i}^2 \mid \FF_{t_i} } = X_{t_i}^2$

Putting this together we have:


 * $\expect {\paren {X_{t_j} - X_{t_i} }^2 \mid \FF_{t_i} } = \expect {X_{t_j}^2 \mid \FF_{t_i} } - 2 X_{t_i}^2 + X_{t_i}^2 = \expect {X_{t_j}^2 \mid \FF_{t_i} } - X_{t_i}^2$

Since $\expect {X_{t_i}^2 \mid \FF_{t_i} } = X_{t_i}^2$, we have:


 * $\expect {\paren {X_{t_j} - X_{t_i} }^2 \mid \FF_{t_i} } = \expect {X_{t_j}^2 - X_{t_i}^2 \mid \FF_{t_i} }$

from Conditional Expectation is Linear.

Setting $i = 0$, $j = n$, we have:


 * $\expect {X_t^2 - X_s^2 \mid \FF_s} = \expect {\paren {X_t - X_s}^2 \mid \FF_s}$

We can also compute:

completing the proof.