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

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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.


Corollary

$\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2} = \expect {X_t^2 - X_s^2} = \expect {\paren {X_t - X_s}^2}$


Proof

We have, for $i < j$:

\(\ds \expect {\paren {X_{t_j} - X_{t_i} }^2 \mid \FF_{t_i} }\) \(=\) \(\ds \expect {X_{t_j}^2 - 2 X_{t_j} X_{t_i} + X_{t_i}^2 \mid \FF_{t_i} }\)
\(\ds \) \(=\) \(\ds \expect {X_{t_j}^2 \mid \FF_{t_i} } - 2 \expect {X_{t_j} X_{t_i} \mid \FF_{t_i} } + \expect {X_{t_i}^2 \mid \FF_{t_i} }\) Conditional Expectation is Linear

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:

\(\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_s}\) \(=\) \(\ds \sum_{i \mathop = 1}^n \expect {\paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_s}\) Conditional Expectation is Linear
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \expect {\expect {\paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_{t_{i - 1} } } \mid \FF_s}\) Tower Property of Conditional Expectation
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \expect {\expect {X_{t_i}^2 - X_{t_{i - 1} }^2 \mid \FF_{t_{i - 1} } } \mid \FF_s}\) by previous computation
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \expect {X_{t_i}^2 - X_{t_{i - 1} }^2 \mid \FF_s}\) Tower Property of Conditional Expectation
\(\ds \) \(=\) \(\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i}^2 - X_{t_{i - 1} }^2} \mid \FF_s}\) Conditional Expectation is Linear
\(\ds \) \(=\) \(\ds \expect {X_{t_n}^2 - X_{t_0}^2 \mid \FF_s}\) the sum telescopes
\(\ds \) \(=\) \(\ds \expect {X_t^2 - X_s^2 \mid \FF_s}\)

completing the proof.

$\blacksquare$


Sources

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