Given $0\leq s <t,$ why is $E[(W(t)-W(s))^2 |F(s)] = E[(W(t)-W(s))^2]?$

Question

Let $(W(t))_t$ be the Brownian motion. When showing$W^2(t)-t$ is a martingale, it is quite common to do the following:

Given any $0\leq s<t$, since \begin{align*} E[(W^2 (t)-t)│F(s) ] & = E[(W(t)-W(s))^2+2W(t)W(s)-W^2 (s)|F(s)]-t\\ & =E[(W(t)-W(s))^2 |F(s)]+2W(s)E[W(t))|F(s)]-E[W^2 (s))|F(s)]-t \\ & =E[(W(t)-W(s))^2 ]+ 2W^2 (s)-W^2 (s)-t \\ & =(t-s)+W^2 (s) \\ & =W^2 (s)-s. \end{align*}

However, there is one part which puzzles me:

Why is $E[(W(t)-W(s))^2 |F(s)] = E[(W(t)-W(s))^2]?$

I know that if $W(t)-W(s)$ is independent of $F(s)$ due to property of Brownian motion. But why is its square $(W(t)-W(s))^2$ also independent of $F(s)$?

Answer 1

If a random variable $Y$ is independent of a sigma algebra $\mathcal G$ then $f(Y)$ is independent of $\mathcal G$ for any Borel measurable function $f:\mathbb R \to \mathbb R$. Since $f(x)=x^{2}$ is continuous it is Borel measurable.

By definition $f(Y)$ is independent of $\mathcal G$ iff $P(f(Y)^{-1}(A) \cap B)=P(f(Y)^{-1}(A)) P(B)$ for all Borel sets $A$ and $B$. This is true because $P(f(Y)^{-1}(A) \cap B)=P(Y^{-1}(f^{-1}A) \cap B)$ and $f^{-1}(A)$ is also a Borel set.

Answer 2

We can prove it using the following fact:

$E[B^2_t -B^2_s | \mathscr{F}_s] = E[(B_t -B_s)^2 | \mathscr{F}_s] \tag*{(*)}$

Proof of (*): \begin{align*} E[(B_t-B_s)^2|\mathscr{F}_s] &= E[B_t^2-2B_tB_s + B_s^2|\mathscr{F}_s] \\ &= E[B_t^2|\mathscr{F}_s]-E[2B_tB_s|\mathscr{F}_s] + E[B_s^2|\mathscr{F}_s] \\ &= E[B_t^2|\mathscr{F}_s]-2B_sE[B_t|\mathscr{F}_s] + B_s^2 \\ &= E[B_t^2- B_s^2|\mathscr{F}_s]. \end{align*}

Now we prove that $W_t^2-t$ is a martingale: \begin{align*} E[B^2_t -B^2_s | \mathscr{F}_s] &= E[(B_t -B_s)^2 | \mathscr{F}_s] \tag*{(*)} \\ &= t-s. \end{align*} So $E[B^2_t -t | \mathscr{F}_s]=B^2_s-s.$