Quadratic variation process of some transformation of the Wiener process

Question

I try to compute $\mathbb{E}((B_t^2-t)^2|\mathcal{F}_s)$ where $B_t$ is a $(\mathcal{F}_t)$-Wiener process on some filtered probability space. It is well known that $B_t^2-t$ is a Martingale hence $(B_t^2-t)^2$ should be a square integrable Submartingale and hence can be decomposed into a sum of a Martingale and an adapted increasing process $(A_t)$, with $A_0=0$. Therefore when computing the above term I should get something like $\mathbb{E}((B_t^2-t)^2|\mathcal{F}_s)=(B_s^2-s)^2+\mathbb{E}(A_t-A_s|\mathcal{F}_s)$ a.s.

But all I get are some nasty terms that don't vanish. I tried already the trick with $B_t=B_t-B_s+B_s$ but it seems not to work properly.

Any suggestions from someone who tried this before?

Answer 1

For $s\le t$ we have $$E((B_t^2 - t)^2|\mathcal{F}_s) = E((B_t-B_s+B_s)^2 - t)^2|\mathcal{F}_s) = E((B_t-B_s)^2+B_s^2 + 2B_s(B_t-B_s) - t)^2|\mathcal{F}_s) = $$ $$=E((B_t-B_s)^4+B_s^4 + 4B_s^2(B_t-B_s)^2 + t^2+2(B_t-B_s)^2B_s^2 +4B_s(B_t-B_s)^3-2t(B_t-B_s)^2 +$$ $$ + 4B_s^3(B_t-B_s) -2tB_s^2 -4tB_s(B_t-B_s)|\mathcal{F}_s).$$ We separately calculate each of the terms:

1. $E((B_t-B_s)^4|\mathcal{F}_s) =E(B_t-B_s)^4 = 3(t-s)^2$,

2. $E(B_s^4|\mathcal{F}_s) = B_s^4$,

3. $E(4B_s^2(B_t-B_s)^2|\mathcal{F}_s) =4B_s^2\cdot E(B_t-B_s)^2 = 4B_s^2(t-s)$,

4. $E(t^2|\mathcal{F}_s) = t^2$,

5. $E(2(B_t-B_s)^2B_s^2|\mathcal{F}_s) = 2B_s^2(t-s)$,

6. $E(4B_s(B_t-B_s)^3|\mathcal{F}_s) = 4B_s E(B_t-B_s)^3 = 0$,

7. $E(2t(B_t-B_s)^2|\mathcal{F}_s) = 2t(t-s)$,

8. $E(4B_s^3(B_t-B_s)|\mathcal{F}_s) = 4B_s^3\cdot 0 = 0$,

9. $E(2tB_s^2|\mathcal{F}_s) = 2tB_s^2 $,

10. $E(4tB_s(B_t-B_s)|\mathcal{F_s}) = 4tB_s\cdot 0 = 0 $.

Finally, we get that $$E((B_t^2 - t)^2|\mathcal{F}_s) = 3(t-s)^2 + B_s^4 + 4B_s^2(t-s) +t^2 +2B_s^2(t-s) -2t(t-s)-2tB_s^2 = $$

$$ =B_s^4 -2sB_s^2+s^2 +3t^2 - 6ts +2s^2 +4B_s^2(t-s) +t^2 -2t^2 +2ts= $$

$$ = (B_s^2-s)^2 + 4B_s^2(t-s) + 2t^2 -4ts+2s^2 = (B_s^2-s)^2 + 4B_s^2(t-s) + 2(t-s)^2. $$ We got an expression of the form $X_s + \underbrace{E(A_t-A_s|\mathcal{F}_s)}_{\ge 0}$, where $X_t = (B_t^2-t)^2$. The decomposition you suggested $$E(X_t|\mathcal{F}_s) = X_s + A_t -A_s \text{ a.s.}$$ is not correct.

Indeed, $X_t$ is submartingale and we have $X_t = M_t + A_t$, where $M_t$ is a martingale and $A_t$ is adapted increasing process. Then $$E(X_t - A_t|\mathcal{F}_s) = E(M_t |\mathcal{F}_s) = M_s = X_s - A_s \text{ a.s.}$$ You are assuming that $$E(X_t|\mathcal{F_s}) = X_s + A_t -A_s \text{ a.s.}$$ But $$E(X_t|\mathcal{F}_s) = X_s - A_s + E(A_t|\mathcal{F}_s) = X_s + \underbrace{E(A_t-A_s|\mathcal{F}_s)}_{\ge 0} \text{ a.s.}$$ So, $E(A_t|\mathcal{F_s})=A_t$ and it follows that $A_t$ is $\mathcal{F}_s$-measurable, but this is not true. For example, if $(\mathcal{F}_t)$ is a natural filtration of Wiener process and if $A_t$ is $\mathcal{F}_s$-measurable, then $A_t$ is $\mathcal{F}_0$-measurable, where $\mathcal{F}_0$ is trivial $\sigma$-algebra and hence $A_t =\text{const}$ a.s. for all $t$.

Answer 2

Greyls' answer is great. Nevertheless, if you want a more straightforward approach, let's recall that the effect of a filtration $\mathcal{F}_s$ on a stochastic process $X_t$ is to make it independent of prior states $X_{t<s}$. In practice, it is done by treating $X_s$ as the "new initial condition", which can thus be seen as a deterministic constant, instead of $X_0$. Applied to the brownian motion $B_t = B_s + (B_t-B_s)$, with $B_t-B_S \sim \mathcal{N}(0,t-s)$, it leads to $B_t|\mathcal{F}_s \sim \mathcal{N}(B_s,t-s)$, which allows to compute the expectation directly.

Answer 3

Using some Ito calculus one can obtain greyls' solution as well:

We know $M_t:=B_t^2-t=2\int_0^tB_s\,dB_s\,.$ Therefore, $$\tag{1} M_t^2-\langle M\rangle_t=M_t^2-4\int_0^t B_u^2\,du $$ is also a martingale. Then,

\begin{align}\tag{2} \mathbb E\big[M^2_t-\langle M\rangle_t\big|{\cal F}_s\big]= M^2_s-\langle M\rangle_s=(B_s-s)^2-4\int_0^s B_u^2\,du\,. \end{align} From integration by parts we get \begin{align} tB^2_t-sB^2_s&=\int_s^tB_u^2\,du+\int_s^tu\,d(B^2)_u\\ &=\int_s^tB_u^2\,du+2\int_s^tuB_u\,dB_u+\int_s^tu\,du\,.\tag{3} \end{align} This implies

\begin{align} \mathbb E\big[\langle M\rangle_t\big|{\cal F}_s\big]&=4\,\mathbb E\Big[ \int_0^tB_u^2\,du \Big|{\cal F}_s\Big]=4\,\mathbb E\Big[tB_t^2-2 \int_0^tuB_u\,dB_u \Big|{\cal F}_s\Big]-2t^2\\[2mm] &=4\,\mathbb E\big[tB_t^2\big|{\cal F}_s\big]-8\int_0^suB_u\,dB_u-2t^2\\ &=4t(B_s^2+t-s)-8\int_0^suB_u\,dB_u-2t^2\,,\tag{4} \end{align} (in the last line I used $\mathbb E\big[B_t^2\big|{\cal F}_s\big]=B_s^2+t-s$ because $M_t$ is a martingale). Putting it all together we obtain from (2) and (4) \begin{align} \mathbb E\big[ M^2_t\big|{\cal F}_s\big]&=E\big[\langle M\rangle_t\big|{\cal F}_s\big]+M_s^2-\langle M\rangle_s\\[2mm] &=4tB_s^2+4t^2-4ts-8\int_0^suB_u\,dB_u-2t^2+(B_s-s)^2-4\int_0^sB_u^2\,ds\,.\tag{5} \end{align} Using (3) this can be simplified to \begin{align} \mathbb E\big[ M^2_t\big|{\cal F}_s\big]&= 4tB_s^2+4t^2-4ts-4sB_s^2+2s^2-2t^2+(B_s-s)^2\\ &=2(t-s)^2+4B_s^2(t-s)+(B_s^2-s)^2 \,.\tag{6} \end{align}