Consider the Brownian bridge process $$B_t = W_t - tW_1,$$ where $W_t$ be a Brownian motion on $[0,1]$. What is the expected quadratic variation of $B_t$?
Definition: The co-variation of two stochastic processes $X_t$ and $Y_t$ is defined by $$[X,Y]_t := \lim_{n\rightarrow\infty}\sum_{i=1}^n(X_{t_i} - X_{t_{i-1}})(Y_{t_i} - Y_{t_{i-1}}),$$ where $[t_i,t_{i-1}]\in \Pi_n$ for $i=1,2,..n$ and $(\Pi_n)_{n\in\mathbb N}$ is a sequence of partitions of $[0,t]$ with mesh converging to zero. The quadratic variation of a stochastic process $Z_t$ is then $[Z,Z]_t$.
My approach: For stochastic processes $X_t,Y_t,Z_t,C_t$, where $C_t = c$ for all $t$, it holds that
- $[X + C, Y]_t = [X,Y]_t$,
- $[CX, Y]_t = C[X,Y]$,
- $[X,Y] = [Y,X]$,
- $[X+Y,Z]_t = [X,Z]_t + [Y,Z]_t$.
Let $I_t = t$. Then \begin{align*} [B,B]_t &= [W - IW_1,W - IW_1]_t \\ &= [W,W]_t - [W, IW_1]_t - [IW_1,W]_t + [IW_1,IW_1]_t \\ &= [W,W]_t - 2[W,IW_1]_t + [IW_1,IW_1]_t \\ &= [W,W]_t - 2W_1[W,I]_t + W_1^2[I,I]_t.\end{align*} Since $[W,W]_t = t$ and $\mathbb E[W_1] = 0$ and $\mathbb E[W_1^2] = 1$, taking expectations yields $$\mathbb E[B,B]_t = t - [I,I]_t.$$
Is this correct?
Edit: Apparently my calculations were not correct. I used that $\mathbb E[W_1[W,I]_t] = \mathbb E[W_1]\mathbb E[[W,I]_t]$. But this relation may not hold. For example, in the case $t=1$. In that case, the term $W_1$ also appears in $[W,I]_1$, and it's a trivial result that a constant random variable is uncorrelated with itself.
The correct approach would be to use the fact that $I$ is continuous with bounded variation, as explained in the comments.
Let $n\in\mathbb N$. Then $$\sum_{i=1}^n(X_{t_i} - X_{t_{i-1}})({t_i} - {t_{i-1}}) \leq \sqrt{\sum_{i=1}^n(X_{t_i} - X_{t_{i-1}})^2}\sqrt{\sum_{i=1}^n({t_i} - {t_{i-1}})^2}$$ by Cauchy-Schwarz inequality. Since the mesh $\sup_{i=1}^n\vert t_i - t_{i-1}\vert$ converges to zero by assumption it follows that the second factor converges to zero as \begin{align*}\sum_{i=1}^n({t_i} - {t_{i-1}})^2 &\leq \left(\sum_{i=1}^n\vert{t_i} - {t_{i-1}}\vert\right)\left(\sum_{i=1}^n\vert{t_i} - {t_{i-1}}\vert\right) \\ &\leq \sup_{i=1}^n\vert t_i - t_{i-1}\vert \left(\sum_{i=1}^n\vert{t_i} - {t_{i-1}}\vert\right) \\ &=\sup_{i=1}^n\vert t_i - t_{i-1}\vert t_n \\ & \leq \sup_{i=1}^n\vert t_i - t_{i-1}\vert t.\end{align*} Hence the terms $[W,I]_t$ and $[I,I]_t$ vanish. The quadratic variation of a Brownian bridge is therefore $$[B]_t = t.$$ In fact, the quadratic variation of a Brownian bridge is deterministic.