Quadratic variation of $X_t=B_t-tB_1$ for $t\in[0,1]$? We define $X_t=B_t-tB_1$ for $t\in[0,1]$, where $B$ is Brownian motion. What is the quadratic variation of this new process?
I tried to calculate it like this (without the limit, $(E_n)$ is a sequence of partitions):
$$\sum_{t_i\in E_n, t_{i+1}\leq t}(B_{t_{i+1}}-t_{i+1}B_1 -B_{t_{i}}+t_{i}B_1)=$$
$$\sum_{t_i\in E_n, t_{i+1}\leq t}((B_{t_{i+1}}-B_{t_i})^2+2(B_{t_{i+1}}-B_{t_i})B_1 + B_1^2(t_i-t_{i+1})^2)$$
I know the quadratic variation for $B_t$, so that can help me with the first part, but I don't know how to compute the other parts.
Is this the wrong way to go about this problem?
This is supposed to be solved without using Itô's lemma or stochastic differential equations (as I saw them used in some other solutions here).
 A: You can use that the quadratic variation is bilinear and that $tB_1$ is a FV process and for a continuous process (semimartingale) $A$ and a FV process $B$ you always have $[A,B]=0$.
To be more precise, you can write
$$[X_t,X_t]=[B_t,B_t]-2[B_t,tB_1]+[tB_1,tB_1]=[B_t,B_t].$$
A: A FV variation is a process with bounded variation on each compact interval. 
But a FV process is not necessarily continuous. For example a simple step process, poisson process or a counting process with finite jumps over each finite interval are discontinuous FV processes.
You can show that a process is a FV process iff it can be represented as the difference of two increasing processes.
And you can show that for a FV process $X$ and a semimartingale $Y$, we have
$$[X,Y]_t=\sum_{s\leq t} \Delta X_s\Delta Y_s,$$
where $\Delta X_s$ denotes the size of the jump of $X$ at time $s$. Now if $Y$ is continuous there are no jumps, hence all jumps have size 0 and thus the quadratic variation of a FV process and a continuous semimartingale is always 0.
