# 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).

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 $$[X_t,X_t]=[B_t,B_t]-2[B_t,tB_1]+[tB_1,tB_1]=[B_t,B_t].$$
• Thank you for your reply. If I understand correctly, FV process is a process with bounded variation (BV)? And it has to be differential (again, if I understand correctly, this is a requirement for BV) with regards to $t$? – Ravonrip Nov 25 '18 at 22:40
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.