Quadratic variation of $B_t=(1-t)W_{\frac{t}{t-1}}$ Let $t\in [0,1), W_t $ Brownian motion and $B_t:=(1-t)W_{\frac{t}{1-t}}$. Show that $\langle B \rangle_t=t$. I tried substituting $s=\frac{t}{1-t}$ and computing the quadratic variation of $B_\frac{s}{s+1}=(1-\frac{s}{s+1})W_s$ with Itô but I got $\langle B \rangle_t=\frac{t}{1+t}$ after resubstitution. This makes me think that i can't use this approach, in particular that using substition to rescale the time changes properties of a stochastic process.
So basicly I have two questions:


*

*Can I substitute the time index of a stochastic process without changing its properties?

*How do I calculate the quadratic variation of $B$?

 A: 
Lemma 1: Let $f: [0,\infty) \to [0,\infty)$ be a continuous function which is stricly increasing and satisfies $f(0)=0$. If $(M_t)_{t \geq 0}$ is a stochastic process with quadratic variation $\langle M \rangle_t$, then the quadratic variation of the time-changed process $N_t := M_{f(t)}$ equals $ \langle M \rangle_{f(t)}$.

Proof: For fixed $t>0$ choose a partition $\Pi = \{0=t_0 < \ldots <t_n=t\}$ of the interval $[0,t]$. Because of our assumptions on $f$, we know that $$s_j := f(t_j)$$ defines a partition $\Pi' := \{0=s_0 < \ldots < s_n = f(t)\}$ of the interval $[0,f(t)]$. Moreover, the continuity of $f$ implies that the mesh size $|\Pi'|$ of the partition $\Pi'$ tends to zero if $|\Pi|$ tends to zero. Therefore we find from the definition of the quadratic variation:
$$\begin{align*} \langle N \rangle_t = \lim_{|\Pi| \to 0} \sum_{j=1}^n (N_{t_j}-N_{t_{j-1}})^2  &= \lim_{|\Pi| \to 0} \sum_{j=1}^n (M(f(t_j))-M(f(t_{j-1})))^2 \\ &= \lim_{|\Pi'| \to 0} \sum_{j=1}^n (M(s_j)-M(s_{j-1}))^2 \\ &= \langle M \rangle_{f(t)}. \end{align*}$$

Lemma 2: Let $g: [0,\infty) \to \mathbb{R}$ be a Lipschitz continuous function and let $(X_t)_{t \geq 0}$ be a stochastic process with quadratic variation $\langle X \rangle_t$. If $(X_t)_{t \geq 0}$ has continuous sample paths, then the quadratic variation of $Y_t := g(t) X_t$ equals $$\langle Y \rangle_t = \int_0^t g(s)^2 \, d\langle X \rangle_s.$$

Proof: Denote by $L$ the Lipschitz constant of $g$, and let $\Pi = \{0=t_0 < \ldots < t_n\}$ be a partition of an interval $[0,t]$ for $t>0$. In order to compute the quadratic variation of $(Y_t)_{t \geq 0}$, we have to consider expressions of the form $$\sum_{j=1}^n (Y_{t_j}-Y_{t_{j-1}})^2 =  \sum_{j=1}^n (g(t_j) X_{t_j}-g(t_{j-1}) X_{t_{j-1}})^2 = I_1+I_2+I_3$$ where $$\begin{align*} I_1 &:= \sum_{j=1}^n (g(t_j)-g(t_{j-1}))^2 X_{t_j}^2 \\ I_2 &:= -2\sum_{j=1}^n  g(t_{j-1}) (g(t_j)-g(t_{j-1})) (X_{t_j}-X_{t_{j-1}}) \\ I_3 &:= \sum_{j=1}^n g(t_{j-1})^2 (X_{t_j}-X_{t_{j-1}})^2. \end{align*}$$ We estimate the terms separately. By the Lipschitz continuity of $g$ and the boundedness of $[0,t] \ni s \mapsto X_s(\omega)$ for fixed $\omega \in \Omega$, we have $$\begin{align*} |I_1| \leq L^2 \sup_{s \leq t} |X_s|^2 \sum_{j=1}^n \underbrace{(t_j-t_{j-1})^2}_{\leq |\Pi| (t_j-t_{j-1})} &\leq L^2 |\Pi| \sup_{s \leq t} |X_s|^2 \underbrace{\sum_{j=1}^n (t_j-t_{j-1})}_{t} \\ &\xrightarrow[\text{a.s.}]{|\Pi| \to 0} 0. \end{align*}$$Similarly, the Lipschitz continuity of $g$ and the uniform continuity of $s \mapsto X_s(\omega)$ on compact sets entails $$\begin{align*} |I_2| &\leq 2L \sup_{s \leq t} |g(s)| \sup_{\substack{u,v \in [0,t] \\ |u-v| \leq |\Pi|}} |X_u-X_v| \underbrace{\sum_{j=1}^n (t_j-t_{j-1})}_{t} \xrightarrow[\text{a.s.}]{|\Pi| \to 0} 0. \end{align*}$$  For the last term we note that $$\lim_{|\Pi| \to 0} \sum_{j=1}^n g(t_{j-1})^2 (X_{t_j}-X_{t_{j-1}})^2 = \lim_{|\Pi| \to 0} \sum_{j=1}^n g(t_{j-1})^2 (\langle X \rangle_{t_j}-\langle X \rangle_{t_{j-1}}) = \int_0^t g(s)^2 \, d\langle X \rangle_s; $$ this follows from the definition of the quadratic variation and basic properties of Riemann-Stieltjes integrals.

Using these two statements it shouldn't be too difficult to calculate the quadratic variation of $B$; if you, however, encounter any problems you are welcome to leave a comment. 
