# 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 $B$?
• In future you should write \langle, \rangle for the quadratic variation instead of <,>. I've edited your question to change this. – Rhys Steele Jan 29 '18 at 12:54
• Which definition of "quadratic variation" do you know/use? – saz Jan 29 '18 at 14:02
• @saz Same as Wikipedia, we also have several Theorems developed but I couldn't find anything helpful. – Andrew Jan 29 '18 at 14:24
• For $t \in [0,1)$, $\frac{t}{t-1} \leq 0$. So how do you make sense of $W_{\frac{t}{t-1}}$? – Calculon Feb 4 '18 at 17:06
• It is a typo, should be $1-t$, thanks for pointing it out. – Andrew Feb 4 '18 at 17:18

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.

• I think I used your Lemmas in my first try (without thinking why I can do it) in my approach but had (and still have) problems with the expression $d \langle W \rangle_{\frac{t}{1-t}}=d\frac{t}{1-t}$. Using the the two Lemmas I get $\langle B \rangle_t=\int_0^t(1-s)^2d \langle W \rangle_{\frac{s}{1-s}}=\int_0^t(1-s)^2d\frac{s}{1-s}\overset{(*)}=\int_0^tdu=t$. Where I used $u=\frac{s}{1-s}, \frac{1}{(1-s)^2}=du$. But I should have changed the bounds of the Integral as well right? How do I get from $d\frac{t}{1-t}$ to something I can compute? – Andrew Jan 29 '18 at 20:12
• @Andrew No, you don't have to change the bounds. Let's set $N_t := W_{t/1-t}$, then $\langle N \rangle_t = t/(1-t)$. In order to apply the second lemma, you have to compute $d \langle N \rangle_t$. By the very definition, $$\langle N \rangle_t = \int_0^t \, d\langle N \rangle_t. \tag{1}$$ On the other hand, we know from basic calculus that $$\langle N \rangle_t = \frac{t}{1-t} = \int_0^t \frac{d}{ds} \left( \frac{s}{1-s} \right) \, ds. \tag{2}$$ Combining $(1)$ and $(2)$ yields $$d \langle N \rangle_s = \frac{d}{ds} \left( \frac{s}{1-s} \right) \, ds = \frac{1}{(1-s)^2} \, ds.$$ – saz Jan 29 '18 at 20:40
• (More generally, if $f$ is a differentiable function, then $df(t) = f'(t) \, dt$.) – saz Jan 29 '18 at 21:07