[Math] Quadratic variation of $X_{t} = tB_{t}$

brownian motionquadratic-variationstochastic-integralsstochastic-processes

Let $X_t = tB_t$ be a process where $B=(B_t)_{t>0}$ is the standard Brownian motion.
Evaluate $\langle X\rangle_t$ the quadratic variation of our process .

I tried to calculate it using :

$d(X_tX_t) = 2X_tdX_t + d\langle X\rangle_t$

$\langle X\rangle_t = X_t^2 – \int\limits_0^t X_t dX_t $

and we have :

$X_tdX_t = t^2B_tdB_t + tB_t^2dt$

so we can write the quadratic variation as :

$\langle X\rangle_t = t^2B_t^2 – 2\int\limits_0^t t^2B_tdB_t – 2 \int\limits_0^t tB_t^2dt$

and I don't know if this is the result wanted or I need to calculate it using different methods ?!

Best Answer

Let $X_t := t \cdot B_t$. By Itô's formula (applied to $f(t,x) := t^2 \cdot x^2$),

$$X_t^2-X_0^2 = 2\int_0^t s^2 B_s \, dB_s + \int_0^t (s^2 + 2s B_s^2) \, ds. \tag{1}$$

On the other hand,

$$X_s dX_s = s^2 B_s \, dB_s + s B_s^2 \, ds. \tag{2}$$

If we combine both equalities, we get

$$\langle X \rangle_t := X_t^2 - 2\int_0^t X_s \, dX_s = \int_0^t s^2 \, ds = \frac{t^3}{3}.$$

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Stochastic Processes – Quadratic Variation of Diffusion Process and Geometric Brownian Motion

Assume that $X$ solves $$ dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t. $$ In integral form, this means $$ X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s. $$ Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is $$ [H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s. $$ You can find this in a special case as Proposition 3.2.17 of Karatzas and Shreve's "Brownian Motion and Stochastic Calculus", or in a general version for continuous semimartingales in Section IV.31 of Rogers and William's "Diffusions, Markov processes and Martingales". Using this, we obtain (with $\mu_t = \mu(t,X_t)$ and $\sigma_t = \sigma(t, X_t)$ as convenient shorthands, and $A_t = t$) $$ [X]_t = [\mu \cdot A]_t + [\sigma\cdot W]_t = \int_0^t \mu_s^2 d[A]_s + \int_0^t \sigma_s^2 d[W]_s \\ = \int_0^t \sigma(s,X_s)^2 ds, $$ where we have used $[A]_t = 0$, since all continuous processes of finite variation have zero quadratic variation, and $[W]_t = t$. As for the geometric Brownian motion, we have that $X$ is a GBM if it satisfies $$ dX_t = \mu X_t dt + \sigma X_t dW_t, $$ which yields $$ [X]_t = \int_0^t \sigma^2 X_s^2 ds. $$ In particular, $[X]$ isn't immediately seen to satisfy any SDE or ODE (as $[X]$ does not figure in the right-hand side of the above). The above is simply a representation of $[X]$ in terms of $X$, but I doubt that there in general exists and SDE or ODE satisfied by $[X]$.

[Math] Quadratic variation of $B_t=(1-t)W_{\frac{t}{t-1}}$

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.

Best Answer

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Stochastic Processes – Quadratic Variation of Diffusion Process and Geometric Brownian Motion

[Math] Quadratic variation of $B_t=(1-t)W_{\frac{t}{t-1}}$

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