7 votes

Quadratic variation of the sum

All càdlàg martingales, and local martingales have well defined quadratic variation, which follows from the fact that such processes are examples of semi- martingales. However, let $I=\{t_0,t_1,\...
Behrouz Maleki's user avatar
7 votes
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Quadratic Variation of a continuous function

If $f$ is continuous and of bounded variation, then any sequence of partitions $\Delta_n$ with $\|\Delta_n\| \to 0$ as $n \to \infty$ will suffice. We have $$\sum_{j=1}^{k_n}|f(t_j) - f(t_{j-1})|^2 ...
RRL's user avatar
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7 votes
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Calculation of the quadratic variation of an Itô process.

Since you only impose mild assumptions on $f$, $g$, the proof is somewhat technical, e.g. we cannot work with Riemann sums because $f^2$ might not be Riemann integrable. Without loss of generality, I ...
saz's user avatar
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7 votes
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Unbounded quadratic variation process for bounded continuous martingale

Consider a one-dimensional Brownian motion $(B_t)_{t \geq 0}$ and the stopping time $$\tau := \inf\{t \geq 0; |B_t| \geq 1\}.$$ By the optional stopping theorem, the stopped process $X_t := B_{t \...
saz's user avatar
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6 votes
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Bound on supremum of local martingale

Define a stopping time $\tau$ by $$\tau := \inf\{t>0; \langle M \rangle_t>b\}.$$ Since \begin{align*} \mathbb{P} \left( \sup_{s \leq t} M_s>a, \langle M \rangle_t \leq b \right) &= \...
saz's user avatar
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6 votes

How many real values of x satisfy the equation?

I would start with the case $x\geq 0$ then we have $$x+x+1=1$$ For $$-1\le x<0$$ we get $$-x+x+1=1$$ and for $x<-1$ we obtain $-x-x-1=1$ and for the second equation i would start with $$x\geq 1$$...
Dr. Sonnhard Graubner's user avatar
6 votes
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Derive an upper bound for the total variation of quadratic covariation [Exercise 1.5.7 in Karatzas and Shreve].

Given a partition $\Pi=\{a=t_0<\ldots<t_n=b\}$ of some interval $[a,b]$ set $$S_{\Pi}(X,Y) := \sum_{t_i \in \Pi} |\langle X,Y \rangle_{t_{i+1}}-\langle X,Y \rangle_{t_i}|.$$ If we add a new ...
saz's user avatar
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5 votes
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Trying to compute the quadratic covariation

There is the following general statement which is helpful for the second approach Let $\psi$ be a progressively measurable function such that $\mathbb{E}\int_0^T \psi(t)^2 \, dt < \infty$ for ...
saz's user avatar
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5 votes
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Quadratic covariation of two Itô processes

You already know that $[Z,Z]_t = \int_0^t \sigma^2(s) \,ds$ for any Itô process $$dZ_t = b(t) \, dt + \sigma(t) \, dW_t.$$ Moreover, the quadratic covariation is defined via the polariation formula, i....
saz's user avatar
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5 votes

Continuous Local Martingale and Quadratic Variation

For the first inclusion note that if $\lim_{t \to \infty} X_t(\omega)$ exists, then it follows from the continuity of the sample paths that $t \mapsto X_t(\omega)$ is bounded. This, in turn, implies $\...
saz's user avatar
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5 votes

Quadratic variation of Brownian motion and almost-sure convergence

If the diameter of the $n$-th partition converges (to zero) fast enough, namely if it is of order less than $1/\log(n)$, then the quadratic variation converges almost surely. If not, then not in ...
AJW's user avatar
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5 votes
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Stopped local martingale as a martingale

Let $(\sigma_n)_{n \in \mathbb{N}}$ be a localizing sequence of the local martingale $M$, i.e. an increasing sequence of stopping times such that $\sigma_n \to \infty$ and $(M_{t \wedge \sigma_n})_{t \...
saz's user avatar
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5 votes
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Quadratic variation of ito integral

Let $X,Y$ be semimartingales and $\xi$ be an $X$-integrable process. Then, $$\left[\int\xi\,dX,Y\right] = \int\xi\,d[X,Y]$$ and $$\left[\int\xi\,dX\right]=\int\xi^2\,d[X]$$ where $[\cdot, \cdot ]$ ...
W. Volante's user avatar
  • 2,254
5 votes
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A function with cubic variation

The paths of fractional Brownian motion with Hurst exponent $H$ admit (almost surely) non-vanishing $\frac{1}{H}$-variation. Since $H \in (0,1)$, then you can find paths with your desired property for ...
Jose Avilez's user avatar
4 votes
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Basic proof of $d\langle X\cdot M , Y\cdot N\rangle=XY\,d\langle M ,N\rangle$

Hint: Show (or recall) that the quadratic covariation satisfies $$\langle M,N \rangle_t = \frac{1}{4} \big( \langle M+N \rangle_t - \langle M-N \rangle_t \big) \tag{1}$$ for any two martingales $M,N \...
saz's user avatar
  • 120k
4 votes
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How does the general stochastic integral generalize the Brownian motion stochastic integral?

No, it's not exactly obvious. Let's prove the following theorem. Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion and $(K_t)_{t \geq 0}$ a progressively measurable process such that $\...
saz's user avatar
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4 votes
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Is a continuous process with finite nonzero quadratic variation a semimartingale?

No, it's not true. There are continuous processes with finite quadratic variation which are not semimartingales: Let $(B_t)_{t \geq 0}$ be a Brownian motion and consider $X_t := |B_t|^{\alpha}$, $ t \...
saz's user avatar
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4 votes
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Quadratic variation of $dX_t$ = $\mu \, dt + \sigma \, dB_t$

Yes, your approach is okay. Not sure why you need (1)-(3) because $\langle X,X \rangle_t = \langle X \rangle_t$ is really just the definition. How to get from (6) to (7)? If you expand the squre in (...
saz's user avatar
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4 votes
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Positive Quadratic Variation implies infinite total variation

This is a general property of continuous functions. Lemma: Suppose $f\not\equiv0$ is continuous on $[a,b]$ and of finite variation, that is $V_f[a,b]=\sup_{P}\sum^{n_p}_{k=1}|f(x_k)-f(x_{k-1})|<\...
Mittens's user avatar
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4 votes
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Bounded Square-Integrable Martingale

Since $(X_n)_{n \in \mathbb{N}}$ is a square-integrable martingale, we know from Doob's decompositon that $X_n^2- \langle X \rangle_n$ is a martingale, where $$\langle X \rangle_n := \sum_{j=1}^n \...
saz's user avatar
  • 120k
4 votes

Quadratic variation of a product

To keep track of things more easly, write $X_t=e^{3Bt}$ and $Y_t=\int_0^t B_s ds$. Then, as you note, $$ dX_t = 3X_t dB_t. $$ Therefore $$ d(X_tY_t) = 3X_tY_t dB_t + V_t, $$ where $V$ is an ...
John Dawkins's user avatar
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3 votes

Quadratic variation of the sum

Two complements to the answer of @BehrouzMaleki: Because $X$ and $Y$ are independent, the product $X_tY_t$ is also a martingale; the semimartingale product rule then tells us that $[X,Y]_t$ is a ...
John Dawkins's user avatar
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3 votes
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Showing a Quadratic Variation is increasing and continuous with respect to time

In fact (and maybe this is what you are saying you want to show), the quadratic variation of $M$ is deterministic; more precisely, $\langle M\rangle_t = E[M^2_t]$ a.s. Here is the outline of a proof. ...
John Dawkins's user avatar
  • 26.1k
3 votes

Quadratic variation of the Ornstein-Uhlenbeck process

You can use it, but first you need to use integration by parts in the integral of $$X_t = \sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s. \tag{$\triangle$}$$ Here, you can regard the integral as a ...
user3417650's user avatar
3 votes

Proof that $p$-th total variation of a brownian motion is $0$ while $p>2$

Note that $2$-variation is not the same as quadratic variation. For the quadratic variation you take the limit as the partition gets finer, whereas for $p$-variation you take the supremum over all ...
EagleOwl's user avatar
3 votes
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$\langle X,Y \rangle$ is the unique process of the form $Z = A - B$ s.t. $XY - Z$ is a martingale

If $X$ and $Y$ are semimartingales you can use Ito's Lemma to write $$X_tY_t = X_0Y_0 + \int\limits_0^t X_{s-}dY_s + \int\limits_0^t Y_{s-}dX_s + \langle X,Y\rangle_t,$$ so you can write the ...
K.Doe's user avatar
  • 116
3 votes
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Quadratic Variation of Brownian Motion Cubed

Your answer is correct. What your friend was attempting was, seemingly to me at least, to use the fact that if $X_t$ is a local martingale, then $X_t^2-\left<X\right>_t$ is also a local ...
hypernova's user avatar
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3 votes
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Quadratic variation and brackets, whats the difference?

The usual notation for quadratic variation is $[\cdot]$. You may also define, for a local martingale $M$, a process that is sometimes called the conditional quadratic variation: $\langle M,M\rangle$ ...
AddSup's user avatar
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3 votes
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On the stochastic exponential in stochastic calculus

Let's consider the first and the second term on the right-hand side of your equation separately: First term: If $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are semimartingales, then $$\int_0^T f(t) \,...
saz's user avatar
  • 120k
3 votes
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Example of a continuous function with discontinuous quadratic variation

An example is given in Appendix 5.2 of Coquet, Jakubowski, Mémin, and Słominski, Natural decomposition of processes and weak Dirichlet processes, pp. 81–116, Springer, Berlin, Heidelberg, 2006. For ...
Jose Avilez's user avatar

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