58 votes
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Angle bracket and sharp bracket for discontinuous processes

Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be an (càdlàg) $L^2$-martingale, i.e. a martingale which satisfies $$\sup_{t < \infty} \mathbb{E}(X_t^2)<\infty.$$ Then it follows from the Doob-Meyer ...
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10 votes
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What is the quadratic variation of compensated poisson process?

Hints: Recall: $\Delta(f(s)-g(s)) = \Delta(f(s))-\Delta(g(s))$ for any two (deterministic) functions $f,g$. Recall: $\Delta(f(s))=0$ if $f$ is continuous at $s$. Using step 1,2 show that $$(\Delta (...
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9 votes
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Quadratic variation of the Ornstein-Uhlenbeck process

You cannot apply the formula $$[G \bullet M]_t = \int_0^t G_s^2 \, d[M]_s \tag{1}$$ because the Ornstein-Uhlenbeck process $X$ is not of the form $$X_t = (G \bullet B)_t,$$ but of the form $$X_t = ...
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8 votes
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quadratic variations of Brownian motion squared

There are several definitions for the quadratic variation: For a "nice" process $(X_t)_{t \geq 0}$ ("nice" means semimartingale), the quadratic variation is defined by $$[X]_t := X_t^2-X_0^2 -2 \...
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7 votes
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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 ...
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7 votes

Quadratic variation of Brownian motion and almost-sure convergence

I would like to comment on your 2nd question. The independence of $\langle M,M\rangle_T$ ($[W,W](T)$) of the partition you chose. Let's see the fundamental steps from the beginning : Let $M$ be a ...
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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 ...
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6 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,\...
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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 $$x\geq 1$$...
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6 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 ...
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6 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 \...
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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 ...
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5 votes
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Quadratic variation of $X_{t} = tB_{t}$?

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 ...
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5 votes

Quadratic variation of $X_{t} = tB_{t}$?

$$\mathrm dX_t=\color{red}{t}\,\mathrm d\color{blue}{B}_t+B_t\,\mathrm dt\implies\mathrm d\langle X\rangle_t=\color{red}{t}^2\,\mathrm d\langle\color{blue}{B}\rangle_t=t^2\,\mathrm dt\implies\langle X\...
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5 votes
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Stochastic Exponential - Protter

Any càdlàg function $f: [0,T] \to \mathbb{R}$ has at most finitely many jumps with jump size $>\epsilon$ for any (fixed) $\epsilon>0$, see e.g. this answer. The estimate "$\leq [X,X]_t$&...
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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 ...
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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 $\...
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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 $[\cdot, \cdot ]$ ...
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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 ...
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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 ...
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4 votes

Quadratic Variation of Brownian Motion

The book Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch contains a (detailed) proof of this fact.
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  • 113k
4 votes
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Quadratic variation of semi-martingale

The trouble is that there are two types of brackets, $\langle X \rangle_t$ (which I call angle bracket) and $[X]_t$ (sharp/square bracket). In this particular case they do not coincide because $(X_t)_{...
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4 votes
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Proof that $p$-th total variation of a brownian motion is $0$ while $p>2$

It is widely known that for $p=2$ the quadratic variation $$S_{\Pi} := \sum_{t_i \in \Pi} |B_{t_{i+1}}-B_{t_i}|^2$$ converges to $T$ in $L^2$ as $|\Pi| \to 0$. Here $\Pi = \{0=t_0<\ldots< t_n \...
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4 votes
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Proving a statement in quadratic variation that ${\langle X \rangle}^{\tau} = \langle X^{\tau} \rangle$

Suppose that $(X_t)_{t \geq 0}$ is a continuous martingale. We know that the compensator $(\langle X \rangle_t)_{t \geq 0}$ is the unique increasing predictable process such that $$X_t^2 - \langle X \...
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4 votes

Computing quadratic variation and criteria for Brownian motion

Your answers to (a) and (b) are essentially correct; just be more demanding about your notation. For instance, explicitly specifying your limits of integration when using integration by parts, and not ...
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4 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....
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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 \...
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4 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 \...
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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 $\...
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4 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) &= \...
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