# Tag Info

Accepted

• 120k

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 $\... • 120k 5 votes Accepted ### Potential Local Martingale property derived from its quadratic variation "$ M $may not be$ o(t) $", the following is an example. Let$ B=\{B_t,t\ge 0\} $be the Brownian motion with$\langle B\rangle_t =t $and \begin{equation*} M_t=B_{t^2}, \qquad t\ge 0. \... • 5,316 5 votes ### Probability of stopping time being finite. Using the hint: $$\mathbb{E}(X_{t\wedge\tau_a})=a\mathbb{P}(\tau_a\leq t)+\mathbb{E}(X_t1_{\tau_a>t}).$$ Now$0\leq X_t1_{\tau_a>t}\leq X_t\to0$almost surely as$t\to\infty$. Additionally$...
• 1,033
Accepted

To answer your first question, the result was published in 1972 by P.A. Meyer in the "Lecture Notes in Mathematics Series". The title is Martingales and Stochastic Integrals I, and the result is ...
• 1,552
Accepted

• 120k

### If $M$ is a local martingale and $τ:=\inf\left\{t\ge0:\left|M_t\right|\ge\varepsilon\right\}$, then $M^τ$ is a martingale

Let $s < t$. We have to show that $E[M_t^\tau \mid \mathcal{F}_s] = M_s^\tau$ almost surely. For each $n$ we know, as you mentioned, that $(M^\tau)^{\sigma_n} = M^{\tau \wedge \sigma_n}$ is a ...
• 97.8k

### Is an $L^2$-bounded continuous local martingale already a strict martingale?

It needn't be the case that an $L^2$-bounded continuous local martingale is a true martingale. Below I outline an example that can be found in "Diffusions, Markov Processes and Martingales: Volume 2" ...
• 19.7k
Accepted

### Is there a process that is not square-integrable but still gives rise to a martingale when integrated w.r.t. Brownian motion?

Yes, that's possible. Example Let $Y$ be a random variable which is independent from $(B_t)_{t \geq 0}$ and satisfies $$\mathbb{E}(|Y|) < \infty \quad \text{and} \quad \mathbb{E}(Y^2)=\infty$$ (e....
• 120k
Accepted

### If $\mathbb{E}[\underset{0 \leq s \leq t}{\sup} \rvert X_s \rvert] < \infty$ for $X_t$ a local martingale, is $X_t$ a proper martingale?

$X_t$ is indeed a martingale. We don't need anything as complicated as Burkholder-Davis-Gundy to see this. Instead, let's go back to the definitions. Let $\tau_n$ be a sequence of stopping times such ...
• 19.7k

### Compare Doob décomposition for stochastic process $(X_t)_{t\geq 0}$ and $(X_n)_{n\in\mathbb N}$

It seems that two distinct notions are being conflated into one in your question. A non-unique semimartingale decomposition, i.e., decomposition into a local martingale and a finite variation process;...
• 804
Accepted

### Local martingale property for stochastic integrals.

This is not exactly an elementary proof. Assuming that you are well versed in the properties of the Ito integral for the class of integrands $\{f: E[\int_a^b f^2(s)\, ds] < \infty \},$ and know ...
• 1,314
Accepted

### Quotient of continuous local martingale with quadratic variation

I believe that you need the further condition that $\langle M,M \rangle_\infty = \infty$ a.s., otherwise you can take $M$ to be something like Brownian motion stopped at $t=1$. Then you can use the ...
• 13.5k
Accepted

### Interpretation of Gisarnovs theorem

Since we also have that $[M,X]=[\hat{M},X]$ are the same for all $X$ both under $\mathbb{P}$ and $\mathbb{Q}$ we know that $M$ and $\hat{M}$ are the same up to a constant. Therefore $X=\hat{M}+B$ is ...
• 1,835
Accepted

### The expected squared increment of a continuous local martingale

While it certain isn't true that $M_{t_{i+1}}^2 - M_{t_i}^2 = (M_{t_{i+1}} - M_{t_i})^2$, their expectations are the same. To see this, first assume that $M$ is a martingale, and use the tower ...
• 12.7k
Accepted

• 1,410