# Tag Info

Accepted

• 113k
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. \...
• 3,745

• 113k
Accepted

### 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) &= \...
• 113k
Accepted

• 113k

### 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 ...
• 90.7k
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....
• 113k
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 ...
• 17.5k
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 ...
• 1,134
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 ...
• 9,057
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 ...
• 1,715
Accepted

### local martingale remains local martingale when lowering the localizing stopping times

This statement is correct and your proof works. We don't need $S_k \le T_k$ to say that $(X^{T_k})^S_k$ is a martingale. A slightly more general statement is that if $(S_k) \rightarrow \infty$ is a ...
• 9,057
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 ...
• 10.3k
Accepted

• 113k
### Showing $E(X_T ^2)\le E([X]_T)$ for bounded stopping times $T$
Hints: Let $T$ be a bounded stopping time. Fix a localizing sequence $(\sigma_n)_{n \in \mathbb{N}}$ of stopping times. Sinc $(X_t)_{t \geq 0}$ has continuous sample paths, we can choose the sequence ...