7 votes
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Uniformly integrable local martingale

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $...
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  • 17.5k
6 votes

Relative entropy for martingale measures

$M_e$ contains $P$, because $P\in M_a$ and $\frac{dP}{dP}=1$ and therefore $H[P|P]=0.$ I will show that $M_e$ contains more than just $P$ in the case where $S$ is a Brownian motion. In that case, ...
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  • 685
6 votes
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Show local martingale

Solution 1: Recall the following two statements. Lemma 1: Let $(Y_t)_{t \geq 0}$ be a supermartingale and $f$ an increasing concave function, then $(f(Y_t))_{t \geq 0}$ is a supermartingale. Lemma 2: ...
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  • 113k
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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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. \...
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  • 3,745
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 $...
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4 votes
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About local martingales..

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 ...
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  • 1,432
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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  • 113k
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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  • 113k
4 votes
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An example to be a local martingale but not a martingale

Take some random variable $Y$ which is not integrable (e.g. Cauchy distributed) and which is independent of $(W_t)_{t \geq 0}$. Define $\varphi_s(\omega):=Y(\omega)$, then $$\int_0^t \varphi_s \, dW_s ...
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  • 113k
4 votes
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Inequality on expectation of exponential martingale

Be careful. Uniform integrability is not sufficient to apply optional stopping theorem to local martingales. Counter-example: if $R$ is a Bessel process of dimension 3 starting at 1, the local ...
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3 votes

Exponential martingales and changes of measure

You are right to be suspicious. The recipe you describe specifies a finitely additive set function $Q$ on $\cup_t\mathcal F_t$ such that $Q|_{\mathcal F_t} = M_t\cdot P|_{\mathcal F_t}$ for each $t\ge ...
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  • 22.9k
3 votes

Uniformly $L^p$ continuous local martingale

As the following example shows, $(X_t)_{t \geq 0}$ is, in general, not uniformly integrable: Let $(X_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Then $$\mathbb{E}(|X_t| 1_{\{|X_t| \geq R\}}...
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  • 113k
3 votes

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 ...
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3 votes
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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....
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  • 113k
3 votes
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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 ...
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  • 17.5k
3 votes
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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 ...
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  • 1,134
3 votes
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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 ...
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  • 9,057
3 votes
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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 ...
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  • 1,715
3 votes
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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 ...
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  • 9,057
3 votes
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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 ...
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  • 10.3k
3 votes
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Show that a local martingale is a true martingale if and only if it is a process of class DL

Suppose that the $M$ is a local martingale with respect to the filtration $\{ {\mathcal F}_s\}$ and the stopping times $S_n \uparrow \infty$. Then for $s<t$, we have $$\forall n, \quad E[M_{t \...
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2 votes
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Question regarding local martingales.

In answer to the question in your comment on my comment: Let $X$ be a martingale, $T$ a stopping time, and $Y_t=X_{t\wedge T}$ the stopped process. By Doob, for $0\le s<t$, $E[Y_t\,|\,\mathcal F_{s\...
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  • 22.9k
2 votes
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continuous local martingale brownian motion

Ito's lemma is one way to prove it. Using the definition of $f_{1-t}$, we have $$X_t = f_{1-t}(B_t) = \frac{1}{\sqrt{2\pi (1-t)}} \exp\left\{\frac{-B_t^2}{2(1-t)}\right \}:=F(t,B_t),$$ and according ...
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  • 1,350
2 votes

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" ...
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  • 17.5k
2 votes

Is a local continuous martingale square integrable.

If you are asking whether a continuous local martingale is square integrable process the answer is NO, as the counterexample of @Jochen demonstrated. But if you are asking whether a continuous local ...
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  • 2,268
2 votes
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Show that $X \in \mathcal{M}^{loc}_c$ implies: $X$ is locally square summable and locally bounded.

Hints: Since $X$ is a local martingale, there exists a sequence of stopping times $(\tau_k)_{k \in \mathbb{N}}$ such that $\tau_k \uparrow \infty$ and $(X_{t \wedge \tau_k})_{t \geq 0}$ is a ...
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  • 113k
2 votes
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Show that $M-M_0$ is an $L^2$ bounded martingale if $E([M]_\infty)<\infty$.

WLOG $M_0=0$ (otherwise replace $M_t$ by $M_t-M_0$). By the very definition of the quadratic variation, $$N_t := M_{t \wedge T_n}^2-[M]_{t \wedge T_n}$$ is a martingale. In particular, $$\mathbb{E}(...
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  • 113k
2 votes
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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 ...
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