4 votes
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If $\frac{dX_{t}}{X_{t}} = dL_{t}$, where $L_{t}$ is a local martingale, then is $X_{t}$ a local martingale?

There is essentially only one process $X$ satisfying $dX_t = X_t dL_t$, namely $X_t := C\exp(L_t - \frac 12 \langle L,L\rangle_t)$ where $C \in \mathbb{R}$ can be arbitrary. More precisely, if $Y$ is ...
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  • 8,767
3 votes
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Lévy's characterization of Brownian motion: right-continuous processes

Cool question. Proposition Let $(X_u)_{u}$ be right-continuous martingale with $X_0=0$, such that $(X^2_u-u)_u,(X_u^3-3uX_u)_u,(X_u^4-6uX_u^2+3u^2)_u$ are martingales. Then for every integer $M \ge 1$,...
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  • 13.5k
3 votes

Two-dimensional random walk

Fleshing my comment out into an answer: this is not an extension of the 1D problem; it is, in fact, precisely the same as the 1D problem. You are considering a process $(R_n, S_n)$, where $R_n, S_n$ ...
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3 votes
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Martingale and independence

Your idea is correct. Is it true that $\mathcal{H}_u=\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$ Of course. Indeed, $\sigma(X_v,0 \leq v \leq u)\subset\sigma(X_v,Y_v,0 \leq ...
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  • 3,429
2 votes
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Showing that a martingale is constantly zero

Consider the Martingale $M_k:=W(Y_{k \wedge T})$, where $T$ is the first time $Y$ hits $i_0$, and $Y_0=i$. From $E[(M_k-M_0)^2]<\infty$, the Martingale is bounded in $L^2$, so it is uniformly ...
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  • 13.5k
1 vote
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Does $\|f_n-f\|_1 \to 0$ where $f_n$ is the mean of $f$ over a uniform partition in $n$-bins of the domain?

Hints: If $f$ is continuous then $f_n (t) \to f(t)$ for each $t$ and $(f_n)$ is uniformly integrable. This implies that $f_n \to f$ in $L^{1}$. For the general case choose a continuous function $g$ ...
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  • 2,789
1 vote

Brownian motion and uniform distribution

By translation, we may assume the starting point $x$ is $0$. Next, observe that the distribution of $W_T$ is rotation invariant, i.e, if $A$ is a Borel set on the sphere $\partial B(0,\delta)$ and $M$ ...
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1 vote
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Can martingale be finite

Yes, finite a martingale is a useful thing. If your particular textbook does not cover the case of a finite martingale $(X_1,X_2,\dots,X_n)$, then extend it by repeating the last value: $(X_1,X_2,\...
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  • 97.6k
1 vote
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Sum of r.v's are martingales then $E[X_iX_j]=0$

Hint: $\mathbb{E}(X_{n+1} X_i) = \mathbb{E}(\mathbb{E}(X_{n+1}X_i|\sigma(X_1,...,X_n))$ Where $1\leq i\leq n$. Then, $$\mathbb{E}(\mathbb{E}(X_{n+1}X_i|\sigma(X_1,...,X_n))=\mathbb{E}(X_{i}\mathbb{E}(...
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1 vote

Brownian motion and harmonic function

A harmonic function is one for which the mean value property holds; i.e. the value of $f$ at the centre of any ball contained in the domain is the mean of $f$ on its surface. In the language of your ...
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  • 714
1 vote
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How do I construct an example of a martingale which is bounded in $L^1$ but not in $L^2$

Simply take $M_n=\mathbb E\left[X\mid\mathcal F_n\right]$, where $\left(\mathcal F_n\right)_{n\geqslant 1}$ is a filtration, $X$ is non-negative, integrable, $\sigma\left(\bigcup_{n\geqslant 1}\...
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1 vote
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When random walk is a martingale

The typical definition of a martingale is that $$\mathbb E[X_{n+1} \mid \mathcal F_n] = X_n$$ where $X_n$ is adapted to the filtration $\mathcal F_n$. This is equivalent to $$\mathbb E[X_{n+1} - X_n \...
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1 vote
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Proof checking on the usage of the optional martingale theorem

Ok, I look at the book to see if I can see something useful, and indeed I find it. The exercise seems to rely in the definition of stopping time given in the book, what says that $T$ is a stopping ...
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