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3

The bound 2 is arbitrary in the sense that it can be replaced by any real number larger than 1 - the real requirement is that you set some bound on the summed conditional variances of the array. Note that you already assume that $V_{n,nt} \xrightarrow{P} t$ - what does this condition entail on the variances? One reason that this is allowed in general is that ...


3

If $M = \int h dX$ where $X$ is a continuous square integrable martingale, you don't need $E\int_0^\infty h_s^2d\langle X\rangle_s < \infty$ for $M$ to be a martingale, you just need $E\int_0^t h_s^2d\langle X\rangle_s < \infty$ for all $t$. You did show this, and hence your $M$ is a martingale and you can say $E M_t = E M_0$ to solve your problem.


2

These still aren't the most general definitions for martingales. The most general definition I know of is: Let $I$ be a poset. A Family $F$ of sigma-algebras is said to be a filtration if for all $s, t \in I$ with $s \le t$ the inclusion $F_s \subset F_t$ holds. A family $X_i$ of (quasi)integrable random variables is called a martingale with respect to $F$, ...


2

Note, for this all you need is that $\tau$ is a stopping time, $c_n$ are non increasing, $X_n \geq 0$ and in $L^1$. $$ c_nX_n = \sum^n_{i=1} [c_i X_i - c_{i-1}X_{i-1}] \leq \sum^n_{i=1} c_i (X_i-X_{i-1}) = \sum^\infty_{i=1} 1_{i \leq n}\ c_i(X_i - X_{i-1}) . $$ Now substitute $\tau \wedge n$ for $n$. $$ c_{\tau \wedge n}X_{\tau \wedge n} \leq ...


2

You just need to look at more stopping times. For any $t\geq 0$, you have $${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$ This gives $$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) ...


2

You can generalize your derivation from: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ to: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$ where $t > s$. As you mentioned, using the fact ...


1

Let $\epsilon > 0$ and consider the stopping time $$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$ Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$. Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write $$E[X_{\tau_\epsilon} ; \tau_0 < T] ...


1

Yes. To show it, first see that it is sufficient to prove it in the case when $X_n$ takes the values $0$ or $1$. (Because then you get simple functions, and from there you get everything.) So consider the collection of sets $\theta_n$ which are those sets for which its indicator function is $h_n(Y_1,\dots,y_N)$ for some Borel measurable $h_n:\mathbb R^n ...


1

For reference what you have is an exponential martingale. For A) I would just go about it that way $$ E[M_\alpha(t)]= e^{-\alpha^2/t}E[e^{\alpha W_t}]=e^{-\alpha^2/t}\int_Re^{\alpha \sqrt tZ} \phi(z) dz$$ where $\phi(.)$ is the normal pdf, compute the integral, done. B) is a generalization of the preceding steps. Second question: $$E[M_\alpha(t)^p]= ...


1

No, not in general. For a discrete time example, let $S_n$ be a simple random walk, and let $Z = 0$ or $1$ with probability $1/2$ and independent of $S_n$. Set $X_n = Z S_n$; this is a martingale. Intuitively, $X_n$ corresponds to the following gambling game: flip a coin at time 0. If it comes up tails, bet $1 on coin flips forever. If it comes up ...


1

Processes that are not right-continuous are a gigantic pain to deal with. Most likely they include right-continuous in their definition because they always intend to work under that assumption, and don't want to have to keep writing it. As with any mathematical term, there doesn't have to be a universal "correct" definition of martingale; authors are free to ...


1

Define $Y_{n,m}=\sum_{k=1}^m\left[t_{n,k}-\mathbb{E}(t_{n,k}\mid\mathcal{F}_{n,k-1})\right]$. $Y_{n,m}$ is a MTG because $$\mathbb{E}[Y_{n,m}\mid\mathcal{F}_{n,m-1}]=\sum_{k=1}^{m-1}\left[t_{n,k}-\mathbb{E}(t_{n,k}\mid\mathcal{F}_{n,k-1})\right]=Y_{n,m-1}$$ and $$t_{n,m}-\mathbb{E}(t_{n,m}\mid\mathcal{F}_{n,m-1})=Y_{n,m}-Y_{n,m-1}$$ By Theorem 5.4.6 ...


1

By assumption, $$P[|H\cdot M|^*_t\ge c]\le P[|H\cdot U|^*_t\ge c/2]+P[|H\cdot V|^*_t\ge c/2]\le 18/c(\|U_t\|_1+\|V_t\|_1)$$ The $M=U-V$ decomposition as described by Writing a martingale as the difference of two non-negative martingales satisfies $\|M_t\|_1=\|U_t\|_1+\|V_t\|_1$.


1

To add onto nullUser's response. Of course $\int_0^t 1 dW_s=W_t$ is a martingale right? Well: $$ \int_0^{\infty} 1 ds=\infty$$ This is just to give a solid counterexample, the rest was explained in his post.



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