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4

As the pointwise limit of the sequence $(X_k)_{k\leqslant-n}$ which is measurable with respect to $\mathcal F_{-n}$, the random variable $X_{-\infty}$ is measurable with respect to $\mathcal F_{-n}$. This holds for every $n$, hence $X_{-\infty}$ is measurable with respect to $\bigcap\limits_n\mathcal F_{-n}=$ $______$.


3

A continuous martingale is a Brownian motion if and only if its quadratic variation over each interval $[0, t]$ is equal to $t$. Thus, if $M$ is a continuous martingale and $\langle M\rangle_t=u(t)$ for some deterministic function $u$, then $M$ is a time change of a Brownian motion in the sense that $M_t=B_{u(t)}$ for every $t$, where $B$ is a Brownian ...


3

Let me answer questions 2) and 3) before dealing with 1). For second question: Martingales are tools which we can use to reduce complicated computations (example computing conditional probabilities), into appealing to a plethora of results about Martingales which you learnt for example in Stochastic process courses. In the research world, probabilists try ...


3

Regarding $\mathbb{E}(X_j \mid \mathcal{G}_n) = \mathbb{E}(X_j \mid S_n)$. Note that $$\mathcal{G}_n := \sigma(S_n,S_{n+1},\ldots,) = \sigma(S_n,X_{n+1},X_{n+2},\ldots).$$ Since, by assumption, $(X_{n+1},X_{n+2},\ldots)$ are independent from $(X_n,S_n)$ you can apply the mentioned theorem. Regarding $\mathbb{E}(X_j \mid S_n) = \mathbb{E}(X_1 \mid S_n)$: You ...


2

Your friend's answer in case 1. is wrong because, according to this "definition", the independence of $\mathfrak{F}_1$ and $\mathfrak{F}_2$ would depend on the choice of $F_1 \in \mathfrak{F}_1$ and $F_2 \in \mathfrak{F}_2$. Thus, this "definition" corresponds to some sort of independence of $\mathfrak{F}_1$ and $\mathfrak{F}_2$ related to $(F_1,F_2)$ in ...


2

Note that the equality $$\mathbb{E}(\mathbb{E}(X \mid Y)) = X$$ does not hold. Instead it should read $$\mathbb{E}(\mathbb{E}(X \mid Y))= \mathbb{E}X.$$ In fact, this holds not only for $\sigma$-algebras generated by a random variable, but for any $\sigma$-algebra, i.e. $$\mathbb{E}(\mathbb{E}(X \mid \mathcal{F})) = \mathbb{E}X$$ for any ...


2

Writing $$W_t^3 = ((W_t-W_s)+W_s)^3 = (W_t-W_s)^3+ 3 W_s (W_t-W_s)^2 + 3 W_s^2 (W_t-W_s) + W_s^3$$ we find using the independence of the increments $$\begin{align*} \mathbb{E}(W_t \mid \mathcal{F}_s) &= \underbrace{\mathbb{E}((W_t-W_s)^3)}_{\mathbb{E}(W_{t-s}^3)=0} + 3W_s \underbrace{\mathbb{E}((W_t-W_s)^2}_{\mathbb{E}(W_{t-s}^2)=t-s} + 3W_s ...


2

The random variable $\color{red}{Z_n}$ is measurable with respect to the $\sigma$-algebra $\sigma(Z_0,\dots,Z_{n-1},\color{red}{Z_n})$. If $\mathcal G$ is a $\sigma$-algebra and $X$ is $\mathcal G$-measurable then $\mathbb E[X\mid\mathcal G]=X$.


2

You simply have that $$Z_n=E[Z_n|Z_n]$$ or in other words, given that at time $n$ you know $Z_n$ then the expected value of the random variable $Z_n$ will be $Z_n$. Note that the expectation on the right hand side is a random variable and not a number! Obviously more knowledge does not change that so that also $$Z_n=E[Z_n|Z_0,Z_1,\ldots,Z_n]$$ The best (and ...


2

From your first equality, \begin{align*} E[(M_T - M_S)^2] &= E[E[(M_T - M_S)^2 | \mathcal{F}_S]] \\ &= E[E[M_T^2 - M_S^2|\mathcal{F}_S]] \\ &= E[M_T^2]-E[M_S^2]. \end{align*} You should also mention the martingales are right-continuous so that we may apply optional stopping.


2

There are 3 things to check: $X_n$ is mesurable with respect to $F_n$: This is ok by definition of $F$. $X_n$ is integrable: as $X_n$ is bounded, so that is ok. (the big part) $E[X_{n+1}|X_n] = X_n$: conditionally to $X_n$, there are two options: with probability 1/2, $X_{n+1} = 2X_n$ with probability 1/2, $X_{n+1} = 0$ hence $$E[X_{n+1}|X_n] = ...


2

In general, 3 does not imply 2 Suppose $g$ is $\mathcal{F}$-measruable and $h=g$ almost surely, it does not imply $h$ is also $\mathcal{F}$-measurable. Take a set $A$ which is contained in $B \in \mathcal{F}$ with $P(B) = 1$, but $A$ is not itself in $\mathcal{F}$. By definition, $h = 1_{A}$ is alomst surely equal to the constant variable $g \equiv 0$. ...


2

Counterexample 1: Let $\xi_j$ be independent identically distributed random variables such that $$\mathbb{P}(\xi_j = 0) = \mathbb{P}(\xi_j =2)=\frac{1}{2}.$$ Then $$X_n := \prod_{j=1}^n \xi_j$$ is a non-negative martingale satisfying $\mathbb{E}(X_n)=1$. On the other hand, it is not difficult to see that $X_n \to 0$ almost surely. As $$\mathbb{E}(X_{\infty} ...


1

Lets assume that we are in the regime in which $V_t > 0$ for every $t>0$. Consider the sequence of stopping times $$\tau_n = \inf \left\{ t: V_t > n \right\}, \quad n \in \mathbf{N},$$ then, as you mentioned before, if we can show that $\tau_n \to \infty$ as $n \to \infty$ in probability the result follows by standard localization: the expectation ...


1

Hi as mentioned in my comment the proof given does not allows us to conclude by localization. I have finally found a rigorous and elementary proof of the fact that a CIR process possesses moments of all orders which is necessary to get the result (order 1 is enough for our need) . First two obersvations : by Yamada-Watanabe's theorem that the CIR's SDE ...


1

If you compute the derivative of $Y_t$ using Ito's lemma, you have $Y'_t=s'(X_t) dX_t + 0.5 s''(X_t)<dX_t> = [s'(X_t) b X_t+ 0.5 s''(X_t) X_t^2\sigma^2] dt + s'(X_t) \sigma X_t dW_t$ $Y_t$ is a martingale if and only if its drift, $s'(X_t) b X_t+ 0.5 s''(X_t) X_t^2\sigma^2$, is $0$. Thus, unless $X_0=0$, $s(x)$ must be such that $$bs'(x) + 0.5 ...


1

For fixed $x \in \mathbb{R}$, we choose $f(x) := e^{\imath \, x \xi}$. By assumption, $$(t,\omega) \mapsto e^{\imath \, \xi W_t(\omega)} - \frac{\xi^2}{2} \int_0^t e^{\imath \, \xi W_r(\omega)} \, dr$$ is a martingale, i.e. $$\mathbb{E}\left( e^{\imath \, \xi W_t} + \frac{\xi^2}{2} \int_0^t e^{\imath \, \xi W_r} \, dr \mid \mathcal{B}_s \right) = ...


1

This follows from the following theorem: Theorem: Let $(M_t,\mathcal{F}_t)_{t \geq 0}$ be a martingale such that the sample paths are (almost surely) continuous and of bounded variation. Then $M_t = M_0$ almost surely. (For a proof see e.g. René Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes.) Since $$M_t := ...


1

Multiplying by a constant doesn't change the Martingale property: for $t>s$, $$ E[M_t|M_s]=E[-B_t|M_s]=E[-B_t|B_s]=-E[B_t|B_s]=-B_s=M_s. $$


1

If $t>s$, then ${\rm E}[S_t\mid \mathcal{F}_s]-S_s\geq 0$ a.s. by the sub-martingale property. Since this random variable also has mean zero, we may conclude that it is zero a.s.


1

Hints Recall that the independence of two stochastic processes $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$ is equivalent to the independence of the corresponding canonical $\sigma$-algebras $\mathcal{F}_{\infty}^M$ and $\mathcal{F}_{\infty}^N$, $$\mathcal{F}_{\infty}^M := \sigma(M_s; s \geq 0).$$ Let $f$ be an $\mathcal{F}_t^M$-adapted process such that ...


1

Hint: consider the martingales $$ M(t) = \int 1_{[0,t]}(s) a(s) dB(s) $$with $a\in L^2(R^+)$. Then $$ \langle M\rangle(t) = \int 1_{[0,t]}(s) a(s)^2 ds $$


1

The above statement is not true. Semimartingales are a quite general class of processes containing processes of both bounded and unbounded variation. For example a constant process $(X_t) \equiv 0 $ is a semimartingale.


1

A martingale is not a process that has zero mean for all time. The defining feature of a martingale process $M_t$ is: $E(M_t|M_1,...,M_{t-1})=M_{t-1}$ Therefore, any stationary process will violate this property. Take a stationary gaussian process, which is about as simple as you can get: Let $X_t \sim \mathcal{N}(0,1)$ Therefore, ...



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