Tag Info

Hot answers tagged

2

Every continuous local martingale is a continuous time change of a Brownian Motion in a possibly extended probability space. So, modulo some technicalities, all the properties obtained from integration w.r.t. Brownian Motion translates, one way or another, to properties when integrating w.r.t. to continuous time local martingales.


2

First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write $$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$ (I don't ...


2

Let $f(x) := x^a$ for some fixed $a>0$. Then $$f'(x) = a x^{a-1} \qquad f''(x) = a (a-1) x^{a-2}.$$ Since by Itô's formula $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s+ \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s$$ we get $$\begin{align*} Y_t - Y_0 &= f(X_t)-f(X_0) \\ &= a \int_0^t X_s^{a-1} \,d X_s + \frac{1}{2} a (a-1) \int_0^t ...


2

For every $n$, let $X_n=\max_{n\leq s\leq n+1}\left|B_{s}-B_{n+1}\right|$, then $(X_n)$ is identically distributed. Furthermore, $X_1$ is square integrable hence $$\lim\limits_{n\to\infty}\frac1{n^2}E(X_n^2)=\lim\limits_{n\to\infty}\frac1{n^2}E(X_1^2)=0.$$ Note that the quantity $\lim\limits_{\color{red}{s\to\infty}}\frac1{n^2}E(X_n^2)$ is undefined since ...


1

$$ Y_t = \mathrm{e}^{2B_t-\alpha t} $$ using Ito $$ dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t $$ thus martingale requires a driftless SDE thus $\alpha = 2$. Was this the way you performed it?


1

I will try to give an intuitive explanation. Suppose that $X_1,X_2,...$ are independent and identically distributed random variables with zero means and unit variances. Define $$ \zeta_n(t)=\frac1{\sqrt n}\sum_{k=1}^{\lfloor nt\rfloor}X_k $$ for $t\in[0,1]$ and $n\ge1$, where $\lfloor\cdot\rfloor$ is the floor function. It is known that $\zeta_n$ converges ...


1

The two sets $\{T_a<t\}$ and $\{M(t)>a\}$ are equal to up to a null set, because the event $M(t)=a$ has probability zero. Indeed, both $T_a$ and $M(t)$ are continuously distributed random variables: Distribution of hitting time of line by Brownian motion.


1

This is a consequence from the Clarke-Ocone Theorem, and uses Malliavin derivative. See also the Clarke-Ocone formula paragraphe here. If you want a technical reference, see this introductory course, mainly theorem 1 p. 18.


1

No, $T_n<\infty$ does not hold true. Proof: Suppose that $T_n<\infty$ almost surely. Since $(X_t)_{t \geq 0}$ is a martingale, $X_0 = 1$, the optional stopping theorem yields $$\mathbb{E}X_{T_n \wedge t} = 1$$ for any $t \geq 0$. Since $|X_{T_n \wedge t}| \leq n$ for all $t$, we get $$\mathbb{E}X_{T_n}=1 \tag{1}$$ by the dominated convergence ...


1

I have a feeling that there's a much simpler reasoning for this, but nevertheless, here's a counterexample. Let $F(x)=1-e^{-x}$ and $G=F^{-1}$. Let $W_t$ be a Brownian motion starting at zero on some probability space $(\Omega_1,\mathcal{F}_1,P_1)$. For $T=G(1/2)*2$ denote $A=\{W_T-W_{T/2}>0\}\subset\mathcal{F}_1$. Define the random variable $\eta=I_A$, ...


1

See here for the fact that $\int_0^t B_u du$ follows the normal distribution with zero mean and variance $\dfrac{t^3}{3}$. See here for moment generating function(MGF) of normal distribution


1

Hint: By the optional stopping theorem, both $$(B_{t \wedge T_{a,b}})_{t \geq 0} \tag{1}$$ and $$(B_{t \wedge T_{a,b}}^2- (t \wedge T_{a,b}))_{t \geq 0} \tag{2}$$ are martingales. Use the dominated convergence theorem to conclude from (2) that $T_{a,b} \in L^1$. Then let $t \to \infty$ to get $$\mathbb{E}T_{a,b} = \mathbb{E}(B_{T_{a,b}}^2) \qquad ...


1

First note that sets of the form $\{t\in\mathbf{R}_{+}, B_t = x\}$ for $x\in\mathbf{R}$ are non empty by limit properties of the brownian motion. I assume more generally that $B$ starts at $x$, not $0$, but that $a < x < b$. (I may have switch your $a,b$ with mine, but who cares.) Let us first recall three "optional sampling" results (théorème d'arrêt ...


1

Hint: It suffices to calculate expectations of the form $$\mathbb{E}(X_t \cdot X_s)$$ where $s \leq t$. Setting $X_t = W_t^2$, we have $$\begin{align*} \mathbb{E}(X_t X_s) &= \mathbb{E}\big( ((W_t-W_s)+W_s)^2 W_s^2 \big) \\ &= \mathbb{E}((W_t-W_s)^2 W_s^2) + 2 \mathbb{E}((W_t-W_s) W_s^3) + \mathbb{E}(W_s^4). \end{align*}$$ Now use that $W_s \sim ...


1

Hence it suffices to show that $\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|$ converges a.s. to $0$ [when $t\to\infty$.] Here is a direct approach. For every $n\geqslant1$, let $$X_n=\max_{n \leqslant s\leqslant n ...



Only top voted, non community-wiki answers of a minimum length are eligible