Tag Info

Hot answers tagged

5

Let $X_1,X_2,X_3,...$ be a sequence of i.i.d. random variables which are distributed uniformly over $[0,1]$. Then we know that $$\int_0^1\int_0^1\cdots\int_0^1\sin\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n \;\;= \;\;E \bigg[\sin \bigg( \frac{X_1+...+X_n}{n} \bigg) \bigg]$$ By the law of large numbers, we know that $\frac{X_1+...+X_n}{n}$ ...


5

I think it's pretty hard to find a book which covers martingale theory; usually, books either give just an introduction or they focus on one particular aspect of martingale theory. I'll list some books which might be of interest and sketch (roughly) which parts they cover: David Williams: Probability with Martingales (Basic properties, optional stopping, ...


4

We have to consider three cases separately: $S_{k-1}(\omega)<0$: Then $S_k(\omega) \leq 0$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$ $S_{k-1}(\omega)=0$: Then $|S_k(\omega)|=1$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$ $S_{k-1}(\omega)>0$: Then $S_k(\omega) \geq 0$ and therefore ...


4

Brownian motion, Solution I Since $W_t \sim N(0,t)$, we have $$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$ and therefore $$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$ Brownian motion, ...


3

Using the martingale property, $$ \begin{align} \text{E}(W_n) &= \text{E}[\text{E}(W_n | W_{n - 1})] \\ &= \text{E}(W_{n - 1}) \\ &= ... \\ &= \text{E}(W_1) , \end{align} $$ and so you have a uniform bound on both $\text{Var}(W_n)$ and $\text{E}(W_n)^2$.


2

Let $T=\inf\{n:X_n>L\}$. Define $\mathbb{E}_x[\cdot]=\mathbb{E}[\cdot|X_0=x]$, $\mathcal{F}_n=\sigma(X_0,\dots,X_n)$, and $\mathcal{F}_T=\{A:A\bigcap\{T=n\}\in\mathcal{F}_n\text{ for all $n$}\}$. Then since $$\{X_n>L\text{ for some $0\le n\le m$ }\}=\{\max_{0\le n\le m}X_n>L\}=\{T\le m\}$$ we have $$P_1\{\max_{0\le n\le ...


2

For $-a \leq i \leq b$ we define a "shifted" martingale by $S_n^i := i+S_n$ and set $$T^i := \inf\{n \geq 0; S_n^i \in \{-a,b\}\}.$$ Moreover, $u(i) := \mathbb{P}(T^i< \infty)$, i.e. $u(i)$ is the probability that the the stopping time is finite if the martingale starts at $S_0^i=i$. It follows directly from the definition that $$u(-a) = u(b) = 1. ...


2

Recall that for any function $f \geq 0$ and any measure $\mu$ we have $$\int f \, d\mu = 0 \implies f=0 \quad \mu\text{-a.s.} $$ Therefore, as $(\langle X \rangle_t)_{t \geq 0}$ is an increasing process, $$\mathbb{E} \left( \int_0^t e^{X_s} \, d\langle X \rangle_s \right)=0$$ implies $$\int_0^t e^{X_s} \, d \langle X \rangle_s =0$$ $\mathbb{P}$-almost ...


2

You're missing an additional condition: $S_n=\sum_{k=1}^nX_k$ with $\mathbb{E}X_k=0, |X_k|\le K$. Let $s_n^2=\sum_{k=1}^n \mathbb{E}X_k^2$ and $T=\inf\{k:|S_k|>x \text{ or } k=n\}$ ($T$ is a stopping time). Since $S_n^2-s_n^2$ is a MTG and wp1 $T\le n$ $$0=\mathbb{E}[S_T^2-s_T^2]\le (x+K)^2\cdot P\{\max_{1\le k\le n}|S_k|>x\}+(x^2-var(S_n)) \cdot ...


2

You have $Y_t = f(X_t, t) = \exp(X_t - t^3/6)$. Assume $dX_t = m(t) dt + s(t) dW_t$ so $(dX_t)^2 = s(t)^2 dt$. Now apply Ito's lemma to $Y_t$: $$ \begin{split} dY_t &= \exp(X_t - t^3/6) \frac{-t^2dt}{2} + \exp(X_t - t^3/6) (dX_t) + \exp(X_t - t^3/6) (dX_t)^2/2 \\ &= Y_t \left[ \frac{-t^2dt}{2} + ...


1

The following answer is based entirely on ideas suggested by TheBridge in his answer. Lemma 1 A bounded local martingale is a martingale. Proof Let $X = (X_t)_{t \in [0,\infty)}$ be a bounded local martingale w.r.t. the filtered probability space $(\Omega, \mathcal{A}, P; \mathcal{F})$. Let $\sigma_1, \sigma_2, \dots$ be a sequence of real-valued ...


1

Take $\tau_n$ as in your definition and let's look at the process $X^{\tau_n}$. For all $n$ this is not only a local martingale (easy to this point) but a bounded one. The fact is that a bounded local martingale is a martingale follows from the fact that a local martingale is a martingale if and only it is of class (DL) (see theorem 1). Being a local ...


1

As @muaddib pointed out, you have simply rewritten the definition of $\hat{W}_t$ - but this doesn't show that $(\hat{W}_t)_{t \geq 0}$ is a martingale. Hints: Show that $(X_t)_{t \geq 0}$ is a martingale with respect to its canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t) = \sigma(W_s; s \leq e^{\beta t}-1).$$ Conclude that $(\hat{W}_t)_{t ...


1

Since $V_t(X_t-X_{t-1})\ge 0$ a.s. it suffices to show that $\mathbb{E}[V_t(X_t-X_{t-1})]=0$. But (assuming that $X_t$ is $(\mathcal{F}_t)$-adapted) $$\mathbb{E}[V_t(X_t-X_{t-1})]=\mathbb{E}[V_t\cdot \mathbb{E}[X_t-X_{t-1}|\mathcal{F}_t]]=0$$


1

Let $(M_t)_{t \geq 0}$ be a uniformly bounded local martingale. Then there exists a sequence of stopping times $(\tau_k)_k$ such that $\tau_k \uparrow \infty$ and $(M_{t \wedge \tau_k})_t$ is a martingale for each $k \in \mathbb{N}$, i.e. $$\mathbb{E}(M_{t \wedge \tau_k} \mid \mathcal{F}_s) = M_{s \wedge \tau_k}$$ for all $ s\leq t$ and $k \in \mathbb{N}$. ...


1

You do have a martingale! \begin{align*} \mathbb{E} [Z_t | \mathcal{F}_{t-1}] &= \mathbb{E} [X_t | \mathcal{F}_{t-1}] + \mathbb{E}[Z_{t-1}|\mathcal{F}_{t-1}] \\ &= 0 + Z_{t-1} \end{align*}


1

Since $(X_t)$ is a Markov process, $\mathbb{E}(g(X_t)\mid {\cal F}_s)=P_{s,t}g(X_s)$ where $P_{s,t}$ is the transition kernel. Therefore $\mathbb{E}(g(X_t)\mid {\cal F}_s)$ is measurable with respect to $\sigma(X_s)=\sigma(Z_s).$


1

First define the upcrossing stopping time as $$T^+_{n + 1} = \inf\{t \geq T^+_n : Y_t - Y_{T^+_n} > a \}$$ Since $Y_t$ is a Martingale, by the Strong Markov Property, the probabilities that $T^+_n$ and $R_n$ are finite are equal: $$P(T^+_n < \infty) = P(R_n < \infty)$$ So $T^+_n$ satisfies the same scaling relationship as $R_n$: ...



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