# Tag Info

### Showing a basic market admits no arbitrage

The idea is actually really simple... presumably the difficulty is in cutting through the abstract formalism. We have an asset whose price is $S_0^1$ and tomorrow it can either go to $\beta S_0^1$ or ...
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### Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

Define $A_t := \int_0^t e^{\lambda s} dB_s$. You are asking if $X_t := e^{-\lambda t} A_t$ is a martingale. By Ito's formula, we have \begin{align*} dX_t &= e^{-\lambda t} dA_t -\lambda e^{-\...
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Accepted

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic. Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $... • 168k 2 votes ### The almost sure event in the law of the iterated logarithm for the Brownian motion: what it looks like The event you have captured in more explicit terms is not$\{\limsup_{t\to 0}W_t/h(t)=1\}$, but rather$\{\limsup_{t\to 0}|W_t/h(t)-1|=0\}$. The placement of the absolute value is crucial! • 26.1k 0 votes ### Optional Stopping Theorem for martingales bounded except at the stopping time My friend pointed out a simple counterexample:$X_0 = 0, $$X_{n+1} = \begin{cases} X_n & \text{with probability 1/2} \\ X_n - 2^n & \text{with probability 1/4} \\ X_n + 2^n & \text{... • 3,727 1 vote Accepted ### Example of a process that yields a non-martingale Easy example: C_k = 1_{X_k > X_{k-1}}. This is adapted because X is adapted, but \sum_{k=1}^n C_k (X_k-X_{k-1}) = \sum_{k=1}^n (X_k-X_{k-1})^+ is non-decreasing and hence cannot be a ... • 13.5k 0 votes ### \mathcal{F}_n measurable In general no.$$ \begin{align*} &\{T\in\{(m,k)\mid m>n\}\}\\ =&\left(\bigcap_{m\le n}\bigcap_{0\le\ell\le K}\{S_{m,\ell}\le\lambda\}\right)\cap\bigcup_{m\ge n+1}\left(\left(\bigcap_{0\le\... • 6,982 1 vote ### Definition of Left-Closable Martingale To build on Jacob Maibach's answer on the positive integers being open on the right, later on in Resnick's book he discusses reversed/backwards martingales: given a DECREASING family of\sigma$-... 1 vote Accepted ### Optional Stopping Theorem and Stopped$\sigma$-fields For real$x$you have $$\{X_T\le x\}\cap\{S\le n\} = \cup_{k=0}^n\{X_k\le x, T=k, S\le n\}.$$ which is clearly$\mathcal F_n$-measurable, for each$n$. It follows that$X_T$is$\mathcal F_S$-... • 26.1k 2 votes Accepted ### Definition of Left-Closable Martingale I haven't run into this terminology before, but here is my guess. The term closed probably refers to the set of indices$n$where the martingale is defined. That is, consider the domain $$\{ n : X_{n}... • 2,522 0 votes Accepted ### Is (e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0} a martingale? \mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)} does hold for c\in\mathbb{C}. The proofs are correct. 1 vote Accepted ### Calculate \mathbb{E}(\exp (- \lambda T_x )) for \lambda > 0 where X is a Brownian Motion with drift We have$$E[e^{\theta B_t+\theta ct-\lambda t}]=e^{\theta^{2}t/2+\theta c t-\lambda t}$$and so we indeed need \theta^{2}/2+\theta c-\lambda=0\Rightarrow \theta=-c\pm \sqrt{c^{2}+2\lambda} to get ... • 3,641 1 vote Accepted ### Expected value of the square of a stopping time \def\={\mathrel{\phantom=}}Your calculation writes\begin{gather*} E( (a^2 - S_a) I_{\{ B_{S_a} = a \}} ) + E( ((-a)^2 - S_a) I_{\{ B_{S_a} = -a \}} )\\ = a^2 - E( S_a I_{\{ B_{S_a} = a \}} ) + (-a)^... • 34.3k 1 vote ### Expectation of the indicator function \{k\} is not independent of \mathcal F_n. Since \mathcal F_n is generated by a partition,$$ \mathbb E\left[X_{n+1}\mid\mathcal F_n\right]=\sum_{k=1}^n \mathbb E\left[X_{n+1}\mathbf{1}_{\{k\}}\... • 173k 1 vote Accepted ### Expected value of the exponential of a stopping time All good. Perhaps, you can add more details on the "by symmetry" Independence of$T$and$B_T$i.e. we have$B_t\stackrel{d}{=}-B_{t}$and$S_{a}$is only a function of$|B_{t}|$, which is ... • 3,641 0 votes Accepted ### Stirling approximation of the probability that the stopping time is finite Consider$x=0$. Using Stirling's approximation you find $$P(S_{2m}=0)=\binom{2m}{m}2^{-2m}\sim \frac{1}{\sqrt{\pi m}},$$ which yields $$\sum_{m=0}^\infty P(S_{m}=0)= \infty.$$ Define$\$\tau_{0}^{(m)}:=...
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