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## Hot answers tagged brownian-motion

6

That should not be true. The problem is that you're not considering the square of the stochastic integral, i.e. the random variable $J= (\int_0^1 {W_s} ds)^2$, which would indeed be the square of a normally distributed random variable, but the r.v. $S= (\int_0^1 {W_s}^2 ds)$. A note in the direction of finding a solution would be a discretization attempt, ...

4

NOTE: Referring to the note at the end of the General Proof Section of THIS ARTICLE, L'Hospital's Rule states that given two functions $f$ and $g$ that are differentiable in an open neighborhood of $\xi$ with $g'(x)\ne 0$ in that neighborhood, such that $\lim_{x\to \xi}|g(x)|=\infty$, then if the limit $\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$ exists, ...

4

You can use Levy's characterization of Brownian Motion, that is: a real valued stochastic process $(B_t)_{t\in[0,\infty)}$ with continuous sample paths and $B_0=0$ is a Brownian Motion if and only if it is a centered Gaussian process ($\mathbb{E}[B_t]=0$ $\forall t \ge0$) with covariance $$Cov[B_s,B_t]=\min(s,t).$$

4

Change variables $s=tu$: $$\int_0^t e^{B_s} \, ds = \int_0^1 e^{B_{t u} }t \, du=(*)$$ Use scaling: $(B_{tu}:u \ge 0) \overset{\mbox{dist}}{=}\sqrt{t}( B_{u}:i \ge 0)$. Therefore $$(*) \overset{\text{dist}}{=} \sqrt{t} \int_0^t e^{\sqrt{t} B_u} \, du$$ Take $\sqrt{t}$-th root: $$t^{1/\sqrt{t}} \left(\int_0^1 e^{\sqrt{t} B_u} \, ... 4 The distribution of \int_0^t W^2_s\,ds was found by Cameron and Martin in the 1940s. Mark Kac (1949, Transactions of the AMS) applied his method to find the Laplace transform of this integral:$$ \Bbb E\left[\exp(-u\int_0^t W^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}. $$3 The following links appear to be relevant -- also I need to ask my professor again where he found the proof of the Riemann mapping theorem using Brownian motion, because I don't remember the author's last name. http://mathoverflow.net/questions/133921/constructing-riemann-maps-using-brownian-motion ... 3 \text{Cov} \, (X_t,X_r)=\text{Cov} \, (\int_0^ts^3W_sdW_s,\int_0^rs^3W_sdW_s)=\int_0^t s^6 W^2_s ds This identity does not hold true. Note that the left-hand side is a fixed real number whereas the right-hand side is a random variable. If you apply Itô's isometry correctly, you find$$\text{cov} \, (X_t,X_r) = \color{red}{\mathbb{E} \bigg( }\int_0^t ...

3

I am omitting any algebra details or anything from probability that you should know by now. Observe: \newcommand{\P}[1]{\mathbb{P}\left[#1 \right]} \begin{align*} \P{S(3) > 2S(1)} &= \P{Ae^{B(3)}>2Ae^{B(1)}} \\ &= \P{e^{B(3)}>2e^{B(1)}} \text{ since }A > 0 \\ &= \P{\dfrac{e^{B(3)}}{e^{B(1)}} > 2} \\ &= \P{e^{B(3)-B(1)} > ... 3 We want to show that X_t = tW_t - \int_0^t W_s \mathrm{d}s is a martingale. Let r\leq t then \mathbb{E}[X_t\ | \ \mathcal{F}_r]= \mathbb{E}[tW_t\ |\ \mathcal{F}_r] - \mathbb{E}\left[\int_0^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right]. $$As you noticed$$\mathbb{E}[tW_t\ |\ \mathcal{F}_r] = \mathbb{E}[t(W_t-W_r) + tW_r\ |\ \mathcal{F}_r] = t ...

3

This is Ito's lemma. If $$dS_{t}=a_{t}S_{t}dt+b_{t}S_{t}dW_{t},$$ then $$d(\log S_{t})=\frac{1}{S_{t}}dS_{t}-\frac{1}{2}\frac{1}{S^{2}}d[S,S]_{t}=\left(a_{t}-\frac{1}{2}b_{t}^{2}\right)dt+bdW_{t}.$$ Therefore, $$\log S_{T}-\log S_{0}=\int_{0}^{T}a_{t}-\frac{1}{2}b_{t}^{2}dt+\int_{0}^{T}b_{t}dW_{t}.$$ Or, moving terms around and exponentiating, ... 2 That's interesting. Seems to me the fact that you can replace (3) by something weaker follows from the Central Limit Theorem, or at least by a CLT-ish argument. I wouldn't know what the fanciest result in this direction would be - there are all sorts of versions of things like the CLT, but it seems clear that at the very least one could replace (3) by the ... 2 \begin{align*} \text{cov}\left[x\left(3\right)-2x\left(2\right),x\left(4\right)\right] & =\text{cov}\left[x\left(3\right),x\left(4\right)\right]-2\text{cov}\left[x\left(2\right),x\left(4\right)\right]\\ & =3-2\cdot2=-1 \end{align*} 2 So continuing the computation gives: \begin{align} \frac{2a}{\sqrt{2 \pi t}}\mathrm{e}^{-\frac{a^2t}{2}}\int_{-\infty}^{m} \mathrm{e}^{a\omega}{\mathrm{e}^{-\frac{(2m-\omega)^2}{2t}}}d\omega &=\\ \frac{2a}{\sqrt{2 \pi t}}\int_{-\infty}^{m} \mathrm{e}^{-\frac{(\omega-at)^2}{2t}}\mathrm{e}^{\frac{-2m^2+\omega m}{t}}d\omega. \end{align} Now the change of ... 2 Think about what the reflection principle says intuitively. It says "if at some time you were at x, then at a later time, you are equally likely to be above or below x." Here you're conditioning on M_{t_1, t_2} > x. That does not imply that at some time between t_1 and t_2 you were at x. You might have started above x (i.e. B_{t_1} > ... 2 Given that \sigma(s) is a deterministic function, then the process (X(t))_{t \geq 0}, whereX(t) = \int_{0}^{t} \sigma(s) dW_s,is a Gaussian process with zero mean and covariance function \rho(s,t) = \displaystyle \int_{0}^{\min(s,t)} \sigma(u)^2 du. A proof of this theorem can be found in Schreve's stochastic calculus for Finance II. Hence, ... 2 Let X_{t} and W_{t} be defined as above. Let's apply Ito with f(x,y)=xy^{2} to get the semimartingale decomposition of f(X_{t},W_{t})=X_{t}W_{t}^{2}. \begin{align*} X_{t}W_{t}^{2}&=\int_{0}^{t}W_{s}^{2}\text{sgn}(W_{s})dW_{s}+2\int_{0}^{t}X_{s}W_{s}dW_{s}\\ ... 2 You need three ingredients: 1. Any continuously differentiable function is a semimartingale, since it is of locally finite total variation, 2. for a Riemann-Stieltjes / Lebesgue-Stieltjes integral you have d\phi(s) = \phi'(s) ds and 3. the covariation is defined in terms of the martingale parts, hence we have [X,\phi]_t = [X,0]_t = 0. Hence combining it ... 2 Yes, that's correct. If X \sim N(0,\sigma^2) is Gaussian with mean 0 and variance \sigma^2, then the random variable Y:= X/\sigma is also Gaussian and\mathbb{E}(Y) = \frac{1}{\sigma} \mathbb{E}(X)=0$$and$$\mathbb{E}(Y) = \frac{1}{\sigma^2} \mathbb{E}(X^2)=1.$$Hence, Y \sim N(0,1). This shows$$\underbrace{X}_{\sim N(0,\sigma^2)} = ...

2

See §10.6. Intermission: Harmonic Functions and Brownian Motion of Ulrich's Complex Made Simple.

2

Generally speaking, in the Black-Scholes model, the stock price process $\{S(t), t \geq 0\}$ is modelled as $$S(t) = S(0) e^{\displaystyle \left(r-\frac{1}{2}\sigma^2\right)t + \sigma W(t)}.$$ Now, comparing the above with $S(t) = 6e^{-2t+2W(t)}$ (note that I wrote $-2t$ instead of $2t$, otherwise it does not hold, so I assume it is a typo) and given that ...

2

I think you misunderstood the meaning there. They talked about "contingent" claims, these are random variables given some other process. A simple example will be something like this: you have a simple SDE, $dX = dB$, with $X_0 = a$, the (strong) solution is, of course, $X_t = a+ B_t$. Now, you have another process on the same Brownian Filtration: $dY ... 1 Here is an example in discrete time explaining why the Markov property is crucial to guarantee that solutions do not cross, in the sense that if two solutions coincide at some given time then they coincide at any later time. Consider the AR2 process$(x_n)_{n\geqslant0}$defined by some initial conditions$(x_0,x_1)$and by the recursion ... 1 It's enough that$f$be Lebesgue measurable and$\int_0^t [f(s)]^2\,ds<\infty$for each$t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set$A_t:=\int_0^t[f(s)]^2\,ds$, then$Y$is a (mean zero) continuous martingale with quadratic variation$\langle Y\rangle_t=A_t$. Let$\tau(t):=\inf\{s:A_s>t\}$. Then ... 1 1) Yes, check if they have the same characteristic function$\forall t$. Else, generally not. But you may be able to do that by some different characterization theorems for specific processes (e.g. Brownian Motion).$\\$2a) From a theorem on weak convergence, you have $$X_n\overset{D}\to X \iff \lim_{n\to \infty} \mathbb{E}[g(X_n)]=\mathbb{E}[g(X)]$$ for ... 1 Using the integration by parts formula we have $$\phi(t)B_t=0+\int_0^t\phi(s)dB_s+\int_0^tB_s d\phi(s)+\langle B_s,\phi(s)\rangle_t=\int_0^t\phi(s)dB_s+\int_0^tB_s \phi'ds$$ using the fact that$B_0=0$for a Brownian Motion and$\langle B_s,\phi(s)\rangle_t=0$since$\phi$is a deterministic function of time. 1 Suppose$p$is$(1,0)$- if it is not then simply rotate the answer - and that the underlying Brownian motion is a standard Wiener process wrapped around the unit circle A standard Wiener process has$W_t$normally distributed with mean$0$and variance$t$, i.e. standard deviation$\sqrt{t}$. Wrap this round the unit circle and you get$X_t = \cos W_t$... 1 I would like to add something to the excellent answer of Siron: although it is good to know the Ito Integral of a deterministic function is Gaussian, we can solve my problem without that property. Let$\quad g \colon [0,T] \longmapsto \mathbb{R}$a deterministic function. Then if we define:$\quad Y_t = \int_0^t g(s) \mathrm{d} W_s \iff \mathrm{d} Y_t = ...

1

This part which you quoted: This is so because the increments of a Wiener process are indipendent an normally distributed almost answers your question. The next step is to recognize that a Wiener process, that is, an $\mathscr{F}_t$-adapted Brownian Motion $W_t$, is a Markov Process and as such possesses the so called Markov Property. This states ...

1

It should be distribution of $B(2)$ most likely, and thus is $N(0,2)$. $1/2$, since this is $P(B(2)-B(1)>0)$ and $B(2)-B(1)$ is a normal distribution with zero mean and is thus >0 with probability $1/2$

1

With probability $1$, $\tau<1$, and $\hat B$ is nonpositive on the whole interval $[0,1-\tau]$. As a Brownian motion takes positive and negative values arbitrarily close to $0$, $\hat B$ is not a Brownian motion.

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