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How to prove an adapted Feynman Kac Formula for $v_t + \frac{1}2 \sigma ^2 (t,y) v_{yy} + b(t,y) v_y - \delta(t,y) v + h(y) = 0$ using SDE techniques?

We see $$\begin{aligned}dv(t,Y_t)-\delta(t,Y_t)v(t,Y_t)dt&=-h(Y_t)dt+v_y(t,Y_t)\sigma(t,Y_t)dW_t\\ \implies d(v(t,Y_t)e^{-\int_0^t\delta(s,Y_s)ds})&=-h(Y_t)e^{-\int_0^t\delta(s,Y_s)ds}dt+v_y(t,...
Snoop's user avatar
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1D Brownian motion: stopping time and stopping state not independent

Given $X_{T}=-3$, it means that state 1 has not been hit by time $T$. Hence, by the reflection principle, $$P(T\le t|X_{T}=-3, X_0=0)=1-P(hitting\; 1\; by\; time\; t)\\ =1-2\int_1^\infty \frac{1}{\...
toronto hrb's user avatar
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Limit for Brownian local time

I think I have an answer: can you please let me know if you spot any mistake? Here we can write for every $n\in \mathbb{N}$, $L^{(n)}(t)=\frac{L(nt)}{\sqrt{n}}$ and $L^{(n)}(t)\overset{(d)}{=}L{(t)}$, ...
Randomwandering's user avatar
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The almost sure event in the law of the iterated logarithm for the Brownian motion: what it looks like

Because $\sup_{0<s\le t}W_s(\omega)/h(s)$ is monotone in $t$, you have $$ \{\omega: \limsup_{t\to 0}W_t(\omega)/h(t)\le 1 \} = \cap_{\epsilon\in\Bbb Q_+}\cup_{\delta\in\Bbb Q_+}\cap_{s\in\Bbb Q\cap(...
John Dawkins's user avatar
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4 votes
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Computing the quadratic covariation $\langle B,B^2\rangle_t$ of a Brownian motion and its square

See first this discussion on how to obtain the quadratic variation of two Itô processes. By Itô's formula we obtain that $B_{t}^{2}$ follows the dynamics $$ B_{t}^{2} = \int_{0}^{t}ds + \int_{0}^{t}...
minginator's user avatar
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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!
John Dawkins's user avatar
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$W_{10t}^2 - W_{5t}$ is not gaussian

Set $$W_{10t }-W_{5 t} =a, W_{5t}=b$$ and determine the moments of $$(a+b)^2 - b$$ with $a,b$ independent Gaussians with the speciality, that the 2n-moments are powers of the second moment with ...
Roland F's user avatar
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1 vote

Using Ito's Lemma to take a stochastic integral

Please avoid to ask too many questions at the same time in future posts. Let's tackle the side questions first. First side question. I mean, why not ? If you have a stochastic variable $X_t$ itself ...
Abezhiko's user avatar
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2 votes
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How can I show that the first exit time by a planar Brownian motion is a.s. finite, i.e. $\mathbb{P}_z(\tau_D<\infty)=1$?

My first question is, where do i need that $W_t$ is a complex Brownian motion, I mean why can't I only work with $B_t$ instead of $W$? You could do it with just using neighborhood-recurrence for 2d-...
Thomas Kojar's user avatar
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3 votes
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What is the solution to the SDE $X^x_t = B_t + \int_0^t \frac{x − X^x_s}{1-s} ds$

Setting $Z_t$ as what you did in your work, and then writing (2) in Differential form gives us that $$dX^x_t = xdt + dB_t - \frac{B_t}{(1-t)}dt + Z_tdt \tag{4}\label{eq4} $$ And (2) also gives us that ...
vinayak's user avatar
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Markov Property of a Ito Process

Rather than bringing in $B_{t+h}-B_h$, consider using $B_{t+h}-B_t$: Because $X^x_{t+h} = X^x_t\cdot\exp(ch+\alpha(B_{t+h}-B_t))$, and $B_{t+h}-B_t$ is independent of $\mathcal F_t$ with the same ...
John Dawkins's user avatar
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Compute the $n$th stochastic integral of Brownian motions

Alternatively, Define $$I_t^{(n)}=\int_0^tdW_{s_1}\int_0^{s_1}dW_{s_2}\cdots\int_0^{s_{n-1}}dW_{s_n}$$ Show that this integral satisfies the following recurrence relation for $n\ge 2$ $$I_t^{(n)}=\...
Fellow InstituteOfMathophile's user avatar
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Is $\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ a stopping time?

(Edited) Useful theorems Assuming nothing about the filtration: If $X=(X_t)_{t\geq 0}$ is a continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq ...
Wilfred Montoya's user avatar
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Question related to canonical construction of Brownian Motion

Yes, $\Omega$ is the set of continuous functions from $[0,\infty)$ to $\mathbb{R}$. The fact that $\Omega$ is not a measurable set of some other $\sigma$-algebra is not at all relevant. It's true ...
user6247850's user avatar
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1 vote
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An (a.s.) continuous process $(X_t)_{t\geq 0}$ is a Brownian motion if $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale

The main idea is to use the following fact: The conditional characteristic function $\mathbb{E}\left[e^{i\lambda X_s}\vert\mathcal{F}_s\right]$ of $X_s$ is equal to $e^{i\lambda\mu - \frac{1}{2}\...
Wilfred Montoya's user avatar
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2 Brownian Motions with Non Zero Correlation and NOT jointly normal?

Consider the following stochastic differential equation (SDE): $dB_t = \rho(B_t, W_t)dW_t + \sqrt{1 - \rho^2(B_t, W_t)}dZ_t$, where $\{W_t\}_t$ and $\{Z_t\}_t$ are two independant Brownian motions, ...
Skeubo's user avatar
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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.
Wilfred Montoya's user avatar
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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 ...
Thomas Kojar's user avatar
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1 vote

An (a.s.) continuous process $(X_t)_{t\geq 0}$ is a Brownian motion if $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale

By using martingale-property and taking derivative at $\lambda=0$, we get martingale property for $X$ $$E[X_{t}|\mathcal{F}_{s}]=X_{s},$$ where we can take derivative due to dominated convergence ...
Thomas Kojar's user avatar
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0 votes
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Expected value of the Brownian motion to the power of $n$

Let $s<t$. From $B_t-B_s$ being independent of $\mathcal{F}_s$ and $B_t-B_s \sim \mathcal{N}\left(0, t-s\right)$ follows: \begin{align*} \mathbb{E}\left[\left(B_t - B_s\right)^n\vert \mathcal{F}_s\...
Wilfred Montoya's user avatar
0 votes

Rate of increase of maximum process of Brownian Motion

This indeed just follows from reflection principle and Borel-Cantelli $$\sum_n P[M_{t_{n}}\frac{1}{g(t_{n})}\geq \epsilon]=\sum_n P[|B_{t_{n}}|\geq g(t_{n})\epsilon]\leq c \sum_{n}exp(-\frac{(g(t_{n}))...
Thomas Kojar's user avatar
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1 vote

Two definitions for $\mathcal F_{\infty}$

$\mathcal F_{\infty} \subseteq \mathcal F'_{\infty}$ because RHS is a $\sigma-$ field and each $B_t$ is measurable w.r.t. it. $\mathcal F'_{\infty} \subseteq \mathcal F_{\infty}$ because each $\...
geetha290krm's user avatar
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1 vote

Running maximum for Geometric Brownian Motion

For the geometric Brownian motion $d X_t = \mu X_t dt + \sigma X_t d B_t$ being started at unity $X_0 = 1 $ the probability density function of the running maximum $M_T := sup(0<s<T: X_s)$ is ...
Przemo's user avatar
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2 votes
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Asymptotic form of solution to biased random walk

I think that you skipped some steps when doing you did the approximations in the unbiased case. If you do it rigorously, the same method applies. More formally, you want to estimate large deviations ...
LPZ's user avatar
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4 votes

Is $(B_t^2 - t^2)_{t\geq 0}$ a local martingale?

If $B^2_t-t^2$ were a local martingale the so too would $t-t^2=(B^2-t)-(B^2-t^2)$ be a (continuous) local martingale, but also a process of finite variation. Such a process must be constant in time, ...
John Dawkins's user avatar
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1 vote
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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)^...
Ѕᴀᴀᴅ's user avatar
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3 votes
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Is $(B_t^2 - t^2)_{t\geq 0}$ a local martingale?

Note that the process $(B_t^2 - t)_{t\ge 0}$ is a martingale. This implies (by Doob's Theorem) that for any finite stopping time $\tau$, the stopped process $X_t = (B_{t \wedge \tau}^2 - t\wedge \tau)...
Presage's user avatar
  • 8,196
0 votes

Posterior probability of Wiener process within interval

$$\begin{aligned}P(B_h\leq x||B_h|<\rho)&:=\frac{P(B_h\leq x,|B_h|<\rho)}{P(|B_h|<\rho)}\\ &=\frac{\Phi(\max(\min(x,\rho),-\rho)/\sqrt{h})-\Phi(-\rho/\sqrt{h})}{2\Phi(\rho/\sqrt{h})-1}...
Snoop's user avatar
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5 votes

Generator of the joint process $(X_t,Y_t)$ where $Y_t= e^{-t}W(e^{2t})$ and $X_t = \int^t_0 Y_sds$.

To find the generator I will look for the two-dimensional SDE that is solved by $(X_t,Y_t)\,.$ It is not hard to see that $$\tag{1} B_t:=\int_0^te^{-s}\,dW_{e^{2s}} $$ is a continuous martingale with ...
Kurt G.'s user avatar
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1 vote
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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 ...
Thomas Kojar's user avatar
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First hitting time of biased random walk in 2 dimensions

Let $B_t = (B_t^x, B_t^y)$ be a 2-dimensional Brownian motion. In particular $B_t^x$ and $B_t^y$ are independent $1$-dimensional Brownian motions. Define the first hitting time of the line $L_{x_0} = \...
Jose Avilez's user avatar
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Distribution of Increments of a fractional Brownian Motion

I messed up the indices. It should be: $(B_{t+h}^H-B_{\mathbf{h}}^H)_{t\geq 0} \stackrel{d}{=}(B_t)_{t\geq 0}$ Now we can derive the desired result: $Cov(B_{t+h}^H-B_h^H,B_{s+h}^H-B_h^H) \\ =E((B_{t+h}...
Gragon's user avatar
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Potential density of killed brownian by local time

Yes. In fact $$ u_k^0(x,y) = u^0_0(x,y)+c^{-1},\qquad \qquad (1) $$ where $u^0_0$ is the Green's function for the Brownian motion killed upon first hitting $0$, and the constant $c$ is given by $$ c=\...
John Dawkins's user avatar
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2 votes

Joint law of Brownian motion maximum and its values at different points

It seems that Thomas Kojar's suggestion does help to solve the problem. Let $0 < u < v < T$. Then, since $$ M_T = \max \{ M_u, B_u + \max_{t \in [u, T]} ( B_t - B_u ) \} = \max \{ M_u, B_u + ...
tsnao's user avatar
  • 258
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Proving the process $(Y_t)_{t\in[0,\infty)}$ is a standard Brownian motion

First, lets clarify what independent processes means Independent stochastic processes and independent random vectors. The definition for the two processes to be independent is given by PlanetMath: ...
Thomas Kojar's user avatar
  • 3,611
2 votes

Proving the process $(Y_t)_{t\in[0,\infty)}$ is a standard Brownian motion

The quadratic variation of $Y$ is $t\,.$ (The rest of the things you need to apply for Levy characterisation I leave to you.) Proof. We only have to consider partitions $0<t_1<\dots<t_n$ of ...
Kurt G.'s user avatar
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basic complex analysis, confusion about complex exponential and its modulus

Let's clarify the misunderstanding here. Your reasoning is almost correct, but there's a small mistake in the application of Euler's equation. Let's go through it step by step: Starting from Euler's ...
Kasi's user avatar
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