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4

Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a $\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair, $\textbf{X}=(X, \mathbb{X})$ where $X\colon [0,T]\to V$ where $V$ is some Banach space (typically $\Bbb{R}$) and a ...

4

If $W_t:=\rho W^{(1)}_t+\sqrt{1-\rho^2} W^{(2)}_t$ then we can show that, $W_t$ is a Brownian motion. Proof Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F_t}\})$ be a probability space . Clearly, $W_t$ has continuous sample paths and $W_0=0$. $$\mathbb{E}[W_t|\mathcal{F_s}]=\rho\,\mathbb{E}[W^{(1)}_t|\mathcal{F_s}]+\sqrt{1-\rho^2}\,\mathbb{E}[W^{(2)... 3 For any t>0, we have W_t^{(1)},W_t^{(2)}\stackrel{\mathrm{i.i.d.}}\sim \mathcal N(0,t), so \rho W_t^{(1)}\sim\mathcal N\left(0,\rho^2 t\right) and \sqrt{1-\rho^2}W_t^{(2)}\sim\mathcal N\left(0,(1-\rho^2)t \right), from which$$W_t \sim \mathcal N\left(0, t \right). $$Therefore \{W_t:t\in\mathbb R_+\} is a Gaussian process, and from independence ... 3 As Did mentioned, [W_1,W_2](t)=0. Let \{t_i\}_{i=1}^{n} be a partition of [0,t]. We want to consider$$A_n=\sum_{i=0}^{n-1}(\,W_1(t_{i+1})-W_1(t_{i})\,)(\,W_2(t_{i+1})-W_2(t_{i})\,)$$Using independent of W_1 and W_2 , thus \mathbb{E}[A_n]=0. Since increment of Wiener process are independent, the variance of sum is sum of variance , and we have ... 2 That can be written as$$ \int_{0}^{+\infty} x^2 \cdot\frac{d^2}{dx^2}\log(f(x))\cdot f(x)\,dx < 1 $$that is a constraint that depends on minimizing a Kullback-Leibler divergence. It essentially gives that your distribution has to be close to a normal distribution (in the KL sense). 2 Set f(x)=-\frac{1}{x}-\tan^{-1}x. we have$$f'(x)=\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1}{x^2+x^4}$$and$$f''(x)=-\frac{2x+4x^3}{(x^2+x^4)^2}$$By application of Ito's lemma we have$$f(B_t)=f(B_1)+\int_{1}^{t}f'(B_s)dB_s+\frac{1}{2}\int_{1}^{t}f''(B_s)ds$$therefore$$-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)=-\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)+\...

1

I doubt that this identity holds true. Just consider $f:=1$, then the assertion reads $$B_t = 0,$$ which is obviously not correct. As @Did pointed out in a comment, Itô's formula shows that the identity $$\int_0^t f(s) \, dB_s = f(t) B_t - \int_0^t f'(s) B_s \, ds$$ holds for $f \in C^1(\mathbb{R})$.

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If $\{B_t:t\in\mathbb R_+\}$ is a standard Brownian motion, then $$X_t := X_0\exp\left(\left(\mu-\frac12\sigma^2\right)t+\sigma B_t\right)$$ (where $X_0$, $\mu$, and $\sigma$ are constants) defines a geometric Brownian motion. It is clear that if $X_0=0$ then $X_t$ is identically zero, and similarly if $\sigma=0$ then $X_t = X_0 e^{\mu t}$ with probability ...

1

Sasha's answer is so nice. I want to offer other way. Other way Let $\alpha\in \mathbb{R}^+$. It is well known that $X_t=\exp\left(\alpha W_t-\frac{1}{2}\alpha^2 t\right)$ is a martingale, therefore $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}[X_0]=1$$ so $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}\left[\exp\left(\alpha W_{T_{a,b}}-\frac{1}{2}\alpha^2 T_{a,b}\right)\... 1$$\mathbb{E}[f(X_t)]=\mathbb{E}[X_t^2-2KX_t+K^2]=\mathbb{E}[X_t^2]-2K\,\mathbb{E}[X]+K^2\tag 1$$set Y=\ln X_t, therefore$$Y=\ln X_t\sim N\left( \ln x_0+\left( \mu -\frac{1}{2} r\sigma^2 \right)t\,\,,\,\,r^2 \sigma^2t \right)$$We have$$\mathbb{E}[X_t^2]=\mathbb{E}\left[e^{2\ln X_t}\right]=\mathbb{E}\left[e^{2Y}\right]=M_{Y}(2)=\exp\left(2\mu_Y+\frac 12\...

1

There are many ways to show this. Perhaps, this is the simplest one: there is a constant $C$ such that $b(x) > -C(x+1)$ for all $x\in \mathbb R$. Consider the equation $$dY_t = -C(Y_t +1) + dB_t\tag{1}$$ with the initial condition $Y_0 = y<x$. It is not hard to show that $Y_t <X_t^x$ for all $t\ge 0$. But equation (1) is easy to solve: $$Y_t = ... 1 Since X(0) = Y(0) = 1, the process obviously cannot live on a unit circle. But it does live on a circle of radius \sqrt{2}. In fact, it is easy to see that the solution to your system is given by X(t) = \sqrt{2}\cos (W(t)+\frac\pi4), Y(t) = \sqrt{2}\sin (W(t)+\frac\pi4). Indeed, X(0)=Y(0)=1 and, by Itô's lemma, dX(t) = -Y(t) dW(t) - \frac12 ... 1$$ dX_t=\left(r\mu+\frac{1}{2}r(r-1)\sigma^2\right)X_tdt+r\sigma X_t dB_t $$By application of Ito lemma, we have$$d(\ln X_t)=\frac{1}{X_t}dX_t+\frac{1}{2}\left(\frac{-1}{X_t^2}\right)d[X_t,X_t]$$therefore$$d(\ln X_t)=\left(r\mu+\frac{1}{2}r(r-1)\sigma^2\right)dt+r\sigma dB_t-\frac{1}{2}r^2\sigma^2dt$$in other words$$d(\ln X_t)=\left(r\mu-\frac{1}{2}r\...

1

We should look for a solution of the form $$X(t)=U(t)V(t)$$ where $$dU_t=a\,U_tdt+\sigma\,U_t\,dW_t$$ and $$dV_t=\alpha_t\,dt+\beta_t\,dW_t$$ $U$ is a geometric Brownian motion, therefore $$U(t)=U(0)\,\large e^{(a-\frac{1}{2}\sigma^2)t+\sigma W_t}$$ let $U(0)=1$, this yields $V(0)=X(0)$. Now we should find $\alpha_t$ and $\beta_t$. $$dX_t=U_tdV_t+V_tdU_t+... 1 In any manner, if$$L(p)=\prod_{i=1}^n f_{X_i} (x_i; p)$$then$$\log\left(L(p)\right)=\sum_{i=1}^n\log\left(f_{X_i} (x_i; p)\right)\frac{L'(p)}{L(p)}=\sum_{i=1}^n \frac{f'_{X_i} (x_i; p)}{f_{X_i} (x_i; p)} and since you want $L'(p)=0$, the rhs seems (at least to me) simpler to manipulate.

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