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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 $f \in C^1(\mathbb{R})$.


1

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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