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## Hot answers tagged brownian-motion

6

let $f(x)=x^2$ by application of Ito's lemma, we have $$f(B_t)=f(B_s)+\int_{s}^{t}f'(B_u)\,dB_u+\frac{1}{2}\int_{s}^{t}f''(B_u)du$$ as a result $$B_t^2=B_s^2+2\int_{s}^{t}B_u\,dB_u+\int_{s}^{t}du$$ let $t=2$ and $s=1$, thus $$B_t^2=B_s^2+2\int_{1}^{2}B_u\,dB_u+\int_{1}^{2}du$$ in the other words $$\int_{1}^{2}B_u\,dB_u=\frac{B^2(2)-B^2(1)-1}{2}$$

5

First, note that if the Brownian motion is started from $(x_0,y_0,z_0)$, then the probability to hit the cylinder $\{y^2+z^2= \delta^2\}$ before the cylinder $\{y^2+z^2=1\}$ equals$$\frac{\log(y_0^2+z_0^2)}{\log \delta^2}.$$ Let $$\Theta_{\delta;r}:=\{y^2+z^2\leq \delta^2;x\geq r\}.$$ Stopping the Brownian motion at the first time it hits the sphere of ...

3

We can compute the covariance between $W_t$ and $I_t$ as follows: \begin{align*} covar(W_t, I_t) &=E\left(W_t \int_0^t W_s ds \right)\\ &= E\left( \int_0^t W_t W_s ds \right)\\ &= \int_0^t E\left(W_t W_s \right) ds\\ &= \int_0^t \min(t, s)\, ds\\ &=\frac{t^2}{2}. \end{align*} Alternaticely, note that \begin{align*} d(tW_t) = W_t dt + t ...

2

The Brownian Motion $(W_t)_{t \geq 0}$ has (almost surely) continuous sample paths. Consequently, we have by the extrem value theorem $$M(T,\omega) := \sup_{t \leq T} |W_t(\omega)|<\infty$$ for all $T \geq 0$ and (almost) all $\omega \in \Omega$. This implies $$X_s^2(\omega) = e^{2a W_s(\omega)^2} \leq e^{2a M(T,\omega)}$$ for all $s \in [0,T]$. Thus, ...

2

$dX_t = (\beta-\alpha X_t) \, dt + \sigma \, dB_t$ Forget a moment about the SDE and consider the associated ordinary differential equation $$dx_t = (\beta- \alpha x_t) \, dt \tag{1}$$ instead. If I would ask you to solve this ODE, you would (hopefully) first solve the homogeneous equation $$dx_t = -\alpha x_t \, dt$$ and find that the solution of this ...

2

To get $E[X(s)E(t)]$, you could do \begin{align*} E[X(s)E(t)]& = E[X(s)\{X(t)-X(s)+X(s)\}]\\ &=E[X(s)\{X(t)-X(s)\}]+E[\{X(s)\}^2]\\ &=E[X(s)]E[X(t)-X(s)]+\text{Var}(X(s))+\{E[X(s)]\}^2\tag 1\\ &=0\cdot 0+ s+0^2\\ &=s \end{align*} where $(1)$ is true since $X(s)$ and $X(t)-X(s)$ are independent.

2

You have to prove the convergence of the integral $$\int_{-\infty}^{+\infty}\left|e^{i\xi \sqrt tu}\right|e^{-u^2/2}\mathrm du$$ for each complex number $\xi$. Since we can write $\xi=a+ib$ where $a$ and $b$ are real numbers, it suffices to prove the convergence of the integral $$\int_{-\infty}^{+\infty}e^{b\sqrt tu}e^{-u^2/2}\mathrm du$$ for each real ...

1

Yes, you can take $b_u = B_u - u B_1$ (note the minus sign). The formulation you have stated is equivalent to the Brownian bridge, in that it has the same distribution. See for example: https://en.wikipedia.org/wiki/Brownian_bridge and Brownian Bridge equivalence of definitions

1

First of all, note that $$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \mathbb{P}\left( \sup_{t \in [0,T]} (B_t+c)^2 \geq (\lambda+c)^2 \right).$$ Since $M_t := (B_t+c)^2$ is a non-negative submartingale, it follows from Doob's maximal inequality that $$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \frac{1}{(\lambda+c)^2} \mathbb{E}((B_T+c)^2).$$ As $\mathbb{E}(... 1 Easy way: we know that$B_t$is a Gaussian process, i.e. for any$t_1, \dots, t_n$the random vector$(B_{t_1}, \dots, B_{t_n})$has a jointly Gaussian distribution. Since any linear transformation of a jointly Gaussian vector is again jointly Gaussian, it follows that$W_t$is also a Gaussian process, and likewise so is$Y_t$. Then it's easy to compute ... 1 Hint: The Moment Generation Function for the sum of independent random variables has an interesting property.$\$\begin{align}\mathsf M_X(z) :=& ~ \mathsf E(\mathsf e^{zX})\\[2ex] \mathsf M_{X+Y}(z) =&~ \mathsf E(\mathsf e^{z(X+Y)}) \\[1ex] \mathop{=}^{\small X\perp Y}&~ \mathsf E(\mathsf e^{zX})~\mathsf E(\mathsf e^{zY}) \\[1ex]=&~ \mathsf ...

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