0

Compute $P(B_{t}>0,B_{s}>0)$ for $t >s$ where $B$ is a standard BM

My attempt:

$P(B_{t}>0,B_{s}>0)=P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)$ and now considering $X:=B_{t}-B_{s}\sim \mathcal{N}(0,t-s)$ and $Y:=B_{s}\sim\mathcal{N}(0,s)$ such that for $\xi$, $\eta$ standard normal iid random variables we have $X=\sqrt{t-s}\xi$ and $\sqrt{s}\eta$ such that we obtain:

$$ P(B_{t}-B_{s}+B_{s}>0,B_{s}>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\sqrt{s}\xi>0)=P( \sqrt{t-s}\xi+\sqrt{s}\eta>0,\xi>0)$$

Now if I want to switch the above into polar coordinates I have:

$$\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,r\cos(\theta)>0\}=\{(r,\theta):\; \sqrt{t-s}r\cos(\theta)+\sqrt{s}r\sin(\theta)>0,\theta \in (-\pi/2,\pi/2)\}\\ = \{(r,\theta):\; -\frac{\sqrt{t-s}}{\sqrt{s}}<\tan(\theta),\theta \in (-\pi/2,\pi/2)\} = \{(r,\theta):\; \arctan(-\frac{\sqrt{t-s}}{\sqrt{s}})<\theta,\theta \in (-\pi/2,\pi/2)\}$$

This looks rather ugly, am I on the right path?

SABOY
  • 1,828

2 Answers2

2

Use the following fact about the bivariate normal distribution: If $X$ and $Y$ are bivariate normal with mean $0$ and correlation $\rho$, then $$P(X >0,Y>0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).$$ The link assumes $X$ and $Y$ have mean zero and standard deviation $1$, but the same result holds with arbitrary standard deviations. So your problem reduces to finding the correlation between $B_t$ and $B_s$. (Your calculation replicates the derivation of the above equation, but obscures the generality of this fact.)

grand_chat
  • 38,951
  • Could you develop the calculation? Because I don't think you can apply your proposed method (which is exactly what the OP used). This method works only if the $\theta$ is independant of $r$. – NN2 Mar 11 '21 at 15:02
  • Note that $\xi$ and $\eta$ are IID standard normal, which means in polar coordinates $\theta$ and $r$ are independent. The conversion to polars isn't using the original $B_t$ and $B_s$. – grand_chat Mar 11 '21 at 16:42
1

I provide a solution without using changing of variables to polar coordinate (which doesn't lead to a solution IMHO) \begin{align} P(B_t >0, B_s>0) &= E(\mathbb{I}_{B_s>0}\mathbb{I}_{B_t>0}) \\ &= E(\mathbb{I}_{B_s>0}\mathbb{I}_{(B_t - B_s)+B_s>0}) \\ &= E\left(E(\mathbb{I}_{B_s>0}\mathbb{I}_{(B_t - B_s)+B_s>0}|\mathcal{F}_s)\right)\\ &= E\left(\mathbb{I}_{B_s>0}E(\mathbb{I}_{(B_t - B_s)>-B_s}|\mathcal{F}_s)\right)\\ &= E\left(\mathbb{I}_{B_s>0}P(\sqrt{t-s}\mathcal{N}(0,1)>-B_s|\mathcal{F}_s)\right)\\ &= E\left(\mathbb{I}_{B_s>0}\Phi(\frac{B_s}{\sqrt{t-s}})\right)\\ &= E\left(\mathbb{I}_{X>0}\Phi(\sqrt{s}\frac{X}{\sqrt{t-s}})\right)\\ &= \color{red}{\int_0^{+\infty}\Phi\left( \sqrt{\frac{s}{t-s}}x \right)\varphi(x)dx}\\ \end{align}

where $X \sim \mathcal{N}(0,1)$, $\Phi(x)$ and $\varphi(x)$ are CDF and density function of $\mathcal{N}(0,1)$.

NN2
  • 15,892