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I want to find the following probability :

$\mathbb{P}\{W_1>0, W_5 <0\}$

Where $W_t$ is a Wiener process, so it follows the law $\mathcal{N}(0, t)$.

My question is : can say that $\{W_1>0\}$ and $\{W_5 <0\}$ are uncorrelated ? In that case :

$\mathbb{P}\{W_1>0, W_5 <0\} = \mathbb{P}\{W_1>0\}\mathbb{P}\{W_5 <0\}= (1-\frac{1}{2})\frac{1}{2} = \frac{1}{4}$

As these are both is the cummulative distribution function of a normal law evaluated in $0$.

However, I'm not sure about the assumption I just made. Can someone bring some clarifications ? Thank you.

Gabriel dLN
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  • They are certainly correlated. You have to compute the joint pdf of $W_1$ and $W_5$. – geetha290krm Nov 22 '23 at 07:21
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    Since $W_1$ and $W_5$ are not independent, you can't compute the probability that way. One way of computing it is to realize the joint distribution in a more comprehensible way. Noting $(W_1,W_5)$ isnidentically distributed as $(X,X+2Y)$ for i.i.d. standard normals, the desired probability is $$\mathbf{P}(X>0,X+2Y<0),$$ which is the probability that $(X,Y)$ lies in the infinite wedge of the form $D={(x,y):x>0,x+2y<0}$. Using the rotationa symmetry of the joint distribution, the probability is $$\frac{\arctan 2}{2\pi}.$$ Note that this is less than $\frac{1}{4}$. – Sangchul Lee Nov 22 '23 at 07:45

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