Recent content by Tetef

I don't know if it's any use but, I've thought of this : P\{T_{a}<T_{b}|W(t)=x\}=\frac{p\{T_{a}<T_{b}\cap W(t)=x\}}{p\{W(t)=x\}} where capital P are probabilities and small p are probability density functions. we know : p\{W(t)=x\}=\frac{1}{\sqrt{2\pi t}}exp( -\frac{x^2}{2t}) we don't know ...

I did mean what I wrote, A standard brownian motion has W(0)=0 I use t in the definition T_{a}=inf\{t:W(t)=a\}to say the first hitting times are defined as : T_{a} is the smallest time t where the brownian motion does hit the level a i.e. W(t)=a In the expression of the probability...

Hi, Letting W be a standard brownian motion, we define the first hitting times T_{a}=inf\{t:W(t)=a\} with a<0 and T_{b}=inf\{t:W(t)=b\} with b>0 The probability of one hitting time being before an other is : P\{T_{a}<T_{b}\}=\frac{b}{b-a} I'm looking for this probability in the case...