A question about brownian motion

[tennishaha]

for a brownian motion W(t)
W(t_i+1)-W(t_i) is normal distribution with mean 0 and variance t_i+1-t_i

so this means var(W(t_i+1)-W(t_i))=var(W(t_i+1))-var(W(t_i))=t_i+1-t_i

I don't think the above equation satisfies because W(t_i+1) and W(t_i) are not independent. Any comment? thanks

 

[wayneckm]

I think you miss out the covariance term in $$Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y)$$ and note that in Brownian motion $$Cov(W_{s},W_{t}) = \min(s,t)$$

 

[hamster143]

Yeah, you can't just subtract variances like that.