# A question about brownian motion

1. Jun 28, 2010

### 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

Last edited: Jun 28, 2010
2. Jun 28, 2010

### 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 $$Cov(W_{s},W_{t}) = \min(s,t)$$

3. Jun 29, 2010

### hamster143

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