# Tag Info

The question is not 100% clear. If you set $X = W_t-W_s$ where $t-s = 1$ then this is equal in distribution to $W_1-W_0$ and the defining property of Brownian motion is that increments are normally distributed. In the general case $W_t-W_s$ is $N(0,t-s)$, where the second parameter is variance. If you set $dW_t = W_{t+dt}-W_t = Z \sqrt{dt}$, where ...