# Why does Euler-Maruyama method use a square root of the time step

Euler-Maruyama method is supposed to be an extension of the Euler method for ODE, but applied to SDE. This means that if we have an equation:

$$dY_t = Y_t dW_t$$

where $W_t$ is the Wiener process, discretizing the total time T into $N$ intervals ($T = N \cdot h$), we could approximate the change in $Y$ from $t=0$ to $t=1$:

$$d Y_1 = Y_1 - Y_0 \approx Y_1 \cdot (W_1 - W_0)$$

Now to evaluate this (implicit) equation for $Y_1$ I would have to generate a random number for $W_1 - W_0$ which is a normal distribution $~ \mathcal{N}(0,1)$, as in the definition of the Wiener process.

However in the wikipedia page (https://en.wikipedia.org/wiki/Euler%E2%80%93Maruyama_method) as well as in other sources the E-M method seems to dictate a random number from $\mathcal{N}(0,1)$ TIMES $\sqrt{h}$.

Why is the square root of the timestep $h$ there? I suspect it has something trivial to do with the standard deviation, but I wasn't able to derive this result, since the expactation value of $W_t - W_s$ is $0$. Would you help a fellow out?

Second, the difference you are considering is actually $W_{t_1}-W_{t_0}=W_h-W_0$, and the increments of the Wiener process follow the law $$W_s-W_t\sim N(0,|s-t|)\sim N(0,1)·\sqrt{|s-t|}$$ which should explain where the square root comes from.
• Because it contradicts the idea of a filtration adapted stochastical process. Note that $Y_{t+dt}=Y_t/(1-dW_t)$ contains the possibility of division by zero. By Ito formalisms, this is also process-equivalent to $Y_{t+dt}=Y_t·(1+dW_t+dt)$ which is not compatible with the original SDE. – LutzL May 3 '16 at 12:59