Simpler for of Euler-Maruyama or am I missing something

Question

The stochastic part of the Euler-Maruyama method is often written with random increments with zero mean and variance $\Delta t$. For example for the SDE (drift diffusion) $$ dX(T)=\delta dt + sdW(t) $$ one could simulate this as $$ X(t+\Delta t)=X(t) + \delta\Delta t+\Delta W, $$ where $\Delta W\sim Normal(0,s^2\Delta t)$.

However, wouldn't this be equivalent to the (maybe simpler form) $$ X(t+\Delta t) = X(t) + \Delta X(t)\Delta t $$ with $\Delta X(t) \sim \cdot Normal(\delta, s^2)$, since here the $\Delta t$ scales both the variance as well as the mean. Or is there something that I am missing and these two forms are not equivalent? I feel, that this is much closer to the original Euler method on which the Euler-Maruyama method is based.

Answer 1

They are not equivalent. The whole point of separating the study of classical ODEs from SDEs is that the term $\Delta W$ has size in the order of $\sqrt{\Delta t}$ and not $\Delta t$ (see the standard deviation of $\Delta W$), so you cannot write the discretization using the classical Euler method with $\Delta X$ independent of $\Delta t$.

Now, let me clarify where your mistake is. You basically say that $$ \Delta X\Delta t = \delta\Delta t+\Delta W $$ which is incorrect. Since $\Delta X \sim \mathcal{N}(\delta,s^2)$ (as you defined) then $\Delta X = \delta + w$ where $w\sim \mathcal{N}(0,s^2)$. Now $$ \Delta X\Delta t = \delta \Delta t+ w\Delta t $$ and we are left to compute the variance of $w\Delta t$ which is $\mathbb{E}\{(w\Delta t)^2\}=\mathbb{E}\{w^2\}\Delta t^2 = s^2\Delta t^2$ and not $s^2\Delta t$ as $\Delta W$ had. This is where the $\sqrt{\Delta t}$ vs $\Delta t$ matters.

Hope this helps!