Assuming $T = N∆t$ fixed as $∆t → 0$ and taking $∆t^{1/2}\sum_{k=0} ξ_k = W(T)$.
I started with the equation $dX = aX dW, X(0) = 1$
I then figured out the Euler–Maruyama and Milstein discretisations of the above which are
$X_{k+1}=X_k+aX_k∆W_k$
$X_{k+1}=X_k-1/2a^2X_kh+aX_k∆W_k+1/2a^2X_k(W_k)^2$
Respectively. After $N$ steps these equations can be written as the following and are what I need to expand as a Taylor series.
$X_n= \prod^{k=0}_{N-1} (1 + a∆t^{1/2}ξk)$ (EM)
$X_n= \prod^{k=0}_{N-1} (1 + a∆t^{1/2}ξk+1/2a^2∆t(ξ^2_k − 1))$ (Mi)
I think I can use the fact that:
$\log(1 + x) = x−x^2/2 +x^3/3−x^4/4 +O(x^5)$
But I'm not sure how any answers would be much appreciated!
The expansion of the step from $X_k$ to $X_{k+1}$ of the exact solution is \begin{align} X_{k+1}&=\exp(aΔW_k-\tfrac12a^2Δt)X_k \\ &=(1+aΔW_k-\tfrac12a^2Δt+\tfrac12(aΔW_k-\tfrac12a^2Δt)^2+\tfrac16(aΔW_k-\tfrac12a^2Δt)^3+\dots)X_k \\ &=(1+aΔW_k+\tfrac12a^2(ΔW_k^{\;2}-Δt)+\tfrac16a^3(ΔW_k^{\;3}-3ΔW_kΔt)+O((Δt+ΔW^2)^2) \end{align} There you see that the first error term of the Euler–Maruyama method is $\tfrac12a^2(ΔW_k^{\;2}-Δt)$ where $ΔW_k^{\;2}-Δt$ has expectation $0$ and variance $2 Δt^2$ which sums up to a global error of scale $\sqrt{t\,Δt}$.
In the Milshtein method this term is contained in the formula, so that the first error term is $\tfrac16a^3(ΔW_k^{\;3}-3ΔW_kΔt)$ where $ΔW_k^{\;3}-3ΔW_kΔt$ again has expectation $0$ and variance $6Δt^3$ so that the global error from these contributions is of scale $\sqrt{t}Δt$.