I am confronted with the following problem: For $W$ being a one-dimensional brownian motion and $\alpha\in[0,1]$, what are the conditions for the numerical scheme
$X_{n+1}=X_n+(1-\alpha)\mu X_n\Delta t+\alpha \mu X_{n+1}\Delta t + \sigma X_n\Delta W$
(which I understand to be the semi-implicit Euler scheme for a geometric brownian motion) such that $\lim_{n\rightarrow\infty}\mathbb{E}(X^2_n)=0$ holds?
I have already proven that $2\mu+\sigma^2<0$ is the condition such that $\lim_{t\rightarrow\infty}\mathbb{E}(X^2(t))=0$ holds for a geometric brownian motion $dX=\mu X dt + \sigma X dW_t$ so naturally I assume that the answer must also depend on the time steps $\Delta t$ and the parameter $\alpha$?
My inital approach to the problem was to express the expectation in terms of the mean square error of the scheme but I have not been able to do so yet.
Any clues or ideas would be greatly appreciated
Sort the terms of the dynamic $$ (1-αμΔt)X_{n+1}=(1+(1−α)μΔt+σΔW)X_n $$ Take the square and the expectation of it. Remember that the increment of the Brownian motion is independent of the history so far. \begin{align} (1-αμΔt)^2E[X_{n+1}^2]&=E[(1+(1−α)μΔt+σΔW)^2]\,E[X_n^2] \\ &=\Bigl((1+(1−α)μΔt)^2+σ^2Δt\Bigr)\,E[X_n^2]. \end{align} So for the convergence to zero you want $$ (1+(1−α)μΔt)^2+σ^2Δt<(1-αμΔt)^2 \\\iff\\ 2(1−α)μ+σ^2+(1−α)^2μ^2Δt<-2αμ+α^2μ^2Δt \\\iff\\ 2μ+σ^2<(2α-1)μ^2Δt $$