I am really confused about a step in the solution of this problem. I would really appreciate it if someone could explain to me what is in bold below. Part A is OK, but I don't understand something in part B. Thanks in advance!
Question: Consider the geometric Brownian motion equation $$dX=aXdW,\quad X(0)=1,\quad a>0.$$ A) Write down the Euler-Maruyama scheme for this equation.
B) Obtain a recurrence relation for the moment $\mathbb{E}X_k^q$ and deduce that $$\mathbb{E}X_k^q=\alpha^k(ah^{1/2},q),$$ where $$\alpha(a,q)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(1+ax\right)^q e^{-x^2/2}dx$$
Solutions:
A) $$X_{k+1}=X_k+aX_k\Delta W_k=(1+a\Delta W_k)X_k$$ B) Taking the $q$th power and expectation of the Euler-Maruyama scheme for this equation gives $$\mathbb{E}X_{k+1}^q=\mathbb{E}\left(1+a\Delta W_k\right)^q\mathbb{E}X_k^q$$ Using the initial condition, we find that
Now, this is what confuses me. How does my professor use the initial condition $X(0)=1$? $$\mathbb{E}X_k^q=\left(\mathbb{E}(1+a\Delta W_k)^q\right)^k=C^k,$$ where $$C=\mathbb{E}(1+a\Delta W_k)^q=\frac{1}{\sqrt{2\pi h}}\int_{-\infty}^{\infty}(1+ax)^qe^{-x^2/(2h)}dx.$$
Introducing the change of variables $x\mapsto h^{1/2}x$ leads to $C=\alpha(ah^{1/2},q)$ and hence to the result.