# Simple question about expected value of brownian motion

I would appreciate some help with the math in this paper : High Frequency Trading in a Limit Order Book

Specifically, I would like to understand how the authors calculated the expected value of price at terminal time T at current time t. What substitution was made to arrive at equation (3) on page 2 (from the value function directly above?).

The equation can easily be derived from the characteristic function of the geometric Brownian motion. As stated in the footnote, the authors use $$\frac{dS_t}{S_t} = \sigma dW_t$$ as the underlying model. The change in stock price $X_T = S_T - S_t$ is therefore normally distributed with mean 0 and variance $\sigma^2 (T-t)$. The characteristic function of the stock price change then follows as: $$\phi(u) = \mathbb{E}\left[ e^{iuX_T} \right] = e^{-\frac{1}{2} u^2 \sigma^2 (T-t)}.$$ The expression they evaluate can then be transformed as: \begin{align} \mathbb{E} \left[ -\exp\left\{-\gamma(x+qS_T)\right\}\right] =&\ -e^{-\gamma x} \mathbb{E} \left[ -\exp\left\{-\gamma q (S_T - S_t + S_t)\right\}\right] \\ =&\ -e^{-\gamma x} e^{-\gamma q s} \mathbb{E} \left[ -\exp\left\{-\gamma q X_T\right\}\right] \\ =&\ -e^{-\gamma x} e^{-\gamma q s} \phi(i\gamma q) = -e^{-\gamma x} e^{-\gamma q s} e^{\frac{1}{2} \gamma^2 q^2 \sigma^2 (T-t)} \end{align}
$$\exp(\sigma W_t - \frac{1}{2}\sigma^2 t)$$ is a martingale, and then with the remaining independent increment, calculate directly using the pdf of the normal density.