Deriving a Terminal-Price Expectation with Brownian Increments
Summary
The document explains an expectation used in a limit-order-book model with an underlying stock-price process driven by Brownian motion. It focuses on how to transform an exponential utility expression involving the terminal price into a form that can be evaluated from the distribution of the price change between the current time and the terminal time. The derivation separates the current stock price from the future increment, then uses the increment’s normal distribution and characteristic function.
Under the model stated in the paper, the increment has zero mean and variance determined by volatility and the remaining time. Evaluating the characteristic function at an imaginary argument yields the exponential term in the displayed expression. A second reply offers a martingale and normal-density route, though it characterizes the process differently. The answers are brief, and the derivation depends on the precise price-process assumptions in the paper; the replies do not resolve that modeling distinction.
Key ideas
- Separate the current price from the future price increment before evaluating the exponential expectation.
- Under the stated zero-drift Brownian price-change model, the increment is normally distributed with variance set by volatility and elapsed time.
- The normal increment’s characteristic function evaluates the exponential expectation at an imaginary argument.
- The alternative martingale explanation depends on the process specification, which the brief responses describe inconsistently.
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# Simple question about expected value of brownian motion
# 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?).
## Answer by pbr142 (score 1, accepted)
https://quant.stackexchange.com/a/10857
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}
## Answer by quasi (score 1)
https://quant.stackexchange.com/a/10856
I only glanced at it for a second. But, looks like the underlying process is arithmetic Brownian motion, and they're computing conditional expectation of it's expectation at time t. So, basically, use the fact that
$$ \exp(\sigma W_t - \frac{1}{2}\sigma^2 t)$$ is a martingale, and then with the remaining independent increment, calculate directly using the pdf of the normal density.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.