I am reading paper High-frequency trading in a limit order book by Marco Avellaneda and Sasha Stoikov. Please help me to understand how they rewritten and obtained function

$$ v(x,s,q,t)= -\exp(-{\gamma}x)exp(-{\gamma}qs)exp(\frac{{\gamma}^2q^2{\sigma}^2(T-t)}{2}) (3) $$


$$ v(x,s,q,t)= E_t[-\exp(-{\gamma}(x+q{S_T}))] $$

And also in appendix they rewritten another utility function $$ V(x,s,q,t)= E_t[(x+q{S_T})-\frac{\gamma}{2}(q{S_T}-qs)^2] $$

into $$ V(x,s,q,t) = x+qs-\frac{{\gamma}q^2s^2}{2}(e^{{\sigma}^2(T-t)}-1) $$

I understand that they're trying to integrate Normal Distribution's PDF (=Brownian motion), but I can't achieve the same results during transformation of integrals


1 Answer 1


First part of your question:

The solution comes from solving the SDE specified in the paper and then using the moment generating function of a normal random variable.

Formally, let us define the SDE specified in the paper:

$$dS_u = \sigma dW_u,$$ with initial value $S_t = s$. To get the solution to the above SDE, we can integrate from $t$ to $T$ on both sides:

\begin{align*} \int_{t}^T dS_u &= \int_t^T\sigma dW_u.\\ &\Updownarrow\\ S_T - S_t &= \sigma (W_T - W_t) \\ &\Updownarrow\\ S_T &= S_t + \sigma (W_T - W_t) \end{align*} Since Brownian increments are normally distributed with $W_T - W_t \sim N(0, T-t)$ and we further have that $S_t = s$, we get the following distribution for $S_T \sim N(s, \sigma^2 (T-t))$.

Now, reformulating the expectation in your second formula gives us:

$$\mathbb{E}_t\left[-e^{-\gamma(x + qS_T)}\right] = -e^{-\gamma x}\mathbb{E}_t\left[e^{-\gamma q S_T}\right],$$

where $x$ is the initial wealth in dollars and is just a known constant.

Remember that the moment generating function of a Normal random variable with distribution $X \sim N(\mu, \sigma^2)$ is given by (see here for formula):

$$\mathbb{E}\left[e^{tX}\right] = e^{t\mu}e^{\frac{1}{2}\sigma^2t^2}. $$

Using this formula, we get the desired result:

$$-e^{-\gamma x}\mathbb{E}_t\left[e^{-\gamma q S_T}\right] = -\exp(-\gamma x)\exp(-\gamma qs) \exp\left(\frac{\gamma^2 q^2 \sigma^2 (T-t)}{2}\right)$$

Second part of your question:

We need to observe that the authors now use a driftless GBM as SDE:

$$\frac{dS_u}{S_u} = \sigma dW_u,$$

with initial value $S_t = s$. This has the solution:

$$S_T = S_t e^{-\frac{\sigma^2}{2}(T-t) + \sigma (W_T - W_t)}.$$

Calculating first and second moment:

Let us redefine $S_T$ as $$S_T \overset{d}{=} S_t e^{-\frac{\sigma^2}{2}(T-t) + \sqrt{T-t}\cdot x},$$

where $x$ is normal distributed with $x \sim N(0, \sigma^2)$ and furthermore its squared counterpart is given by:

$$S_T^2\overset{d}{=} S_t^2 e^{-\sigma^2(T-t) + 2\sqrt{T-t}\cdot x}.$$

Calculating the moments follows from the moment generating function defined above and we conclude that:

\begin{align} \mathbb{E}_t\left[S_T\right] &= \mathbb{E}_t\left[S_t e^{-\frac{\sigma^2}{2}(T-t) + \sqrt{T-t}\cdot x}\right] \\ &= S_t e^{-\frac{\sigma^2}{2}(T-t)} \mathbb{E}_t\left[e^{\sqrt{T-t}x} \right]\\ &= S_t e^{-\frac{\sigma^2}{2}(T-t)}e^{\frac{\sigma^2}{2}(T-t)}\\ &= S_t \end{align} \begin{align} \mathbb{E}_t\left[S_T^2\right] &=\mathbb{E}_t\left[ S_t^2 e^{-\sigma^2(T-t) + 2\sqrt{T-t}\cdot x}\right]\\ &= S_t^2 e^{-\sigma^2(T-t)} \mathbb{E}_t\left[e^{2\sqrt{T-t}x}\right]\\ &= S_t^2 e^{-\sigma^2(T-t)}e^{2\sigma^2(T-t)}\\ &= S_t^2 e^{\sigma^2(T-t)}, \end{align} and remember that $S_t = s$.

Now, rewriting your third equation and inserting the above moments, we get the answer as depicted in your fourth equation: \begin{align} \mathbb{E}_t\left[x + qS_T - \frac{\gamma}{2} \left(qS_t - qs\right)^2\right] &= \mathbb{E}_t\left[x + qS_T - \frac{\gamma}{2} q^2(S_t - s)^2\right]\\ &= x + q \mathbb{E}_t\left[S_T\right] - \frac{\gamma}{2} q^2 \mathbb{E}_t\left[(S_T-s)^2\right]\\ &= x + qs - \frac{\gamma q^2}{2} \mathbb{E}_t\left[S_T^2 + s^2 - 2 S_T\cdot s\right]\\ &= x + qs - \frac{\gamma q^2}{2} \left(\mathbb{E}_t\left[S_T^2\right] + s^2 - 2s^2\right)\\ &= x + qs - \frac{\gamma q^2}{2} \left(s^2 e^{\sigma^2 (T-t)} - s^2\right)\\ &= x + qs - \frac{\gamma q^2 s^2}{2} \left(e^{\sigma^2 (T-t)} - 1\right)\\ \end{align}

I hope this provides a bit of help.

  • $\begingroup$ @Mike, If you found this answer helpful, could you please consider accepting it by clicking the green tick-mark below the score? Otherwise, write back to me and tell me if you have any problems with the above derivations, then I can explain further :-) $\endgroup$
    – Pleb
    Commented Jun 22, 2023 at 9:24

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