Solving the HJB Market-Making Model with Matrix Exponentials

# Market Making Strategies Found by Hamilton-Jacobi-Bellman Equation


# Market Making Strategies Found by Hamilton-Jacobi-Bellman Equation












Im working my way through the book "Algorithmic and High-Frequency Trading" (AHFT) by Cartea, Jaimungal and Penalva and i'm curious to see how the market making model with an exponential utility function compares to the similar model published by Avelaneda and Stoikov, but im having some problems understanding the result from the book.

In both models the bid and ask prices entered into the order book are calculated from an inventory dependent function, $\theta$ in 2 and $g(t,q)$ or $h(t,q)$ in 1.

To me the main differences seem to be that in in the paper by A&S, this inventory dependent function is expanded in terms of the inventory parameter $q$. And importantly A&S makes a linear approximation of the exponential terms in the exponential part of the HJB equation for $\theta$ (see eq 26 in 2).

In 1 the authors never explicitely calculate the inventory dependent function, $h(t,q)$, but state that one can find it via:

However, I cant figure out how to do the product between $e^\mathbf{A}$ and $\mathbf{z}$ in order to calculate $\omega(t,q)$ and $h(t,q)$ for a given inventory $q$.

Can you help me understand this better?

## Answer by Danny (score 3, accepted)

https://quant.stackexchange.com/a/41087

At the terminal time $T$, the terminal condition is $g(T, q) = -\alpha q^2$, this implies, $$ \begin{aligned} g(T, q) &= \frac{1}{\kappa} \log{\omega(T, q)} = -\alpha q^2\\ \Rightarrow \omega(T,q) &= e^{-\kappa\alpha q^2} \end{aligned} $$ Therefore, $\mathbf{z}$ is given by, $$ \mathbf{z} = \boldsymbol{\omega(T)} = \begin{bmatrix} e^{-\alpha\kappa \bar{q}^2}\\ e^{-\alpha\kappa (\bar{q}-1)^2}\\ \vdots\\ e^{-\alpha\kappa (\underline{q}+1)^2}\\ e^{-\alpha\kappa \underline{q}^2}\\ \end{bmatrix} $$ Next we turn our attention to the term $e^{\mathbf{A}(T-t)}$. The matrix $\mathbf{A}(T-t)$ is a real symmetric matrix, so we know that we can perform eigen-decomposition of $\mathbf{A}(T-t)$ into, $$ \mathbf{A}(T-t) = PDP^{-1} $$ with $D$ the diagonal matrix with elements being the eigenvalues $\lambda_j$, and $P$ is a matrix whose columns are the corresponding eigenvectors $v_j$.

Finally, we use the standard result, $$ \begin{aligned} e^{\mathbf{A}(T-t)} &= Pe^{D}P^{-1}\\ &=P\begin{bmatrix} e^{\lambda_1} & 0 & \ldots & 0\\ 0 & e^{\lambda_2}& \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\ldots & e^{\lambda_{(\bar{q} - \underline{q}+1)}} \end{bmatrix}P^{-1} \end{aligned} $$ which would enable us to work out the final expression for $\boldsymbol{\omega(t)}$.