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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:

enter image description here

enter image description here

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?

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  • $\begingroup$ I'm having some problem deriving their result as well. $\endgroup$
    – Danny
    Jul 31 '18 at 8:07
  • $\begingroup$ @Danny nice knowing I’m not the only one struggling with this. As my concrete problem is solving the matrix equation, I’m going to do a post the question to math stackexchange once I’m back from vacation. Hopefully they can help. $\endgroup$
    – Mesalas
    Jul 31 '18 at 18:51
  • $\begingroup$ Ok i managed to figure out what went wrong with my derivation. Will go on to your actual question. $\endgroup$
    – Danny
    Aug 1 '18 at 13:46
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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 $\boldsymbol{\omega(t)}$.

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  • $\begingroup$ Awesome @Danny. Didn’t know about the toeplitz matrix. I’ll give it a thorough read once I’m back from vacation next week. 👍 $\endgroup$
    – Mesalas
    Aug 3 '18 at 15:22
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    $\begingroup$ See rendezvouswithdestiny.me/finance/stoch_ctrl_mm.html for my python numerical implementation at the bottom of the notes. $\endgroup$
    – Danny
    Aug 5 '18 at 15:42
  • $\begingroup$ Thats very thorough! And Its clever use of the eigen decomposition! Im going to accept your answer. Now its "just" finding expressions for the eigen values and eigen vectors of A. $\endgroup$
    – Mesalas
    Aug 6 '18 at 8:17
  • $\begingroup$ Thank you! Glad that it helps! I'm working through the Cartea stuff as well so I might have new questions later too! $\endgroup$
    – Danny
    Aug 7 '18 at 9:09
  • $\begingroup$ not sure where you get the information from that A is symmetric? as far as i see there are off-diagonal terms with lambda_plus and terms with lambda_minus that make the matrix asymmetric, unless you assume that the lambda's are equal, which is certainly not the case. or am i somehow mistaken in the matrix construction? best $\endgroup$
    – user101893
    Jul 28 at 15:32

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