Abstract
We solve the problem of optimal inventory management for a CARA market-maker who faces proportional transaction costs and marking to market. Our model accommodates inventory shocks following an arbitrary compound Poisson process. We show that the no-trading region is always wider in the presence of inventory shocks.
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Notes
Marking to market is a common practice for most over-the-counter (OTC) derivatives such as credit default swaps (CDS), as well as for futures contracts and currency forwards. Our model is particularly relevant for currency (Forex) markets that are highly decentralized, with smaller dealers managing inventories of their clients and infrequently rebalancing using larger dealers (typically, major investment banks) as liquidity providers. Trades with liquidity providers often imply significant transaction costs, and hence the optimal inventory-management problem is important. Smaller dealers (such as, e.g., retail aggregators who cater to retail clients) continuously mark to market their clients’ accounts. Major dealers have accounts for smaller dealers, hedge funds (through the prime brokerage services), etc. These accounts are also continuously marked to market.
As in our model, Liu [33] assumes that the agent maximizes CARA preferences from intermediate consumption. However, in his model there is no marking-to-market, prices follow geometric Brownian motions, and there are no inventory shocks.
Garleanu [14] considers the problem of optimal inventory management in the presence of preference shocks, which are qualitatively equivalent to inventory shocks as they both induce the agent to rebalance his asset position.
The independence assumption is made for technical convenience. One could imagine that, in the real world, order flow does depend on the price level, implying a correlation between \(X_t\) and \(B_t.\) Investigating optimal policies in the presence of such dependencies is an interesting topic for future research.
The assumption of arithmetic Brownian motion is often used when modelling contracts that are marked to market. See, for example, Bichuch and Shreve [4]. In this case, price level is irrelevant, and what matters are the mark-to-market returns, which are given by the increments of the Brownian motion.
This assumption is standard in the literature. See, for example, Vayanos [47], Mania and Schweizer [39], Kondor and Vayanos [30], Gromb and Vayanos [21], and Kyle, Obizhaeva, and Wang [31]. The key simplification used in all these papers (as well as in our paper) is that, with a CARA utility, wealth level has no impact on portfolio choice. The drawback of this approach is that the liquid (cash) wealth can become negative. As is common in the literature, we (partially) circumvent this problem by assuming that the initial wealth, \(W_0,\) is sufficiently high, so that the probability of wealth ever becoming negative in the future is negligible. In addition, the possibility of negative wealth implies the possibility of negative consumption: In this case, we assume that such negative consumption is financed through short-term debt, which is continuously rolled over. Since we are assuming that default is impossible, this debt is risk free. However, this means that we prohibit the agent to exit even when his cash balance, \(W_t\) becomes very low.
As usual, CARA preferences imply that consumption can be negative. In the real world, it would mean that the agent (say, a market maker) is borrowing money, in which case negative consumption can be interpreted as debt coupon payments.
Existence of a solution this equation is provided in the Appendix.
For example, the proof follows directly from Theorem 3 below.
Note that \(q_t^*,\ M_t^{*,\pm }\) are pure jump processes and the existence of a unique solution to (6) is straightforward.
In particular, since a is convex, we have \(|a'(q)|\le \varphi \) for all \(q\in {{\mathbb {R}}}.\)
Indeed, \(e^{-\psi |x|+a(q+x)-a(q)}\ \le \ e^{-(\psi -\varphi )|x|}\) and hence standard arguments imply that differentiating is allowed under the integral sign.
The fact that the no-trade region is always wider in the presence of idiosyncratic shocks has also be established in Garleanu [14] in a different setting with a binary taste shock, without inventory shocks.
The joint concavity follows directly from the convexity of a(q).
The latter follows directly from (7).
This follows directly from the Ito formula.
Note that, while \(\Phi (q)\not \in X,\) the map I(q) is still well defined for \(\Phi (q)\) and we have that \({{\mathcal {I}}}(\Phi (q))\in X.\)
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We thank Remy Chicheportiche, Julien Hugonnier, Haobo Jia, and Johannes Muhle-Karbe for helpful comments and remarks.
Semyon Malamud thanks the Swiss Finance Institute and the Swiss National Science Foundation for financial support.
Proof of Theorem 3
Proof of Theorem 3
Road Map.
The proof of Theorem 3 is organized as follows:
Assuming a smooth convex solution to the HJB equation exists, Lemma 7 proves that the candidate strategy is admissible.
Assuming a smooth convex solution to the HJB equation exists, Lemma 8 shows that the candidate strategy is optimal in the class of admissible strategies.
Lemma 14 proves the existence of a smooth, convex solution to the HJB equation. The proof is based on auxilliary Lemmas 9–13.
Lemma 7
Suppose that a(q) is a convex function solving the HJB equation (4). Then, the conjectured optimal policy is admissible.
Proof
First we notice that the conjectured optimal policy satisfies
and hence \(M_t^{*,+} - M_t^{*,-} + X_t\) stays uniformly bounded.
Define
Then, the fees collected by the agent over the time interval [0, t] are given by \(\alpha Y_t.\) We have that, for the conjectured optimal policy,
Thus,
and where we have used that the Brownian motion is independent on the inventory process.
Since a is convex, we have \(a(q+x)-a(q)\ \ge \ a'(q)x,\) and hence
By definition,
Thus, \(M_0^++M_0^-\le \max \{(q_0-q_H)^+,(q_L-q_0)^+\}\ \le \ |q_0|+q_L+q_H.\) Furthermore, since \(q_t^*\in [q_L,q_H],\) we always have
and
Thus,
Therefore, with probability one, we have
because, with probability one, no jumps occur at time zero. Hence, using the formula for moment generating functions of compound Poisson processes, we get
Thus, by (8), we have
and the admissibility follows from Assumption 1. \(\square \)
Lemma 8
Suppose that a(q) is a convex function solving the HJB equation (4). Then, the corresponding policy is optimal among the set of admissible policies.
Proof
Pick an arbitrary admissible policy \(\{{\hat{c}}_t,\ {\hat{M}}_t^+,{\hat{M}}_t^-\},\) and let \({\hat{q}}_t=X_t+{\hat{M}}_t^+-{\hat{M}}_t^-\) be the corresponding inventory process. Let also \(W_t\) be the corresponding wealth process. Then, by the Ito formula for jump-diffusions, we have
where \(M^{\pm ,c}\) denote the continuous parts of the monotone increasing processes \(M^{\pm }.\) The function \(V(w,q)=-e^{-r\gamma w+a(q)}\) is monotone increasing in w and is jointly concave in (w, q).Footnote 18 Furthermore, \(\Delta {\hat{M}}_t^++\Delta {\hat{M}}_t^->\Delta {\hat{M}}_t^+-\Delta {\hat{M}}_t^-=\Delta M_t,\) where we have defined \(M_t=M_t^+-M_t^-.\)
Furthermore, by (2),
Furthermore, by the admissibility conditions, \(a'(q)\in [-\varphi ,\varphi ].\) Since \(q_t\) always stays bounded under any admissible policy, random variables \(e^{-r\gamma W_t+a(q_t)}\) have a uniformly bounded second moment as a function of t.Footnote 19 It thus follows that we can take the expectation of both sides of the equation, and the martingale parts vanish, while the HJB equation implies
Taking the limit as \(T\rightarrow \infty \) and using the admissibility conditions, we get
because, due to admissibility, \(q_t\) stays bounded. Thus,
and the equality takes place for the conjectured optimal policy by Lemma 7 because the latter is admissible and because for the conjectured optimal policy \(V(W_t,q_t)\) is a martingale.Footnote 20\(\square \)
Let now \(X\subset C({{\mathbb {R}}})\) to be the set of convex, continuously differentiable functions such that there exists a \(Q>0\) such that \(a'(q)=-a'(-q)=\varphi \) for all \(|q|>Q.\) Equip this set with a metric
Importantly, while each function \(a,b\in X\) is unbounded, their difference is always constant for sufficiently large x, and hence is bounded, implying that the metric is well defined.
Lemma 9
The map \({{\mathcal {I}}}:\ X\rightarrow X\) defined via
satisfies
for any \(a_1,a_2\in X.\) Furthermore, the map preserves pointwise inequality: for any two functions \(a_1,a_2\) with \(a_1(q)\le a_2(q),\) we have
for all q. Furthermore, \({{\mathcal {I}}}(a(\cdot ))(q)\) has a derivative bounded in absolute value by \(\varphi \) for any convex function \(a\in X.\)
Proof of Lemma 9
Follows by standard arguments. \(\square \)
Recall that convex functions have left and right derivatives at every point, and hence the Lipschitz constant for a convex function a(q) is the maximum of the suprema of the absolute values of its left and right derivatives.
Lemma 10
Define
Then, \({{\mathcal {Z}}}\) maps X into the set of convex functions with Lipschitz constant less than or equal to \(\varphi ,\) and satisfies
and \({{\mathcal {Z}}}(a(\cdot ))(q)\le a(q)\) for all \(q\in {{\mathbb {R}}}.\)
Proof
The last claim follows because \(a(q+x)-a(q)\le \varphi |x|\) implies
Here, we have used that \(\alpha >\alpha _l\) and, hence, \(\psi >\varphi .\) Now, since all functions \(a\in X\) have a Lipschitz constant less than or equal to \(\varphi ,\) we have
and we get the claim about the Lipschitz constant . To prove the claim about convexity, we note that, since a is convex, we have
where in the last step we have used the Cauchy-Schwarz inequality.
Now, by direct calculation, the Frechet differential of \({{\mathcal {Z}}}\) is given by
and hence
\(\square \)
Lemma 11
Let \(G(z,q):\ {{\mathbb {R}}}^2\rightarrow {{\mathbb {R}}}\) be the function defined implicitly as the unique solution to
Then, the function G is monotone increasing in z and is jointly convex in \((z,q)\in {{\mathbb {R}}}^2.\) Furthermore, if \(z\le \Phi (q),\) then
Proof
The function \(G + r^{-1}\lambda - r^{-1}\lambda e^{-G} e^z\) is monotone increasing in G and converges to \(\pm \infty \) when \(G\rightarrow \pm \infty .\) Hence, the equation \(G + r^{-1}\lambda - r^{-1}\lambda e^{-G} e^z = \Phi (q)\) always has a unique solution G(z, q). The convexity of G(z, q) in (z, q) follows by the implicit function theorem because the function \(r G + \lambda - \lambda e^{-G} e^z\ - r\Phi (q)\) is jointly concave in (G, z, q). The bound for the derivative of G follows because \(G(z,q) = \Phi (q)+r^{-1}\lambda e^{-G} e^z-r^{-1}\lambda \ge \Phi (q)-r^{-1}\lambda .\) Indeed, when \(z\le \Phi (q),\) we have
and therefore we get
and G is increasing because \(G_z\ge 0.\)\(\square \)
By direct calculation, we can rewrite the HJB equation inside the no-trading zone as
That is, for any \(q\in {{\mathbb {R}}},\) we have that a(q) equals G(z, q) evaluated at a \(z={{\mathcal {Z}}}(a(\cdot ))(q).\) Define the map \({\mathcal {F}}:\ a(q)\ \rightarrow {{\mathcal {I}}}\circ G({{\mathcal {Z}}}(a(\cdot )),q)\). Our goal is to show that iterations of this map converge to its unique fixed point.
This will follow from a sequence of technical lemmas. Define \(\tilde{X}\) to be the subset of X consisting of functions satisfyingFootnote 21
Define the map \({\mathcal {F}}:\ a(q)\ \rightarrow {{\mathcal {I}}}\circ G({{\mathcal {Z}}}(a(\cdot ))(q),q) \) on \({\tilde{X}}.\) That is, for any function \(a(q)\in {\tilde{X}},\) we define \({\mathcal {G}}(q)\ \equiv \ G({{\mathcal {Z}}}(a(\cdot ))(q),q)\) and we define \({{\mathcal {F}}}(a(\cdot ))(q)\) to be the function \({{\mathcal {I}}}({\mathcal {G}}(\cdot ))(q).\)
Lemma 12
The map \({\mathcal {F}}:\ a(q)\ \rightarrow {{\mathcal {I}}}\circ G({{\mathcal {Z}}}(a(\cdot )),q)\) maps the set \({\tilde{X}}\) into itself.
Proof
It suffices to prove that
for all \(a\in {\tilde{X}}.\) By Lemma 11, we always have
and hence
because \({{\mathcal {I}}}\) preserves pointwise inequality. Suppose now that \(a(q)\ \le \ \Phi (q)\). Then, by Lemma 10, we have \({{\mathcal {Z}}}(a(\cdot ))(q)\le \ a(q)\ \le \ \Phi (q),\) and hence, by Lemma 11,
and hence
Thus, \( {{\mathcal {I}}}\circ G({{\mathcal {Z}}}(a(\cdot )),q)\ \le \ {{\mathcal {I}}}(\Phi (\cdot ))\) because the map \({{\mathcal {I}}}\) preserves point-wise inequality. Thus, we have that
is convex, and satisfies the required inequalities; it also has derivative bounded in absolute value by \(\varphi .\) Finally, in order to show that \({\tilde{a}}(q)\in {\tilde{X}},\) it remains to prove that it indeed has derivative equal to \(\varphi \) in absolute value, for |q| sufficiently large. This follows directly from the inequality \({{\mathcal {I}}}(\Phi (\cdot ))(q)-r^{-1}\lambda \ \le \ {\tilde{a}}(q):\) Otherwise, this inequality cannot hold for |q| sufficiently large. \(\square \)
Lemma 13
The iterations of the map \({\mathcal {F}}:\ a(q)\ \rightarrow {{\mathcal {I}}}\circ G({{\mathcal {Z}}}(a(\cdot ))(q),q)\) converge to a unique fixed point in \(\tilde{X}.\)
Proof
We have that, for any q,
Since \(a_i\in {\tilde{X}},\) we have \(a_i(q)\le \ {{\mathcal {I}}}(\Phi (\cdot ))(q).\) Furthermore, by Lemma 10, \({{\mathcal {Z}}}(a_i(\cdot ))(q)\le a_i(q).\) Thus, \({{\mathcal {Z}}}(a_i(\cdot ))(q)\le {{\mathcal {I}}}(\Phi (\cdot ))(q),\) and hence \(t{{\mathcal {Z}}}(a_1(\cdot ))+(1-t){{\mathcal {Z}}}(a_2(\cdot ))\ \le \ {{\mathcal {I}}}(\Phi (\cdot ))\,.\) Therefore, by Lemmas 11 and 10,
Thus, the map \(G\circ {{\mathcal {Z}}}\) mapping a(q) to \(G({{\mathcal {Z}}}(a(\cdot ))(q),q)\) is a strict contraction with a contraction index \(\frac{r^{-1}\lambda e^{r^{-1}\lambda }}{1+r^{-1}\lambda e^{r^{-1}\lambda }}.\) Lemma 9 implies that \({{\mathcal {F}}}\) is also a strict contraction with the same index, and the claim follows from the contraction mapping theorem. \(\square \)
Lemma 14
The unique fixed point of \({\mathcal {F}}\) from Lemma 13 is \(C^1\) and solves the full HJB equation (4).
Proof of Lemma 14
Let a(q) be the fixed point of \({\mathcal {F}}\) from Lemma 13. Define \({\tilde{a}}(q)=G({{\mathcal {Z}}}(a(\cdot ))(q),q)\) and let \((q_L,q_H)\) be the largest interval on which \(|\tilde{a}'_{\pm }(q)|<\varphi \), where \({\tilde{a}}'_{\pm }\) are the left and right derivatives of the convex function \({\tilde{a}}.\) Since \(a(q)={{\mathcal {I}}}({\tilde{a}}(\cdot )(q)),\) by the definition of the map \({{\mathcal {I}}},\) we have that \(a(q)={\tilde{a}}(q)\) on \((q_L,q_H)\) and \(|a'(q)|=\varphi \) outside of that interval. Let us now show that \({\tilde{a}}(q)\) is everywhere differentiable and hence, by convexity, is \(C^1.\) Since the function G is smooth, it suffices to show that \({\tilde{{{\mathcal {Z}}}}}(q) =e^{{{\mathcal {Z}}}(a(\cdot ))(q)}\) is everywhere differentiable. The latter follows from the Lebesgue dominated convergence theorem. Indeed, we get
where we have defined
Since a(q) is Lipschitz continuous with degree \(\varphi \) and is almost everywhere differentiable (because it is convex), we have that, for any q, \(f(q,\varepsilon )\) converges almost surely to \(a'(q+x),\) while \(|f(q,\varepsilon )|\ \le \ \varepsilon ^{-1}|e^{\varepsilon \varphi }-1|\ \le \ C\) for some constant \(C>0.\) Thus, Lebesgue dominated convergence theorem applies, with the majorante \(e^{-(\phi -\varphi )|x|}C,\) and we get the required \(C^1\) property of \({\tilde{a}}.\) Since the map \({{\mathcal {I}}}\) preserves the \(C^1\) property, we get that a is also \(C^1\) and satisfies \(a(q)={{\mathcal {I}}}({\tilde{a}}(q))\ \le \ {\tilde{a}}(q)\) for all q outside of \([q_L,q_H].\) Thus, since, by the definition of G,
and the left-hand side is monotone increasing in \({\tilde{a}},\) we get that
Hence, we get that a indeed satisfies the HJB equation (4). \(\square \)
Proof of Theorem 3
By Lemma 14, HJB equation (4) has a convex solution. By Lemma 7, the policy corresponding to this solution is admissible. By Lemma 8, this policy is in fact optimal. The proof is complete. \(\square \)
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Gallien, F., Kassibrakis, S. & Malamud, S. Managing inventory with proportional transaction costs. Math Finan Econ 14, 121–138 (2020). https://doi.org/10.1007/s11579-019-00248-8
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DOI: https://doi.org/10.1007/s11579-019-00248-8