Abstract
We study how trading costs are reflected in equilibrium returns. To this end, we develop a tractable continuous-time risk-sharing model, where heterogeneous mean–variance investors trade subject to a quadratic transaction cost. The corresponding equilibrium is characterized as the unique solution of a system of coupled but linear forward–backward stochastic differential equations. Explicit solutions are obtained in a number of concrete settings. The sluggishness of the frictional portfolios makes the corresponding equilibrium returns mean-reverting. Compared to the frictionless case, expected returns are higher if the more risk-averse agents are net sellers or if the asset supply expands over time.
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Notes
Mean-reverting fundamentals also drive the mean-reverting dynamics in the overlapping-generations model with linear costs studied in [44], for example.
This will be the time-discount rate below; for infinite-horizon models, it needs to be strictly positive.
If one instead assumes that the volatility follows some (sufficiently integrable) stochastic process , the subsequent characterization of individually optimal strategies and equilibrium returns in terms of coupled but linear FBSDEs as in (A.1) and (A.2) still applies. However, the stochastic volatility then appears in the coefficients of this equation, so that the solution can no longer be characterized (semi-)explicitly in terms of matrix power series. Instead, a “backward stochastic Riccati differential equation” appears as a crucial new ingredient already in the one-dimensional models with exogenous price dynamics studied by [31, 6].
The assumption of quadratic rather than proportional costs is made for tractability. However, buoyed by the results from the partial equilibrium literature, we expect the qualitative properties of our results to be robust across different small transaction costs; compare with the discussion in [36].
More general specifications do no seem natural for the tax interpretation of the model. Note, however, that the mathematical analysis below only uses that \(\varLambda\) is symmetric and positive definite.
This means that agents stop trading near maturity when there is not enough time left to recuperate the costs of further transactions. If \(T=\infty\), this terminal condition is replaced by the transversality conditions implicit in for \(\delta>0\).
Several groups of noise traders with different mean positions as considered in [18, Sect. 4] can be treated analogously.
Note that \(\int_{0}^{\infty}e^{-\sqrt{\Delta} s} B \tilde{\xi}_{s} \,ds\) is square-integrable because and all eigenvalues of \(\sqrt{\Delta}\) are at least \(\delta/2\).
Note that the inverses are well defined by Lemma A.3.
If \(A \in\mathbb{R}^{\ell\times\ell}\) and all eigenvalues of \(A\) are real, \(O\) can be taken as an open neighborhood of \(\lambda_{1}, \dots, \lambda_{m}\) in ℝ, provided that \(f\) is also real-valued.
If \(A\), \(\lambda_{1}, \ldots, \lambda _{m}\), \(O\) and \(f\) are all real-valued and \(f\) is defined on the spectrum of \(A\), the Hermite interpolating polynomial is also real-valued.
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Acknowledgements
The authors are grateful to Michalis Anthropelos, Peter Bank, Paolo Guasoni and Felix Kübler for stimulating discussions and detailed comments. Moreover, they thank an anonymous referee for his or her careful reading and pertinent remarks. Parts of this paper were written while the fourth author was visiting ETH Zürich; he is grateful to the Forschungsinstitut für Mathematik and H.M. Soner for their hospitality.
The second and third author were partially supported by KAKENHI Grant number 25245046 and by the Swiss National Science Foundation (SNF) under grant 150101, respectively.
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Appendices
Appendix A: Existence and uniqueness of linear FBSDEs
For the determination of both individually optimal trading strategies in Sect. 4 and equilibrium returns in Sect. 5, this appendix develops existence and uniqueness results for systems of coupled but linear FBSDEs of the formFootnote 9
where \(B \in\mathbb{R}^{\ell\times\ell}\) has only positive eigenvalues, \(\delta\geq0\) and . If \(T<\infty\), (A.2) is complemented by the terminal condition
If \(T = \infty\), we assume that \(\delta> 0\) and the terminal condition is replaced by the transversality conditions implicit in for \(\delta>0\). A solution of (A.1) and (A.2) is a triple \((\varphi, \dot{\varphi}, M)\) for which and \(M\) is a martingale on with finite second moments.
We first consider the infinite-horizon case. In this case, the linear FBSDEs (A.1) and (A.2) can be solved using matrix exponentials similarly as in [18]. To this end, we first establish a technical result stating that the martingale \(M\) appearing in the solution of the FBSDE (A.1) and (A.2) automatically belongs to the space .
Proposition A.1
Let \(T = \infty\). If \((\varphi, \dot{\varphi}, M)\) is a solution to the FBSDE (A.1) and (A.2), then .
Proof
Let \((\varphi, \dot{\varphi}, M)\) be a solution to the FBSDE (A.1) and (A.2), where and \(M\) is a martingale on \([0, \infty)\) with finite second moments. Fix \(t \in(0, \infty)\). Then integration by parts yields
Let \(\Vert\cdot\Vert_{2}\) be the Euclidean norm in \(\mathbb{R}^{\ell }\) and \(\Vert\cdot\Vert_{\max}\) the maximum norm on \(\mathbb {R}^{\ell\times\ell}\). Rearranging (A.4) and using the elementary inequality \((a + b + c)^{2} \leq3(a^{2} + b^{2} + c^{2})\) for \(a, b, c \in\mathbb{R}\), the estimate \(\Vert A x \Vert_{2} \leq \sqrt{\ell} \Vert A \Vert_{\max} \Vert x \Vert_{2}\) and Jensen’s inequality gives
From the definitions of the maximum norm and the Euclidean norm, the estimate \(|[N^{1}, N^{2} ]| \leq\frac{1}{2} ([N^{1} ]+ [N^{2}])\) for real-valued local martingales \(N^{1}\) and \(N^{2}\) and Itô’s isometry, we obtain, for \(t \in(0, \infty)\),
Since , there exists an increasing sequence \((t_{k})_{k \in\mathbb{N}}\) with \(\lim_{k \to\infty} t_{k} =\infty\) such that
Monotone convergence, (A.5)–(A.7) and in turn yield
Thus , as claimed. □
Theorem A.2
Suppose that \(T = \infty\), \(\delta> 0\) and the matrix \(B\) from (A.2) has only positive eigenvalues. Set \(\Delta= B + \frac {\delta^{2}}{4} I_{\ell}\). Then the unique solution of the FBSDE (A.1) and (A.2) is given by
where
Proof
Let \((\varphi,\dot{\varphi}, M)\) be a solution to the FBSDE (A.1) and (A.2) and define
Using that
and the FBSDE (A.1) and (A.2) for \((\varphi ,\dot{\varphi})\), it follows that \((\tilde{\varphi},\dot{\tilde{\varphi}})\) solves the FBSDE
where \(d\tilde{M}_{t} = e^{-\frac{\delta}{2} t} dM_{t}\) and \(\tilde{\xi}_{t} = e^{-\frac{\delta}{2} t} \xi_{t}\). In matrix notation, this equation can be rewritten as
with
Integration by parts shows
and in turn, for \(t< u < \infty\), we obtain
Multiplying (A.11) by the matrix
and setting \(H(t)=\tilde{C}_{2} e^{-C_{2} t}\) yields, for \(t< u < \infty\),
It follows by induction that for \(n \geq0\),
Now the power series for the exponential function allows deducing that
Together with (A.12), we obtain, for \(t< u < \infty\),
By the assumption that , (A.10) and the fact that all eigenvalues of \(\sqrt{\Delta}\) are greater than or equal to \(\delta /2\) (because \(B\) has only nonnegative eigenvalues), there exists an increasing sequence \((u_{k})_{k \in\mathbb{N}}\) with \(\lim_{k \to \infty} u_{k} = + \infty\) along which the left-hand side of (A.13) converges a.s. to zero. Moreover, since by Proposition A.1 and , the martingale convergence theorem and monotone convergence (together with Jensen’s inequality)—also using that all eigenvalues of \(\sqrt{\Delta}\) are at least \(\delta/2\)—imply that for \(u \to\infty\) (and a fortiori along \((u_{k})_{k \in\mathbb{N}}\)), the right-hand side of (A.13) converges a.s. to
Together, these two limits show that (A.14) vanishes. Multiplying (A.14) by \(e^{\sqrt {\Delta} t}\), rearranging and taking conditional expectations (using again that all eigenvalues of \(\sqrt{\Delta}\) are larger than or equal to \(\delta/2\)), we obtain
Now (A.10) and rearranging give
Finally, since \(B = (\sqrt{\Delta} - \frac{\delta}{2} I_{\ell})(\sqrt {\Delta} + \frac{\delta}{2} I_{\ell}) \) commutes with \(e^{-\sqrt{\Delta} (s-t)}\) by Lemma B.2(a), it follows that
By the variations of constants formula, this linear (random) ODE has the unique solution (A.8). If a solution of the FBSDE (A.1) and (A.2) exists, it therefore must be of the form (A.8).
It remains to verify that (A.8) indeed solves the FBSDE (A.1) and (A.2). To this end, we first show that . Denote by \(\Vert\cdot \Vert_{2}\) both the Euclidean norm in \(\mathbb{R}^{\ell}\) and the spectral norm in \(\mathbb{R}^{\ell\times\ell}\). Since all eigenvalues of \(B\) are positive, there is \(\varepsilon> 0\) such that all eigenvalues of \(\sqrt{\Delta} + \frac{\delta}{2} I_{\ell}\) are at least \(\delta+ \varepsilon\). Hence, by Lemma B.2(c) and the definition of the spectral norm, it follows that
Thus by the definition of \(\bar{\xi}\) in (A.9), the fact that \(B = (\sqrt{\Delta} - \frac{\delta}{2} I_{\ell})(\sqrt {\Delta} + \frac{\delta}{2} I_{\ell}) \), Jensen’s inequality and Fubini’s theorem, we obtain
Next we show that . Arguing similarly as above, we have
Thus by the definition of \(\varphi\) in (A.8), Jensen’s inequality and Fubini’s theorem and since by the above arguments, we obtain
By definition, we have \(\varphi_{0}=0\). Next, integration by parts shows that \(\dot{\varphi}\) satisfies the ODE (A.15), and this yields because . Define the \(\mathbb{R}^{\ell}\)-valued square-integrable martingale \((\bar {M}_{t})_{t \in[0, \infty)}\) byFootnote 10
where \(\tilde{\varphi}_{t} := e^{-\frac{\delta}{2}t} \varphi_{t}\) and \(\tilde{\xi}_{t} := e^{-\frac{\delta}{2}t} \xi_{t}\) as before. Then multiplying (A.15) by the matrix \(e^{-(\sqrt{\Delta} + \frac{\delta}{2}I_{\ell})t}\) and using (A.10) as well as \(\Delta- \frac{\delta^{2}}{4} I_{\ell}= B\) gives, after some rearrangement,
Taking differentials, we therefore obtain
Rearranging, multiplying by \(e^{\sqrt{\Delta} t}\) and using that \(\sqrt{\Delta}\) and \(e^{\sqrt{\Delta} t}\) commute, it follows that
Finally, again taking into account (A.10) and defining the martingale \(M\) (which has finite second moments) by
we obtain that \(\varphi\) from (A.8) indeed satisfies (A.1) and (A.2). □
Let us briefly sketch the financial interpretation of the solution; cf. [18] for more details. In the context of individually optimal trading strategies (cf. Lemma 4.1), the ODE (A.15) describes the optimal trading rate. It prescribes to trade with a constant relative speed \(\sqrt {\Delta} - \frac{\delta}{2} I_{\ell}\) towards the target portfolio
In the context of Lemma 4.1, this is an average of the future values of the frictionless optimal trading strategy \(\xi\), computed using an exponential discounting kernel. As the trading costs tend to zero, the discount rate tends to infinity and the target portfolio approaches the current value of the frictionless optimizer, in line with the small-cost asymptotics of [36].
We now turn to the finite-horizon case. In order to satisfy the terminal condition \(\dot{\varphi}_{T}=0\), the exponentials from Theorem A.2 need to be replaced by appropriate hyperbolic functions in the one-dimensional case [5]. In the present multivariate context, this remains true if these hyperbolic functions are used to define the corresponding “primary matrix functions” in the sense of Definition B.1. The first step to make this precise is the following auxiliary result, which is applied for \(\Delta=B+\frac{\delta^{2}}{4}I_{\ell}\) in Theorem A.4 below.
Lemma A.3
Let \(\Delta\in\mathbb{R}^{\ell\times\ell}\). The matrix-valued function
is twice continuously differentiable on ℝ with derivative
and solves the ODE
Moreover, if the matrix \(\Delta\) has only positive eigenvalues, then in the sense of Definition B.1,
for \(\delta\geq0\), the matrix \(\Delta G(t) - \frac{\delta}{2} \dot {G}(t)\) is invertible for any \(t \in[0, T]\), and for any matrix norm \(\Vert\cdot\Vert\),
Proof
Note that \(\sum_{n = 0}^{\infty}\frac{1}{(2n)!} \Vert\Delta\Vert^{n} (T - t)^{2n} < \infty\) for any matrix norm \(\Vert\cdot\Vert\) so that \(G(t)\) is well defined for each \(t \in\mathbb{R}\). By twice differentiating term by term and estimating the resulting power series in the same way, it is readily verified that \(G\) is twice continuously differentiable on ℝ, has the stated derivative and is a solution of (A.16).
Suppose now that \(\Delta\) has only positive eigenvalues and \(\delta \geq0\). Then the first two additional claims follow from Definition B.1 via the fact that \(B = (\sqrt {B})^{2}\) and the series representation of the smooth functions cosh and sinh. The final claim follows from Lemma B.2(c) and (d) since for fixed \(x \in(0,\infty)\),
□
The unique solution of our FBSDE can now be characterized using the function \(G(t)\) from Lemma A.3 as follows.
Theorem A.4
Suppose that \(T < \infty\) and that the matrix \(\Delta=B + \frac {\delta^{2}}{4} I_{\ell}\) has only positive eigenvalues. Then the unique solution of the FBSDE (A.1) and (A.2) with terminal condition (A.3) is given by
whereFootnote 11
Proof
Let \((\varphi,\dot{\varphi})\) be a solution of the FBSDE (A.1) and (A.2) with terminal condition (A.3) and set
Using that
and the FBSDE (A.1) and (A.2) for \((\varphi ,\dot{\varphi})\) with (A.3), it follows that \((\tilde{\varphi},\dot{\tilde{\varphi}})\) solves the FBSDE
with terminal condition
Here \(d\tilde{M}_{t} =e^{-\frac{\delta}{2} t} dM_{t}\) is like \(M\) a square-integrable martingale and \(\tilde{\xi}_{t} = e^{-\frac{\delta}{2} t} \xi_{t}\). In matrix notation, this FBSDE can be rewritten as
with
Integration by parts shows
and in turn
Set \(H(t)=e^{C_{2}(T-t)}\) and note that
for the function \(G(t)\) from Lemma A.3, as is readily verified by induction. Together with (A.21), it follows that
Since \(\dot{\tilde{\varphi}}_{T}=-\frac{\delta}{2} \tilde{\varphi}_{T}\) by (A.20), this in turn yields
Multiplying this equation by \(\Delta\) and taking conditional expectations gives
Now using (A.19) and rearranging, it follows that
After multiplying with the inverse of the matrix \(\Delta G(t) -\frac {\delta}{2} \dot{G}(t)\) (which exists by Lemma A.3) and using that \(\Delta- \frac{\delta^{2}}{4} I_{\ell}= B\), this leads to
By the variations of constants formula, this linear (random) ODE has the unique solution (A.18). If a solution of the FBSDE (A.1) and (A.2) exists, it therefore must be of the form (A.18).
It remains to verify that (A.18) indeed solves the FBSDE (A.1) and (A.2). First, note that by the fact that and the estimate (A.17), and in turn . Moreover, by definition, we have \(\varphi_{0}=0\). Next, integration by parts shows that \(\dot{\varphi}\) satisfies the ODE (A.22), and this yields (because ) and \(\dot{\varphi }_{T}=0\) (because \(G(T)=I\) and \(\dot{G}(T)=0\)). Define the \(\mathbb {R}^{\ell}\)-valued square-integrable martingale \((\bar{M}_{t})_{t \in[0, T]}\) by
where \(\tilde{\varphi}_{t} := e^{-\frac{\delta}{2}t} \varphi_{t}\), \(\tilde{\xi}_{t} := e^{-\frac{\delta}{2}t} \xi_{t}\), as before. Multiplying (A.22) by \(\Delta G(t) -\frac{\delta}{2} \dot{G}(t)\) and using (A.19) as well as \(\Delta- \frac{\delta^{2}}{4} I_{\ell}= B\) gives, after some rearrangement,
Taking differentials, we therefore obtain
Using that \(\ddot{G}(t) = \Delta G(t)\) by the ODE (A.16) and taking into account that \(\Delta\) commutes with both \(G(t)\) and \(\dot{G}(t)\) by Lemma B.2(a), it follows that
Now multiplying with the inverse of the matrix \(\Delta G(t) -\frac {\delta}{2} \dot{G}(t)\) (which exists by Lemma A.3) and using that \(\Delta = B + \frac{\delta^{2}}{4} I_{\ell}\), we obtain
Finally, again taking into account (A.19) and defining the square-integrable martingale \(M\) by
we obtain that \(\varphi\) from (A.18) indeed satisfies the FBSDE dynamics (A.1) and (A.2) with terminal condition (A.3). □
Let us again briefly comment on the financial interpretation of this result in the context of Lemma 4.1. The basic interpretation is the same as in the infinite-horizon case studied in Theorem A.2. However, to account for the terminal condition that the trading speed needs to vanish, the optimal relative trading speed in (A.22) is no longer constant. Instead, it interpolates between this terminal condition and the stationary long-run value from Theorem A.2 that is approached if the time horizon is distant. Analogously, the exponential discounting kernel used to compute the target portfolio in Theorem A.2 is replaced by a more complex version here; compare [5] for a detailed discussion in the one-dimensional case.
To apply Theorems A.2 and A.4 to characterize the equilibrium in Theorem 5.2, it remains to verify that the matrix \(B\) appearing there only has real, positive eigenvalues, since this implies that the matrix \(\Delta:= B + \frac {\delta^{2}}{4} I_{\ell}\) has only real eigenvalues greater than \(\delta ^{2}/4 \geq0\).
Lemma A.5
Let \(\varLambda\in \mathbb{R}^{d \times d}\) be a diagonal matrix with entries \(\lambda^{1}, \ldots, \lambda^{d} > 0\), let \(\varSigma\in \mathbb{R}^{d \times d}\) be a symmetric, positive definite matrix and \(\gamma^{1},\ldots,\gamma^{N}>0\) with \(\gamma^{N} = \max(\gamma ^{1},\ldots,\gamma^{N})\). Then the \(\mathbb{R}^{d (N-1)\times d (N-1)}\)-valued matrix
has only real, positive eigenvalues.
Proof
First, recall that two matrices that are similar have the same eigenvalues. Since the matrix \(\frac{\varLambda^{-1} \varSigma}{2}\) is diagonalizable and has only positive eigenvalues (because it is the product of two symmetric positive definite matrices; cf. [40, Proposition 6.1]), there are an invertible matrix \(P \in \mathbb{R}^{d\times d}\) and a diagonal matrix \(U \in\mathbb {R}^{d\times d}\) with positive diagonal entries \(u^{1}, \ldots, u^{d}\) such that \(\frac{\varLambda^{-1} \varSigma}{2} = P U P^{-1}\). Now define the matrix
A direct computation shows that \(Q\) is invertible with inverse
Therefore \(B\) is similar to \(\bar{B} := P^{-1} B P\), and
To prove that \(\bar{B}\) (and hence \(B\)) only has real and positive eigenvalues, we calculate the determinant of \(V(x) = xI_{d (N-1)} - \bar{B}\) for \(x \in\mathbb{C} \setminus(0, \infty)\) and show that \(\det(V(x)) \neq0\). So let \(x \in\mathbb{C} \setminus(0, \infty)\). Denote by \(\mathcal{R}_{d}\) the commutative subring of all diagonal matrices in \(\mathbb{C}^{d\times d}\) and let \(\aleph^{1}, \ldots, \aleph^{N-1}, \gimel^{1}(x), \ldots, \gimel^{N-1}(x) \in\mathcal {R}_{d}\) be given by
With this notation, the \(\mathbb{R}^{d(N-1) \times d (N-1)}\)-valued matrix \(V(x)\) can also be understood as an element of \(\mathcal {R}_{d}^{(N-1)\times(N-1)}\) (the \((N-1)\times(N-1)\) matrices with elements from the diagonal matrices in \(\mathbb{R}^{d \times d }\)), and we have
Now use that by [41, Theorem 1], we have \(\det(V (x)) = \det( \mathfrak{det}(V(x)))\), where \(\mathfrak{det}: \mathcal {R}_{d}^{(N-1)\times(N-1)} \to \mathcal{R}_{d}\) is the determinant map on the commutative ring \(\mathcal{R}_{d}\). By subtracting the last row in \(\mathcal{R}_{d}\) of \(V(x)\) from the other rows, the problem boils down to calculating the determinant of
As \(x \in\mathbb{C} \setminus(0, \infty)\) and for \(n \in\{1, \ldots , N-1\}\), the eigenvalues of \(\gamma^{n} U\) are \(\gamma^{n} u^{1}, \ldots, \gamma^{n} u^{d} \in(0, \infty)\), it follows that \(\det(\gimel^{n} (x)) \neq0\) and hence \(\gimel^{n} (x)\) is invertible for each \(n\). Now, subtracting \(\aleph^{n} (\gimel^{n}(x))^{-1}\)-times the \(n\)th row from the last row for \(n = 1, \ldots, N-2\), the problem simplifies to calculating the determinant of
As a result,
and in turn
It therefore remains to show that we have \(\det(I_{d} + \sum_{n =1}^{N-1} \aleph^{n} (\gimel^{n}(x))^{-1}) \neq0\). Because \(I_{d} + \sum_{n =1}^{N-1} \aleph^{n} (\gimel^{n}(x))^{-1}\) is a diagonal matrix, we have
where \(\alpha^{n} = -\frac{\gamma^{N} -\gamma^{n}}{N}\), \(n \in\{1, \ldots , N-1\}\). As \(\gamma^{N} = \max(\gamma^{1}, \ldots, \gamma^{N})\), we have \(\alpha^{n} \leq 0\) for each \(n \in\{1, \ldots, N-1\}\). It suffices to show that for \(i \in\{1, \ldots, d\}\),
Now writing \(x = \Re(x) + \mathrm{i} \Im(x)\) and expanding each fraction in (A.23) by the factor \(\Re(x) - \Im(x) \mathrm{i} - \gamma^{n} u^{i}\), we obtain
where \(\overline{x}\) is the complex conjugate of \(x\) and
As each \(\alpha^{n}\) is nonpositive, \(d^{i}(x)\) is nonnegative and \(c^{i}(x)\) is nonpositive. Combining this with \(\overline{x} \in\mathbb {C} \setminus(0, \infty)\), it follows that \(1 + d^{i}(x) + c^{i}(x) \overline{x} \neq0\) so that all eigenvalues of the matrix \(B\) are indeed real and positive. □
Appendix B: Primary matrix functions
In this appendix, we collect some facts about matrix functions from the textbook [25] that are used in Appendix A. First, we recall the definition of a (primary) matrix function.
Definition B.1
Let \(A \in\mathbb{C}^{\ell\times\ell}\) be a matrix with distinct eigenvalues \(\lambda_{1}, \dots, \lambda_{m}\), \(m \leq\ell\). Denote by \(n_{i}\) the algebraic multiplicity of \(\lambda_{i}\), \(i \in\{1, \ldots, m\}\). Let \(O\) be an open neighborhood of \(\lambda_{1}, \dots, \lambda_{m}\) in ℂ and \(f: O \to\mathbb{C}\) a function.Footnote 12
-
(a)
The function \(f\) is said to be defined on the spectrum of \(A\) if it is \((n_{i}-1)\) times differentiable at \(\lambda_{i}\), \(i \in\{1, \ldots, m\} \).
-
(b)
If \(f\) is defined on the spectrum of \(A\), then the primary matrix function \(f(A)\) is defined by
$$ f(A) := p(A), $$where \(p: \mathbb{C} \to\mathbb{C}\) is the unique Hermite interpolating polynomialFootnote 13 satisfying \(p^{(k)}(\lambda_{i}) = f^{(k)}(\lambda_{i})\) for \(k \in\{0, \ldots, n_{i}-1\}\) and \(i \in\{1, \ldots, s\}\).
As a prime example, note that the exponential function is defined on the spectrum of all matrices \(A \in\mathbb{C}^{\ell\times\ell}\) and \(\exp(A)\) is just the matrix exponential. We recall some elementary properties of (primary) matrix functions.
Lemma B.2
Let \(A \in\mathbb{C}^{\ell\times\ell}\) be a matrix with distinct eigenvalues \(\lambda_{1}, \dots, \lambda_{m}\), \(m \leq\ell\) and \(f: \mathbb{C} \to\mathbb{C}\) a function defined on the spectrum of \(A\). Then:
-
(a)
If \(P \in\mathbb{C}^{\ell\times\ell}\) commutes with \(A\), then \(f(A)\) and \(P\) also commute.
-
(b)
If \(P \in \mathbb{C}^{\ell\times\ell}\) is invertible, then \(P f(A) P^{-1} = f(P A P^{-1})\).
-
(c)
The eigenvalues of \(f(A)\) are \(f(\lambda_{1}), \ldots, f(\lambda_{m})\).
-
(d)
\(f(A)\) is invertible if and only if \(f(\lambda_{i}) \neq0\) for all \(i \in\{1, \ldots, m\}\).
Proof
Assertions (a), (b) and (c) are parts of [25, Theorem 1.13]. Finally, (d) follows from (c) and the fact that \(f(A)\) is invertible if and only if zero is not an eigenvalue. □
Finally, we recall a result on the principal square root [25, Theorem 1.29].
Lemma B.3
Let \(A \in\mathbb{R}^{\ell\times\ell}\) be a matrix whose eigenvalues are all real and positive. Then there exists a unique matrix \(P \in\mathbb{R}^{\ell\times\ell}\) with positive eigenvalues such that \(P^{2} = A\). It is given by the primary matrix function \(P = \sqrt{A}\) in the sense of Definition B.1.
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Bouchard, B., Fukasawa, M., Herdegen, M. et al. Equilibrium returns with transaction costs. Finance Stoch 22, 569–601 (2018). https://doi.org/10.1007/s00780-018-0366-6
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DOI: https://doi.org/10.1007/s00780-018-0366-6