Fund Managers’ Competition for Investment Flows Based on Relative Performance

Abstract

N mutual funds compete for fund flows based on relative performance over their average returns, by choosing between an idiosyncratic and a common risky investment opportunities. The unique constant equilibrium is derived in closed form, which implies that funds generally decrease the investments in their idiosyncratic risky assets under competition, in order to lower the risk of the relative performance. It pushes all funds to herd and hurts their after-fee performance. However, the sufficiently disadvantaged funds with poor idiosyncratic investment opportunities or highly risk averse managers may take excessive risk for a better chance of attracting new investments, and their performance may improve comparing to the case without competition and benefit the investors.

Appendix

The proof of Theorem 3.1

The first step is to find the optimal portfolio choice of the fund i, given the investment strategies $\pi _{-i}$ and $\theta _{-i}$ of other funds. Since we focus on constant equilibria, assume that $\pi _{-i}$ and $\theta _{-i}$ are constants. Then, with ${\tilde{X}}_{it} = \exp \left( -\left( r -{\tilde{\psi }}_i\right) t\right) X_{it}$, where ${\tilde{\psi }}_i = \left( 1+\frac{N-1}{N} \alpha _i\right) \psi _i - \frac{\alpha _i}{N} \sum _{j\ne i}^N\psi _j$, Fubini’s theorem implies that

$$\begin{aligned}&\mathbb {E}\left[ \int _0^T e^{-\beta _i t} \frac{(\psi _i X_{it})^{1-\gamma _i}}{1-\gamma _i} dt\right] =\int _0^T e^{\left( -\beta _i + r - {\tilde{\psi }}_i\right) t} \psi _i^{1-\gamma _i} \frac{\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right] }{1-\gamma _i} dt. \end{aligned}$$

For any $(\pi _i,\theta _i)\in {\mathcal {A}}_i\times \varTheta$ ($1\le i\le N$), let $\tilde{\pi }_{it} = \left( 1+\frac{N-1}{N}\alpha _i \right) \sigma _i \pi _{it}$, $\tilde{\theta }_{it} = \left( 1+\frac{N-1}{N}\alpha _i \right) b \theta _{it}$, so that $\pi = (\pi _1,\cdots ,\pi _N) = A_f{\tilde{\pi }}$ and $\theta = A_m{\tilde{\theta }}$, where $\tilde{\pi }$ and $\tilde{\theta }$ are N-dimensional vectors with $(\tilde{\pi })_i = \tilde{\pi }_i$ and $(\tilde{\theta })_i = \tilde{\theta }_i$. Thus,

$$\begin{aligned} \begin{aligned} \frac{d\tilde{X}_{it}}{\tilde{X}_{it}} =\,&\tilde{\pi }_{it}(\lambda _i dt + dW_{it}) +\left( \tilde{\theta }_{it} -\sum _{j\ne i}^N c_{ij} \tilde{\theta }_{j}\right) ( \lambda _m dt + dB_t) \\&\qquad - \sum _{j\ne i}^N c_{ij}\left( \tilde{\pi }_{j}(\lambda _jdt + dW_{jt})\right) . \end{aligned} \end{aligned}$$

With $\phi _{it}= \left[ \begin{array}{c} \tilde{\pi }_{it} \\ \tilde{\theta }_{it} - \sum \limits _{j\ne i}^N c_{ij}\tilde{\theta }_{j} \end{array}\right] , h_i = \left[ \begin{array}{c} \lambda _i \\ \lambda _m \end{array} \right] , F_{it} = \left[ \begin{array}{c} W_{it} \\ B_t \end{array}\right] , \lambda _{-i} = \left[ \begin{array}{cccc} \cdots&\lambda _{i-1}&\lambda _{i+1}&\cdots \end{array}\right] '$, and $W_{-it} = \left[ \begin{array}{cccc} \cdots&W_{(i-1)t}&W_{(i+1)t}&\cdots \end{array}\right] '$, the dynamics of ${\tilde{X}}$ is

$$\begin{aligned} \frac{d\tilde{X}_{it} }{\tilde{X}_{it}} = \phi _{it}' (h_i dt + dF_{it}) -{\tilde{\pi }}_{-i}' C_i (\lambda _{-i} dt+ dW_{-it}), \end{aligned}$$

where $C_i$ is an $(N-1)$-dimensional matrix with diagonal entries $c_{ij}$ for $1\le j \le N$ and $j\ne i$.

Lemma 4.2 shows that $\hat{\phi }_i = \frac{1}{\gamma _i} w_i^{-1} h_i + w_i^{-1} w_{-i}C_i\pi _{-i}$ maximizes $\frac{\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right] }{1-\gamma _i}$. Since it is a constant strategy independent of t, it also maximizes the discounted expected utility from management fees for each manager i

$$\begin{aligned} \int _0^T e^{\left( -\beta _i + r - {\tilde{\psi }}_i\right) t} \psi _i^{1-\gamma _i} \frac{\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right] }{1-\gamma _i} dt. \end{aligned}$$

$\hat{\phi }_i = \frac{1}{\gamma _i} w_i^{-1} h_i + w_i^{-1} w_{-i}C_i{\tilde{\pi }}_{-i}$ for $1\le i \le N$ are 2N equations of constants ${\tilde{\pi }} = ({\tilde{\pi }}_1,\cdots ,{\tilde{\pi }}_N)$ and ${\tilde{\theta }} = ({\tilde{\theta }}_1,\cdots ,{\tilde{\theta }}_N)$:

$$\begin{aligned} P_f \tilde{\pi } = \gamma ^{-1} \lambda _f, \quad P_m {\tilde{\theta }} = \gamma ^{-1} \eta _m + C {\tilde{\pi }}, \end{aligned}$$

of which the solution corresponds to the equilibrium strategies of the N funds. Since Lemma 4.3 shows that $P_f$ and $P_m$ are invertible, there exists a unique solution ${\tilde{\pi }} = P_f^{-1}\gamma ^{-1}\lambda _f$, ${\tilde{\theta }} = P_m^{-1}\left( \gamma ^{-1} \eta _m + C {\tilde{\pi }} \right)$. Therefore, $\pi ^* = A_f P_f^{-1} \gamma ^{-1} \lambda _f$, and $\theta ^* = A_m P_m^{-1} \left( \gamma ^{-1} \eta _m + C A_f^{-1} \pi ^* \right) .$ $\square$

Lemma 4.2

Given constant $\pi _{-i}$ and $\theta _{-i}$, $\mathop {\textrm{arg}\,\textrm{max}}\limits _{\phi _i:(\pi _i,\theta _i)\in {\mathcal {A}}_i\times \varTheta } \frac{\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right] }{1-\gamma _i} = \hat{\phi }_i = \frac{1}{\gamma _i} w_i^{-1} h_i + w_i^{-1} w_{-i}C_i\pi _{-i}$, for every $0\le t\le T$, where $w_i = \left[ \begin{array}{cc} 1 &{} \rho _{im} \\ \rho _{im} &{} 1 \end{array}\right] , w_{-i} = \left[ \begin{array}{c} (\rho _i)_{-i}^\prime \\ (\rho _m)_{-i}^\prime \end{array}\right]$, and $\rho _i$ is the N-dimensional vector with $(\rho _i)_j = \rho _{ij}$.

Proof

We prove the case of $0<\gamma _i <1$ and focus on $\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right]$ because $1-\gamma _i >0$. The case of $\gamma >1$ follows similarly. Define a stochastic process $\xi$ such that $\xi _0 = 1$ and

$$\begin{aligned} -\frac{d\xi _t}{\xi _t} = \left( M_i' w_{-i} + M_{-i}'\rho _{-i} -\lambda _{-i}'\right) C_i \pi _{-i} dt + M_i' dF_{it} + M_{-i}' dW_{-it}, \end{aligned}$$

where $M_i$ and $M_{-i}$ are two constant vectors to be determined later, which satisfy $w_i M_i + w_{-i}M_{-i} = h_i$. Then,

$$\begin{aligned} \frac{d\xi _t \tilde{X}_{it}}{\xi _t \tilde{X}_{it}} =\,&-\left( M_i' w_i' +M_{-i}' w_{-i}' - h_i' \right) {\phi }_{it} dt +({\phi }_{it}'-M_i') dF_{it} \\&\qquad - ({\pi }_{-i}'{C}_i + M_{-i}') dW_{-it}\\ =\,&({\phi }_{it}'-M_i') dF_{it} - ({\pi }_{-i}'{C}_i + M_{-i}') dW_{-it}. \end{aligned}$$

Thus, $\xi _t \hat{X}_{it}$ is a nonnegative local martingale and hence a supermartingale. Therefore (ignoring the positive $1-\gamma _i$), by Hölder’s inequality and noticing that ${\tilde{X}}_{i0} = X_{i0} = 1$,

$$\begin{aligned}&\mathbb {E}\left[ \tilde{X}_{it}^{1-\gamma _i}\right] \le \mathbb {E}\left[ \xi _t \tilde{X}_{it} \right] ^{1-\gamma _i} \mathbb {E}\left[ \xi _t^{\frac{\gamma _i-1}{\gamma _i}}\right] ^{\gamma _i} \le \mathbb {E}\left[ \xi _t^{\frac{\gamma _i-1}{\gamma _i}}\right] ^{\gamma _i}\\&\quad = \exp \left( (1-\gamma _i) \left( \left( M_i' w_{-i} + M_{-i}'\rho _{-i} - {\lambda }_{-i}'\right) {C}_i{\pi }_{-i} +\frac{1}{2\gamma _i}M_i' w_i M_i \right. \right. \\&\qquad + \left. \left. \frac{1}{2\gamma _i} M_{-i}' \rho _{-i} M_{-i} + \frac{1}{\gamma _i}M_i' w_{-i} M_{-i}\right) t \right) , \end{aligned}$$

which is an upper bound for $\mathbb {E}\left[ \tilde{X}_{it}^{1-\gamma _i}\right]$ corresponding to any $(\pi _i,\theta _i)\in {\mathcal {A}}_i\times \varTheta$.

Next we search for the minimum among all such upper bounds corresponding to different choices of $M_i$ and $M_{-i}$, by considering the following constrained minimization problem:

$$\begin{aligned}&\min _{\{ M_i, M_{-i}\}} \frac{1}{2\gamma _i} M_i' w_i M_i + \frac{1}{2\gamma _i} M_{-i}' \rho _{-i}M_{-i} + \frac{1}{\gamma _i} M_i' w_{-i} M_{-i} + \\&\qquad \left( M_i' w_{-i} + M_{-i}' \rho _{-i}\right) C_i \pi _{-i} ,\\&\text { subject to: } w_i M_i + w_{-i}M_{-i} = h_i. \end{aligned}$$

The corresponding Lagrangian function, with Lagrange multiplier l, is

$$\begin{aligned} \begin{aligned} \mathcal {L} =\,&\frac{1}{2\gamma _i} \left( M_i' w_i M_i + M_{-i}' \rho _{-i}M_{-i} + 2M_i' w_{-i} M_{-i}\right) + \\&\quad \left( M_i' w_{-i} + M_{-i}' \rho _{-i}\right) C_i \pi _{-i} + l' \left( h_{i} - w_{i} M_{i} - w_{-i} M_{-i} \right) . \end{aligned} \end{aligned}$$

The first-order conditions for $M_i$, $M_{-i}$ and l are

$$\begin{aligned} M_i =\,&\gamma _i l - w_i^{-1} w_{-i} M_{-i} - \gamma _i w_i^{-1}w_{-i}C_i\pi _{-i}, \end{aligned}$$

(13)

$$\begin{aligned} 0=\,&\frac{1}{\gamma _i} \rho _{-i} M_{-i}+ \frac{1}{\gamma _i} w_{-i}'M_i + \rho _{-i}C_i \pi _{-i} - w_{-i}' l, \nonumber \\ 0=\,&h_i - w_i M_i - w_{-i} M_{-i}. \end{aligned}$$

(14)

Plugging (13) into (14) implies that

$$\begin{aligned} 0 = \left( \rho _{-i} -w_{-i}' w_{i}^{-1} w_{-i} \right) \left( M_{-i} + \gamma _i C_i \pi _{-i} \right) . \end{aligned}$$

Instead of discussing the uniqueness of solutions to the above equation, we pick out one of them $M_{-i} = -\gamma _i C_i \pi _{-i}$, $M_i = \gamma _i{\hat{\phi }}_i$, $l= {\hat{\phi }}_i$, and verify that the candidate strategy ${\hat{\phi }}_i$ can achieve the upper bound corresponding to $M_{-i}$ and $M_i$, which verifies that ${\hat{\phi }}_i$ is indeed the maximizer of $\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right]$.

The upper bound corresponding to $M_{-i} = -\gamma _i C_i \pi _{-i}$ and $M_i = \gamma _i{\hat{\phi }}_i$ is

$$\begin{aligned}&\exp \left( (1-\gamma _i) \left( \left( M_i' w_{-i} + M_{-i}'\rho _{-i} - {\lambda }_{-i}'\right) {C}_i{\pi }_{-i} +\frac{1}{2\gamma _i}M_i' w_i M_i \right. \right. \nonumber \\&\qquad + \left. \left. \frac{1}{2\gamma _i} M_{-i}' \rho _{-i} M_{-i} +\frac{1}{\gamma _i}M_i' w_{-i} M_{-i}\right) t \right) \nonumber \\&\quad = \exp \left( (1-\gamma _i)\left( \left( \gamma _i\hat{\phi }_i' w_{-i} -\gamma _i\pi _{-i}'C_i\rho _{-i} - {\lambda }_{-i}'\right) {C}_i{\pi }_{-i} \right. \right. \nonumber \\&\qquad + \left. \left. \frac{\gamma _i}{2}\left( \hat{\phi }_i' w_i\hat{\phi }_i+\pi _{-i}'C_i \rho _{-i}C_i\pi _{-i} - 2\hat{\phi }_i' w_{-i} C_i\pi _{-i}\right) \right) t\right) \nonumber \\&\quad = \exp \left( \left( -(1-\gamma _i)\lambda _{-i}' C_i\pi _{-i} +\frac{(1-\gamma _i)\gamma _i}{2}\left( \hat{\phi }_i' w_i \hat{\phi }_i-\pi _{-i}'C_i\rho _{-i}C_i\pi _{-i}\right) \right) t\right) . \end{aligned}$$

(15)

On the other hand, for ${\tilde{X}}_i$ corresponding to $\hat{\phi }_i$,

$$\begin{aligned} \tilde{X}_{it} =\,&\exp \left( \hat{\phi }_i' (h_i t + F_{it}) - \pi _{-i}'C_i (\lambda _{-i} t+ W_{-it}) \right. \\&\qquad + \left. \left( -\frac{1}{2} \hat{\phi }_i' w_i\hat{\phi }_i -\frac{1}{2} \pi _{-i}'C_i \rho _{-i}C_i\pi _{-i} + {\hat{\phi }}_i' w_{-i}C_i \pi _{-i}\right) t\right) . \end{aligned}$$

Thus,

$$\begin{aligned}&\mathbb {E}\left[ \tilde{X}_{it}^{1-\gamma _i}\right] \\&\quad =\exp \left( (1-\gamma _i) \left( \hat{\phi }_i' h_i - \pi _{-i}' C_i\lambda _{-i} -\frac{1}{2} \hat{\phi }_i' w_i \hat{\phi }_i -\frac{1}{2} \pi _{-i}' C_i\rho _{-i} C_i\pi _{-i} + \right. \right. \\&\qquad \left. \left. \hat{\phi }_i' w_{-i} C_i \pi _{-i} +\frac{(1-\gamma _i)}{2}\hat{\phi }_i'w_i \hat{\phi }_i' +\frac{(1-\gamma _i)}{2}\pi _{-i}'C_i \rho _{-i} C_i\pi _{-i} \right. \right. \\&\qquad \left. \left. - (1-\gamma _i) \hat{\phi }_i'w_{-i} C_i\pi _{-i} \right) t\right) \\&\quad =\exp \left( \left( -(1-\gamma _i)\pi _{-i}' C_i\lambda _{-i} +\frac{\gamma _i(1-\gamma _i)}{2} \left( \hat{\phi }_i' w_i \hat{\phi }_i' - \pi _{-i}' C_i \rho _{-i} C_i\pi _{-i}\right) \right) t\right) , \end{aligned}$$

which coincides with the upper bound in (15). $\square$

Lemma 4.3

$P_f$ and $P_m$ are invertible.

Proof

$P_f$ and $P_m$ can be rewritten as $P_f = A_1P_{diag} P_1 P_{diag}A_2$, and $P_m = P_{diag}^2A_1P_2A_2$, where $A_1$, $A_2$ and $\textbf{P}_{diag}$ are $N\times N$ diagonal matrices with $(A_1)_{ii} = \frac{1}{N+(N-1)\alpha _i}$, $(A_2)_{ii} = \alpha _i$ and $(P_{diag})_{ii} = \sqrt{1-\rho _{im}^2}$, and $P_1$ and $P_2$ are $N\times N$ matrices with

$$\begin{aligned} (P_1)_{ij} = {\left\{ \begin{array}{ll} \frac{1}{c_{ii}}&{} \text { if } i = j,\\ -\frac{\rho _{ij}-\rho _{im}\rho _{jm}}{\sqrt{1-\rho _{im}^2}\sqrt{1-\rho _{jm}^2}} &{} \text { if } i \ne j, \end{array}\right. } \quad (P_{2})_{ij} = {\left\{ \begin{array}{ll} \frac{1}{c_{ii}} &{} \text { if } i = j,\\ -1 &{} \text { if } i \ne j. \end{array}\right. } \end{aligned}$$

On the other hand, for $i\ne j$, Brownian motions $W_i$ and $W_j$ can be written as

$$\begin{aligned} W_{it} = \rho _{im} B_t + \sqrt{1-\rho _{im}^2} Z_{it}, \quad W_{jt} = \rho _{jm} B_t + \sqrt{1-\rho _{jm}^2} Z_{jt}, \end{aligned}$$

where $Z_{i}, Z_{j}$ are Brownian motions independent of B. Suppose that $\langle Z_i,Z_j\rangle _t = \rho ^z_{ij} t$, and then $\rho _{ij} = \rho _{im}\rho _{jm} + \sqrt{(1-\rho _{im}^2)(1-\rho _{jm}^2)} \rho ^z_{ij}$, which implies that $\frac{(\rho _{ij}-\rho _{im}\rho _{jm})^2}{(1-\rho _{im}^2)(1-\rho _{jm}^2)} = (\rho ^z_{ij})^2 \le 1$. Since $c_{ii} = \frac{\alpha _i}{N+(N-1)\alpha _i} < \frac{1}{N-1}$, both $P_1$ and $P_2$ are strictly diagonally dominated matrices, and hence invertible. Therefore, $P_f$ and $P_m$ are invertible, because the diagonal matrices $A_1$, $A_2$ and $P_{diag}$ are also invertible.$\square$

The proof of Proposition 3.1

• (i) The claim follows from the fact that $\frac{\partial \pi _i^*}{\partial \lambda _i} = \frac{2(1-\rho _{jm}^2)}{(2+\alpha _i)\sigma _i \gamma _i \kappa _1}$ and $\kappa _1 >0$.

• (ii) The claim follow from $\frac{\partial \pi _i^*}{\partial \lambda _j} = \frac{2\alpha _i(\rho _{12}-\rho _{1\,m}\rho _{2\,m})}{(2+\alpha _i)(2+\alpha _j)\sigma _i \gamma _j \kappa _1}$.

• (iii) The claims are direct results of the following derivatives and the fact that $(1-\rho _{1\,m}^2)(1-\rho _{2\,m}^2) \ge (\rho _{12}-\rho _{1\,m}\rho _{2\,m})^2$ from the proof of Lemma 4.3.

  $$\begin{aligned} \frac{\partial \theta _i}{\partial \lambda _i} =\,&-\frac{2(1-\rho _{jm}^2)\rho _{im}}{(2+\alpha _i)b\kappa _2\gamma _i \kappa _1}\left[ \left( 1+\frac{\alpha _1\alpha _2}{(2+\alpha _1)(2+\alpha _2)}\right) (1-\rho _{1m}^2)(1-\rho _{2m}^2)\right. \\&- \left. \frac{2\alpha _1\alpha _2}{(2+\alpha _1)(2+\alpha _2)}(\rho _{12}-\rho _{1m}\rho _{2m})^2\right] , \\ \frac{\partial \theta _i}{\partial \lambda _j} =&- \frac{2\alpha _i(1-\rho _{im}^2)}{(2+\alpha _1)(2+\alpha _2)b \kappa _2 \gamma _j \kappa _1}\left[ \left( 1+\frac{\alpha _1\alpha _2}{(2+\alpha _1)(2+\alpha _2)}\right) \cdot \right. \\&\left. \left( 1-\rho _{1m}^2\right) \left( 1-\rho _{2m}^2\right) \rho _{jm}-\frac{2\alpha _1\alpha _2\rho _{jm}}{(2+\alpha _1)(2+\alpha _2)}\left( \rho _{12}-\rho _{1m}\rho _{2m}\right) ^2 \right. \\&- \left. \left( 1-\frac{\alpha _1\alpha _2}{(2+\alpha _1)(2+\alpha _2)}\right) (1-\rho _{jm}^2) (\rho _{jm}-\rho _{12}\rho _{im})\right] . \end{aligned}$$

$\square$

The proof of Proposition 3.2

The claims follow from the derivatives of $\pi ^*_i$ with respect to $\alpha _i$ and $\alpha _j$.

$$\begin{aligned} \frac{\partial \pi _i^*}{\partial \alpha _i}=\,&\frac{2}{(2+\alpha _i)\sigma _i\kappa _1}\left( \frac{\rho _{12}}{2+\alpha _j}\left( 1-\frac{\alpha _i}{2+\alpha _i}\frac{1-\frac{\alpha _j}{2+\alpha _j}\rho _{12}^2}{\kappa _1}\right) \lambda _{j,\gamma _j}\right. \\&\qquad - \left. \frac{1-\frac{\alpha _j}{2+\alpha _j}\rho _{12}^2}{(2+\alpha _i)\kappa _1}\lambda _{i,\gamma _i}\right) \\ =\,&\frac{2}{(2+\alpha _i)^2(2+\alpha _j)\sigma _i\kappa _1^2}\left( 2\rho _{12}\lambda _{j,\gamma _j}-\left( 2 + (1-\rho _{12}^2)\alpha _j\right) \lambda _{i,\gamma _i}\right) ,\\ \frac{\partial \pi _i^*}{\partial \alpha _j}=\,&\frac{2}{(2+\alpha _i)\sigma _i\kappa _1}\left( \frac{2\alpha _i\rho _{12}^2}{(2+\alpha _i)(2+\alpha _j)^2\kappa _1}\lambda _{i,\gamma _i} \right. \\&\qquad - \left. \frac{\alpha _i\rho _{12}}{(2+\alpha _j)^2}\left( 1-\frac{2\alpha _i\rho _{12}^2}{(2+\alpha _i)(2+\alpha _j)\kappa _1}\right) \lambda _{j,\gamma _j}\right) \\ =\,&\frac{2\alpha _i\rho _{12}}{(2+\alpha _i)^2(2+\alpha _j)^2\sigma _i\kappa ^2_1}\left( 2\rho _{12}\lambda _{i,\gamma _i} - (2 + (1-\rho _{12}^2)\alpha _i)\lambda _{j,\gamma _j}\right) . \end{aligned}$$

$\square$

The proof of Proposition 3.3

Following (7) and (10), with $\rho _{12}\in (-1,1)$ and ${\bar{\lambda }} \le 1$,

$$\begin{aligned} \frac{\pi _1^*}{\pi _1^M} =\,&\frac{2\left( 1 + \frac{\alpha }{2+\alpha } \rho _{12} {\bar{\lambda }}\right) }{(2+\alpha )\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2\rho _{12}^2\right) } =\frac{2(2+\alpha )\left( 1 + \frac{\alpha }{2+\alpha } \rho _{12} {\bar{\lambda }}\right) }{4\alpha + \alpha ^2 (1-\rho _{12}^2)+4}\\ \le&\frac{4 + 4\alpha }{\alpha ^2(1-\rho _{12}^2)+4\alpha + 4}<1.\\ \eta _1^* - \eta _1^M =\,&\lambda _1 - \frac{\psi (2+\alpha )\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2\rho _{12}^2\right) }{2\lambda _{1,\gamma _1}\left( 1+ \frac{\alpha }{2+\alpha } \rho _{12} {\bar{\lambda }}\right) } - \left( \lambda _1 - \frac{\psi }{\lambda _{1,\gamma _1}} \right) \\ =\,&-\frac{\alpha \psi \left( \left( 2+ \alpha (1-\rho _{12}^2)\right) - 2\rho _{12} {\bar{\lambda }} \right) }{2(2+\alpha )\lambda _{1,\gamma _1}\left( 1+ \frac{\alpha }{2+\alpha } \rho _{12} {\bar{\lambda }}\right) }\\ <&-\frac{\alpha \psi \left( 2- 2 {\bar{\lambda }} \right) }{2(2+\alpha )\lambda _{1,\gamma _1}\left( 1+ \frac{\alpha }{2+\alpha } \rho _{12} {\bar{\lambda }}\right) } \le 0. \end{aligned}$$

On the other hand,

$$\begin{aligned} \frac{\pi _2^*}{\pi _2^M} - 1 =\,&\frac{2\left( \frac{\alpha }{2+\alpha } \rho _{12} + {\bar{\lambda }}\right) }{{\bar{\lambda }}(2+\alpha )\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2\rho _{12}^2\right) } -1= \frac{\frac{2\alpha \rho _{12}}{2+\alpha }+ \alpha \left( \frac{\alpha \rho _{12}^2}{2+\alpha }-1\right) {\bar{\lambda }}}{{\bar{\lambda }}(2+\alpha )\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2\rho _{12}^2\right) }. \end{aligned}$$

Thus, $\pi ^*_2 \le \pi ^M_2$ is equivalent to $\frac{2\alpha \rho _{12}}{2+\alpha }+ \alpha \left( \frac{\alpha \rho _{12}^2}{2+\alpha }-1\right) {\bar{\lambda }}\le 0$, which, since $\frac{\alpha \rho _{12}^2}{2+\alpha } <1$, always holds for $\rho _{12}<0$, and is equivalent to ${\bar{\lambda }} \ge \frac{2\rho _{12}}{2+\alpha (1-\rho _{12}^2)}$ if $\rho _{12} \ge 0$. Finally,

$$\begin{aligned} \eta _2^* - \eta _2^M =\,&-\frac{ \psi \alpha \left( \left( 2+ \alpha (1-\rho _{12}^2)\right) {\bar{\lambda }} - 2\rho _{12} \right) }{2(2+\alpha )\lambda _{2,\gamma _2}\left( {\bar{\lambda }}+ \frac{\alpha \rho _{12}}{2+\alpha }\right) }. \end{aligned}$$

Thus, if $\rho _{12} \ge 0$, ${\bar{\lambda }} + \frac{\alpha \rho _{12}}{2+\alpha }>0$, and $\eta _2^* \le \eta _2^M$ if and only if $\left( 2+ \alpha (1-\rho _{12}^2)\right) {\bar{\lambda }} - 2\rho _{12}\ge 0$, or equivalently ${\bar{\lambda }} \ge \frac{2\rho _{12}}{2+ \alpha (1-\rho _{12}^2)}$. If $\rho _{12} < 0$, $\left( 2+ \alpha (1-\rho _{12}^2)\right) {\bar{\lambda }} - 2\rho _{12} > 0$, and $\eta _2^* \le \eta _2^M$ if and only if ${\bar{\lambda }} \ge -\frac{\alpha \rho _{12}}{2+\alpha }$. $\square$

The proof of Proposition 3.4

Following (8) and (9),

$$\begin{aligned} \text {Beta}_1^* -1 =\,&\frac{\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2\rho _{12}^2\right) (1-\bar{\lambda }^2)}{ K_1^*} \ge 0,\\ K_1^*=\,&\left( 1+\frac{\alpha }{2+\alpha }\rho _{12} \bar{\lambda }\right) ^2+2\rho _{12} \left( 1+\frac{\alpha }{2+\alpha }\rho _{12} \bar{\lambda }\right) \left( \frac{\alpha }{2+\alpha } \rho _{12} \bar{\lambda }\right) \\&\qquad + \left( \frac{\alpha }{2+\alpha } \rho _{12} + \bar{\lambda }\right) ^2,\\ \text {Beta}_1^M -1 =\,&\frac{ 1- \bar{\lambda }^2}{1 + 2\rho _{12} \bar{\lambda }^2 + \bar{\lambda }^2} \ge 0 . \end{aligned}$$

Therefore, if ${\bar{\lambda }}=1$, $\textrm{Beta}^*_1 = \textrm{Beta}^M_1 = 1$. Otherwise, since $\bar{\lambda }<1$, the sign of $|\text {Beta}_1^* - 1| -|\text {Beta}_1^M -1| = \textrm{Beta}^*_1 - \textrm{Beta}^M_1$ is the same as that of

$$\begin{aligned}&\left( 1-\left( \frac{\alpha }{2+\alpha }\right) ^2 \rho _{12}^2\right) \left( 1 + 2\rho _{12} \bar{\lambda } + \bar{\lambda }^2\right) -\left( 1+\frac{\alpha }{2+\alpha }\rho _{12} \bar{\lambda }\right) ^2\nonumber \\&\quad - 2\rho _{12} \left( 1+\frac{\alpha }{2+\alpha }\rho _{12} \bar{\lambda }\right) \left( \frac{\alpha }{2+\alpha } \rho _{12} + \bar{\lambda }\right) - \left( \frac{\alpha }{2+\alpha } \rho _{12} + \bar{\lambda }\right) ^2 \nonumber \\&\quad = - \frac{4\alpha (1+\alpha )}{(2+\alpha )^2}\rho ^2_{12}\left( 1 + \left( \frac{2+\alpha }{1+\alpha }\frac{1}{\rho _{12}}+\frac{\alpha }{1+\alpha } \rho _{12}\right) \bar{\lambda } + \bar{\lambda }^2\right) . \end{aligned}$$

(16)

If $\rho _{12} \ge 0$, $1 + \left( \frac{2+\alpha }{1+\alpha }\frac{1}{\rho _{12}}+\frac{\alpha }{1+\alpha } \rho _{12}\right) \bar{\lambda } + \bar{\lambda }^2 \ge 0$, and hence $|\textrm{Beta}_1^* - 1| -|\textrm{Beta}_1^M -1| \le 0$. If $\rho _{12} < 0$, $|\text {Beta}_1^* -1| - |\textrm{Beta}_1^M - 1| \le 0$ is equivalent to

$$\begin{aligned} 1 + \left( \frac{2+\alpha }{1+\alpha }\frac{1}{\rho _{12}}+\frac{\alpha }{1+\alpha } \rho _{12}\right) \bar{\lambda } + \bar{\lambda }^2 \ge 0. \end{aligned}$$

(17)

Since $\frac{\alpha }{1+\alpha } \rho _{12} + \frac{2+\alpha }{1+\alpha } \frac{1}{\rho _{12}}$ is negative and is decreasing in $\rho _{12} \in [-1,0)$, the maximum value at $\rho _{12} = -1$ is $-2$, and $\varDelta \ge 0$. Since $\lambda \le 1$, the two roots of the left hand side of (17) are

$$\begin{aligned} \frac{-\left( \frac{\alpha }{1+\alpha } \rho _{12} + \frac{2+\alpha }{1+\alpha } \frac{1}{\rho _{12}} \right) + \sqrt{\varDelta }}{2} \ge 1 \text { and } 0\le \frac{-\left( \frac{\alpha }{1+\alpha } \rho _{12} + \frac{2+\alpha }{1+\alpha } \frac{1}{\rho _{12}} \right) - \sqrt{\varDelta }}{2}. \end{aligned}$$

Thus, $|\text {Beta}_1^* - 1|-|\text {Beta}_1^M -1| \le 0$ if $\bar{\lambda } \le \frac{-\left( \frac{\alpha }{1+\alpha } \rho _{12} + \frac{2+\alpha }{1+\alpha } \frac{1}{\rho _{12}} \right) - \sqrt{\varDelta }}{2}$, and otherwise the inequality is reversed.

On the other hand, algebraic calculations show that $\textrm{Beta}_2^*-1 = - \left( \textrm{Beta}^*_1 -1\right) \le 0$ and $\textrm{Beta}_2^M-1 = -\left( \textrm{Beta}^M_1-1 \right) \le 0$. Therefore, the sign of $|\text {Beta}_2^* - 1| -|\text {Beta}_2^M -1| = -\textrm{Beta}^*_2 + \textrm{Beta}^M_2$ is the same as (16), and equivalent conditions for Fund 1 still hold. $\square$

The proof of Proposition 3.5

Let $\tilde{X}_{it} = \exp \left( -\left( r- {\tilde{\psi }}_i \right) t\right) X_{it}$. Then, with $\zeta _{it} = \frac{N+(N-1)\alpha _i}{N} \theta _{it} -\frac{\alpha _i}{N} \sum _{j\ne i}^N \theta _{jt}$, ${\tilde{X}}_i$ follows

$$\begin{aligned} \frac{d\tilde{X}_{it}}{\tilde{X}_{it}} = \zeta _{it} (adt+bdB_t), \quad \tilde{X}_{i0} = X_{i0} =1. \end{aligned}$$

We first calculate the optimal $\zeta _i$ (or equivalently the optimal $\theta _i$) given $\theta _j$’s ($j\ne i$) of other funds. With $d\xi _t /\xi _t = - \lambda _m dB_t$ and $\xi _0 =1$, $d(\xi _t\tilde{X}_{it})= \xi _t \tilde{X}_{it} \left( \zeta _{it} b - \lambda _m \right) dB_t$, which is a nonnegative local martingale, and thus a supermartingale. Then, for $0<\gamma _i <1$ (the case of $\gamma _i >1$ follows similarly), by Hölder’s inequality, for any $\theta = (\theta _1,\dots ,\theta _N) \in \varTheta ^N$,

$$\begin{aligned}&\mathbb {E}\left[ \frac{\tilde{X}_{it}^{1-\gamma _i} }{1-\gamma _i}\right] = \frac{\mathbb {E}\left[ \left( \xi _t\tilde{X}_{it}\right) ^{1-\gamma _i} \xi _t^{\gamma _i-1} \right] }{1-\gamma _i} \\&\quad \le \frac{\mathbb {E}\left[ \xi _t \hat{X}_{it} \right] ^{1-\gamma _i} \mathbb {E}\left[ \xi _t^{-\frac{1-\gamma _i}{\gamma _i}}\right] ^{\gamma _i} }{1-\gamma _i} \le \frac{\exp \left( \frac{1-\gamma _i}{2\gamma _i} \lambda _m^2 t \right) }{1-\gamma _i}, \end{aligned}$$

which gives an upper bound of $\mathbb {E}\left[ \frac{\tilde{X}_{it}^{1-\gamma _i} }{1-\gamma _i}\right]$. On the other hand, with $\theta _{it} = \frac{N\left( \frac{\lambda _m}{\gamma _i b} + \frac{\alpha _i}{N}\sum \limits _{j\ne 1}^N \theta _{jt}\right) }{N+(N-1)\alpha _i}$, and thus $\zeta _{it} = \frac{\lambda _m}{\gamma _i b}$, $\mathbb {E}\left[ \frac{1}{1-\gamma _i} \tilde{X}_{it}^{1-\gamma _i} \right] = \frac{\exp \left( \frac{1-\gamma _i}{2\gamma _i} \lambda _m^2 t \right) }{1-\gamma _i},$ which indicates that $\zeta _i$ is the maximizer of $\mathbb {E}\left[ \frac{\tilde{X}_{it}^{1-\gamma _i} }{1-\gamma _i}\right]$. Since $\zeta _{it} = \frac{\lambda _m}{\gamma _i b}$ is a constant strategy, independent of t, it also maximizes manager i’s expected utility

$$\begin{aligned} \int _0^T e^{\left( -\beta _i + r - {\tilde{\psi }}_i\right) t} \psi _i^{1-\gamma _i} \frac{\mathbb {E}\left[ {\tilde{X}}_{it}^{1-\gamma _i}\right] }{1-\gamma _i} dt, \end{aligned}$$

and the optimal strategy given $\theta _j$’s ($j\ne i$) is $\theta _{it} = \frac{N\left( \frac{\lambda _m}{\gamma _ib} + \frac{\alpha _i}{N}\sum \limits _{j\ne i}^N \theta _{jt}\right) }{N+(N-1)\alpha _i}$.

To find the equilibrium, it suffices to solve the system of N equations, each representing the optimal strategy given the portfolio of other funds: $P_I \theta _t = \frac{\lambda _m}{b}\gamma ^{-1}e$, where $\theta _t= (\theta _{1t},\cdots ,\theta _{Nt})'$, e is the N-dimensional vector with all entries equal to 1, $P_I$ the $N\times N$ matrix with $(P_I)_{i,j} = {\left\{ \begin{array}{ll} \frac{N+(N-1)\alpha _i}{N} &{} \text { if } i =j \\ -\frac{\alpha _i}{N} &{} \text { if } i \ne j, \end{array}\right. }$ and $\gamma$ is defined in Theorem 3.1. Since $P_I$ is strictly diagonally dominated and thus invertible, there exists a unique solution $\theta ^* = \frac{\lambda _m}{b}P_I^{-1} \gamma ^{-1}e$ for every $0\le t\le T$. Furthermore, since $P_I = -\frac{1}{N} A_2 (D + ee')$, where $A_2$ is the diagonal matrix defined in the proof of Lemma 4.3, D is an $N\times N$ diagonal matrix with $(D)_{ii} = -\frac{1}{c_{ii}}-1$, $1\le i\le N$, by Sherman–Morrison–Woodbury formula (See equation 2.1.4 in [26]),

$$\begin{aligned} (P_I^{-1})_{i,j} = \left\{ \begin{array}{ll} \frac{1}{1+\alpha _i} + \frac{1+\bar{\alpha }}{N}\frac{\alpha _i}{(1+\alpha _i)^2}&{} \text {, } i = j\\ \frac{1+\bar{\alpha }}{N}\frac{\alpha _i}{(1+\alpha _i)(1+\alpha _j)}&{} \text {, } i \ne j \end{array} \right. , \end{aligned}$$

and this solution reduces to $\theta _i^* = \frac{\lambda _m}{b} \left( \frac{1}{1+\alpha _i}\frac{1}{\gamma _i} + \frac{\alpha _i}{1+\alpha _i} \frac{1}{\bar{\gamma }}\right)$ for $1\le i \le N$. $\square$

The proof of Proposition 3.6

• (i) Both $\theta ^*_i$ and $\theta ^M_i$ are proportion to $\frac{\lambda _m}{b}$, while the coefficient for $\theta ^M_i$ is $\frac{1}{\gamma _i}$ and that for $\theta ^*_i$ is a convex combination between $\frac{1}{\gamma _i}$ and $\frac{1}{{\bar{\gamma }}}$. The claim follows by the comparing the two coefficients.

• (ii) Since $\frac{1}{N}\sum \limits _{i=1}^{N} \left( \frac{1}{1+\alpha _i}\frac{1}{\gamma _i} + \frac{\alpha _i}{1+\alpha _i}\frac{1}{{\bar{\gamma }}}\right) = \frac{1}{{\bar{\gamma }}}$, ${\bar{\theta }}^* = \frac{\lambda _m}{b{\bar{\gamma }}}$, and

  $$\begin{aligned} \bar{\theta }^* - \bar{\theta }^M =\,&\frac{\lambda _m}{b} \left( \frac{1}{\bar{\gamma }} - \frac{1}{N} \sum _{i = 1}^N \frac{1}{\gamma _i}\right) = \frac{\lambda _m}{b} \left( \frac{1+{\bar{\alpha }}}{N} \sum _{i = 1}^N \frac{1}{(1+\alpha _i)\gamma _i}- \frac{1}{N} \sum _{i = 1}^N \frac{1}{\gamma _i}\right) \\ =\,&\frac{\lambda _m}{b} \frac{1}{N} \sum _{i = 1}^N \left( \frac{1+\bar{\alpha }}{1+\alpha _i}-1 \right) \frac{1}{\gamma _i}. \end{aligned}$$

  Since $(\gamma _i-\gamma _j)(\alpha _i-\alpha _j) \ge 0$ for every pair of i and j, $\left( \frac{1+\bar{\alpha }}{1+\alpha _i}-1 \right)$’s and $\frac{1}{\gamma _i}$’s are similarly ordered, and from Tchebychef’s inequality [32, 2.17.1], the above is greater than or equal to

  $$\begin{aligned} \frac{\lambda _m}{b} \frac{1}{N} \sum _{i = 1}^N \left( \frac{1+\bar{\alpha }}{1+\alpha _i}-1 \right) \frac{1}{N} \sum _{i = 1}^N\frac{1}{\gamma _i} =0, \end{aligned}$$

  and the inequality is reversed if $(\gamma _i-\gamma _j)(\alpha _i-\alpha _j) \le 0$ for every pair of i and j.

• (iii) $\bar{\theta }^* = \bar{\theta }^M$ follows from (ii). Furthermore, since $\frac{1}{{\bar{\gamma }}} = \frac{1}{N}\sum \limits _{i=1}^N\frac{1}{\gamma _i}$,

  $$\begin{aligned}&\theta _i^* - \bar{\theta }^* = \frac{\lambda _m}{b}\left( \frac{1}{1+\alpha } \frac{1}{\gamma _i} +\sum _{j=1}^N\left( \frac{\alpha }{N(1+\alpha )} \frac{1}{\gamma _j} \right) - \frac{1}{N}\sum _{j = 1} \frac{1}{\gamma _j} \right) \\&\quad = \frac{\lambda _m}{b}\left( \frac{1}{1+\alpha } \frac{1}{\gamma _i} - \frac{1}{1+\alpha } \frac{1}{N} \sum _{j = 1}^N \frac{1}{\gamma _j}\right) \\&\quad = \frac{1}{1+\alpha }\frac{\lambda _m}{b} \left( \frac{1}{\gamma _i} -\frac{1}{N} \sum _{j = 1}^N \frac{1}{\gamma _j}\right) = \frac{1}{1+\alpha } \left( \theta ^M_i - \bar{\theta }^M\right) . \end{aligned}$$

$\square$

The proof of Proposition 3.7

(i) and (ii) follow the same calculation as for Theorem 3.1. For (iii), notice that $\pi ^{*\alpha }_N - \pi ^{*0}_N = (k_\rho ({\tilde{\lambda }} +\lambda _{N,\gamma _N}) - \lambda _{N,\gamma _N} )/\sigma _1$. Thus, the comparison between $\pi ^{*0}$ and $\pi ^{*\alpha }$ is reduced to

$$\begin{aligned}&k_\rho ({\tilde{\lambda }} +\lambda _{N,\gamma _N}) - \lambda _{N,\gamma _N} \\&\quad =k_\rho \left( \sum \limits _{j = 1}^N \frac{\alpha \rho }{N+(N-1)(1-\rho )\alpha }\lambda _{j,\gamma _j} +\left( 1-\frac{N+(N-1+\rho )\alpha }{N}\right) \lambda _{N,\gamma _N} \right) \\&\quad = \frac{\alpha (N-1+\rho ) k_\rho }{N}\left( \frac{\rho N^2 \frac{1}{N} \sum \limits _{j = 1}^N \lambda _{j,\gamma _j} }{\left( (1+(1-\rho )\alpha )N-(1-\rho )\alpha \right) \left( N-(1-\rho )\right) }- \lambda _{N,\gamma _N} \right) \\&\quad = \frac{\alpha (N-1+\rho ) k_\rho }{N}\left( \varphi \frac{1}{N} \sum \limits _{j = 1}^N \lambda _{j,\gamma _j} - \lambda _{N,\gamma _N}\right) . \end{aligned}$$

Thus, if $\lambda _{N,\gamma _N} \ge \varphi \sum _{j =1}^N \lambda _{j,\gamma _j}/N$, then $\lambda _{N,\gamma _N} \ge k_\rho ({\tilde{\lambda }} + \lambda _{1,\gamma _1})$, and $\pi ^{*0}_i \ge \pi ^{*\alpha }_i$ for every i. $\square$

The proof of Proposition 3.8

(i), (ii) is a direct application of Theorem 3.1 and follows similar arguments to those for Proposition 3.7 (iii).

(iii) Given $\pi ^*_i$’s, $\textrm{Beta}_N^* = \frac{(N+\alpha ){\bar{\lambda }} -(N-1)\alpha }{(1+\alpha )({\bar{\lambda }}-(N-1))}\text {, } \textrm{Beta}_N^M = \frac{N {\bar{\lambda }}}{{\bar{\lambda }}-(N-1)}.$ Thus,

$$\begin{aligned}&\left|\textrm{Beta}_N^M-1 \right|-\left|\textrm{Beta}_N^*-1 \right|= \frac{(1+\alpha )(N-1)({\bar{\lambda }}+1)-(N-1)\left( {\bar{\lambda }}+1\right) }{(1+\alpha )|{\bar{\lambda }}-(N-1)|} \\&\quad =\frac{\alpha (N-1)({\bar{\lambda }}+1)}{(1+\alpha )|{\bar{\lambda }}-(N-1)|} >0. \end{aligned}$$

For fund $i \ne N$, $\textrm{Beta}_i^* = \frac{1}{1+\alpha }-\frac{N}{(1+\alpha )({\bar{\lambda }}-(N-1))}\frac{{\bar{\gamma }}_{-N}}{\gamma _i}$ and $\textrm{Beta}_i^M= -\frac{N}{{\bar{\lambda }}-(N-1)}\frac{{\bar{\gamma }}_{-N}}{\gamma _i}$. If ${\bar{\lambda }} > N-1$, both $\textrm{Beta}$’s are less than 1, and $\left|\textrm{Beta}_i^M-1 \right|\ge \left|\textrm{Beta}_i^*-1 \right|$ because

$$\begin{aligned}&\left|\textrm{Beta}_i^M-1 \right|-\left|\textrm{Beta}_i^*-1 \right|= \frac{\alpha }{1+\alpha } \left( 1+ \frac{N}{{\bar{\lambda }}-(N-1)}\frac{{\bar{\gamma }}_{-N}}{\gamma _i}\right) \ge 0. \end{aligned}$$

(18)

If ${\bar{\lambda }} < N-1$ and $\frac{{\bar{\gamma }}_{-N}}{\gamma _i} \ge \max (\alpha ,1) \frac{N-1-{\bar{\lambda }}}{N}$, both $\textrm{Beta}$s are greater than or equal to 1, and

$$\begin{aligned}&\left|\textrm{Beta}_i^M-1 \right|-\left|\textrm{Beta}_i^*-1 \right|= -\frac{\alpha }{1+\alpha } \left( 1+ \frac{N}{{\bar{\lambda }}-(N-1)}\frac{{\bar{\gamma }}_{-N}}{\gamma _i}\right) \ge 0. \end{aligned}$$

Other cases of ${\bar{\gamma }}_{-N}/\gamma _i$ follow by similar arguments. $\square$