Relative performance concerns among investment managers

Mark Whitmeyer

doi:10.1007/s10436-019-00343-2

Annals of Finance

pp 1–27 | Cite as

Relative performance concerns among investment managers

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Mark Whitmeyer

Research Article

First Online: 07 February 2019

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Abstract

This paper examines the strategic interaction of n portfolio managers with relative performance concerns. We characterize the unique constant Nash equilibrium and derive some compelling results. Surprisingly, in equilibrium, more risk tolerant players do not generally take riskier positions than less risk tolerant players. We derive sufficient conditions under which this relation does hold. We also examine the effects of adding new players to the game on the equilibrium, and look at the equilibrium in the limiting case as the number of players goes to infinity. We show that for a symmetric population, the equilibrium strategy of the players converges pointwise to some limiting equilibrium policy.

Keywords

Behavioral finance Relative concerns Portfolio choice Continuous-time finance

JEL Classification

G11 G40 C72 C73

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Notes

Appendix A

A.1 Theorem 1 proof

We proceed by considering WLOG player 1’s problem. First, apply the Dynamic Programming Principle:

$$\begin{aligned} V(X_{1},\ldots ,X_{n},t; \pi _{2},\ldots ,\pi _{n}) = \max _{\pi }{\mathbb {E}}[u(X_{1T}^{\pi _{1}})|X_{1t} = x_{1},\ldots ,X_{nt} = x_{n}] \end{aligned}$$

where for convenience we write only the first argument of player 1’s utility function instead of $u_{iT}(X_{iT}, \times _{j \ne i} R_{ijT})$.

In infinitesimal form, $\max _{\pi }{\mathbb {E}}(dV) = 0$. By the Principle of Optimality, V has zero drift at $\pi ^{*}$. Now, we use It${\hat{o}}$’s Lemma:

$$\begin{aligned} dV(X_{1t},\ldots ,X_{n},t; \pi _{2},\ldots ,\pi _{n}) = V_{t}dt + \sum _{i=1}^{n}V_{X_{i}}dX_{i} + \frac{1}{2}\sum _{i,j =1}^{n}V_{X_{i}X_{j}}dX_{i}dX_{j}\nonumber \\ \end{aligned}$$

(A1)

We use (1) and (2) to obtain:

$$\begin{aligned} \begin{aligned} (dX_{it})^{2}&= \sigma ^{2}_{X_{i}}\pi _{it}^{2}X_{it}^{2}dt dX_{it}dX_{jt} = \rho _{ij}\sigma _{i}\sigma _{j}X_{it}X_{jt}\pi _{it}\pi _{jt}dt \end{aligned} \end{aligned}$$

and substitute into (A1) to obtain

$$\begin{aligned} \begin{aligned} dV&= \bigg [V_{t} + \sum _{i=1}^{n}V_{X_{i}}X_{i}[r+\pi _{i}\mu _{i}] + \frac{1}{2}\sum _{i=1}^{n}V_{X_{i}X_{i}}X_{i}^{2}\pi _{i}^{2}\sigma _{i}^{2} \\&\quad +\sum _{i=1}^{n}\sum _{j=i+1}^{n}\rho _{ij}V_{X_{i}X_{j}}X_{i}X_{j}\pi _{i}\pi _{j}\sigma _{i}\sigma _{j} \bigg ]dt\\&\quad + \sum _{i=1}^{n}\bigg [ \cdots dW_{it} \bigg ] \end{aligned} \end{aligned}$$

At the optimum, the drift is equal to 0 and so we have:

$$\begin{aligned} \max _{\pi _{1}}\left\{ V_{t} + \sum _{i=1}^{n}V_{X_{i}}X_{i}[r+\pi _{i}\mu _{i}] + \frac{1}{2}\sum _{i=1}^{n}V_{X_{i}X_{i}}X_{i}^{2}\pi _{i}^{2}\sigma _{i}^{2} + \sum _{i=1}^{n}\sum _{j=i+1}^{n}\rho _{ij}V_{X_{i}X_{j}}X_{i}X_{j}\pi _{i}\pi _{j}\sigma _{i}\sigma _{j} \right\} \end{aligned}$$

The HJB Equation (A2) is:

$$\begin{aligned} dV= & {} V_{t}dt + V_{X_{1}}X_{1}r + \sum _{i=2}^{n}V_{X_{i}}X_{i}[r+\pi _{i}\mu _{i}] + \frac{1}{2}\sum _{i=2}^{n}V_{X_{i}X_{i}}X_{i}^{2}\pi _{i}^{2}\sigma _{i}^{2}\nonumber \\&+\, \sum _{i=2}^{n}\sum _{j=i+1}^{n}\rho _{ij}V_{X_{i}X_{j}}X_{i}X_{j}\pi _{i}\pi _{j}\sigma _{i}\sigma _{j}\nonumber \\&+ \max _{\pi _{1}}\left\{ V_{X_{1}}X_{1}\pi _{1}\mu _{1} + \frac{1}{2}V_{X_{1}X_{1}}X_{1}^{2}\pi _{1}^{2}\sigma _{1}^{2} + \sum _{j \ne 1}\rho _{1j}V_{X_{1}X_{j}}X_{1}X_{j}\pi _{1}\pi _{j}\sigma _{1}\sigma _{j} \right\} \nonumber \\ \end{aligned}$$

(A2)

We take the First Order Condition of the HJB Equation (A2):

$$\begin{aligned} V_{X_{1}}X_{1}\mu _{1} + V_{X_{1}X_{1}}X_{1}^{2}\pi _{1}\sigma _{1}^{2} + \sum _{j \ne 1}\rho _{1j}V_{X_{1}X_{j}}X_{1}X_{j}\pi _{j}\sigma _{1}\sigma _{j} = 0 \end{aligned}$$

Rearranging, obtain:

$$\begin{aligned} \pi _{1}^{*} = -\left( \frac{V_{X_{1}}X_{1}\mu _{1} + \sum _{j \ne 1}\rho _{1j}V_{X_{1}X_{j}}X_{1}X_{j}\pi _{j}\sigma _{1}\sigma _{j}}{ V_{X_{1}X_{1}}X_{1}^{2}\sigma _{1}^{2}}\right) \end{aligned}$$

(A3)

Now suppose

$$\begin{aligned} V(X_{1},\ldots ,T) = U(X_{1}) = \frac{f(t)}{1-\gamma _{i}}\left( X_{it}^{{\hat{\theta }}_{i}}\prod _{j \ne i}R_{ijt}^{\theta _{ij}}\right) ^{(1-\gamma _{i})} \end{aligned}$$

(A4)

For computational convenience, set

$$\begin{aligned} {\mathcal {A}} :=\left( X_{it}^{{\hat{\theta }}_{i}}\prod _{j \ne i}R_{ijt}^{\theta _{ij}}\right) ^{(1-\gamma _{i})} \end{aligned}$$

Thus, (A4) can be written as:

$$\begin{aligned} \frac{f(t)}{1-\gamma _{i}}{\mathcal {A}} \end{aligned}$$

(A5)

From (A5), we obtain:

$$\begin{aligned} \begin{aligned} V_{t}&= \frac{f_{t}}{1-\gamma _{1}}{\mathcal {A}} \quad \quad V_{X_{1}} = f(t){\mathcal {A}}X_{1}^{-1}\\ V_{X_{1}X_{1}}&= -\gamma _{1} f(t){\mathcal {A}}X_{1}^{-2} \quad \quad \quad V_{X_{j}} = -\theta _{1j} f(t){\mathcal {A}}X_{j}^{-1} \qquad \text {for }j \ne 1\\ V_{X_{j}X_{j}}&= (\theta _{1j}^{2}(1-\gamma _{1}) + \theta _{1j}) f(t){\mathcal {A}}X_{j}^{-2} \quad \qquad \qquad \qquad \qquad \text {for }j \ne 1\\ V_{X_{1}X_{j}}&= -\theta _{1j}(1-\gamma _{1}) f(t){\mathcal {A}}X_{1}^{-1}X_{j}^{-1} \quad \qquad \qquad \qquad \qquad \quad \text {for }j \ne 1\\ V_{X_{j}X_{k}}&= \theta _{1j}\theta _{1k}(1-\gamma _{1}) f(t){\mathcal {A}}X_{j}^{-1}X_{k}^{-1}\,\, \quad \qquad \qquad \qquad \qquad \text {for }j,k \ne 1 \end{aligned} \end{aligned}$$

(A6)

Substituting (A6) into (A3) we obtain:

$$\begin{aligned} \pi _{1}^{*} = \left( \frac{\mu _{1} + \sum _{j \ne 1}\rho _{1j}\theta _{1j}(\gamma _{1}-1)\pi _{j}\sigma _{1}\sigma _{j}}{ \gamma _{1}\sigma _{1}^{2}}\right) \end{aligned}$$

It remains to verify that our guess does indeed satisfy the HJB equation. Revisiting the HJB Equation,

$$\begin{aligned} 0 = V_{t} + \sum _{i=1}^{n}V_{X_{i}}X_{i}[r+\pi _{i}\mu _{i}] + \frac{1}{2}\sum _{i=1}^{n}V_{X_{i}X_{i}}X_{i}^{2}\pi _{i}^{2}\sigma _{i}^{2} + \sum _{i=1}^{n}\sum _{j=i+1}^{n}\rho _{ij}V_{X_{i}X_{j}}X_{i}X_{j}\pi _{i}\pi _{j}\sigma _{i}\sigma _{j} \end{aligned}$$

Substitute in (A6)

$$\begin{aligned} \begin{aligned} 0 =&\left( \frac{f_{t}}{1-\gamma _{1}}\right) {\mathcal {A}} + f{\mathcal {A}}\left( r+\pi _{1}\mu _{1}\right) - \sum _{j\ne 1} \theta _{1j} f {\mathcal {A}}\left( r+\pi _{j}\mu _{j}\right) - \frac{1}{2}\gamma _{1}f{\mathcal {A}}\pi _{1}^{2}\sigma _{1}^{2} \\&+ \frac{1}{2}\sum _{j\ne 1} (\theta _{1j}^{2}(1-\gamma _{1}) + \theta _{1j})f{\mathcal {A}} \pi _{j}^{2}\sigma _{j}^{2}- \sum _{j\ne 1}\theta _{1j}(1-\gamma _{1})f{\mathcal {A}}\pi _{1}\pi _{j}\sigma _{1}\sigma _{j}\rho _{1j}\\&+ \sum _{i=2}\sum _{j=i+1}\theta _{1i}\theta _{1j}(1-\gamma _{1})f{\mathcal {A}}\pi _{i}\pi _{j}\sigma _{i}\sigma _{j}\rho _{ij} \end{aligned} \end{aligned}$$

Thus, f satisfies

$$\begin{aligned} \frac{f_{t}}{1-\gamma _{1}} + \xi _{1} f = 0 \end{aligned}$$

(A7)

where

$$\begin{aligned} \begin{aligned} \xi _{1} =&\left( r+\pi _{1}\mu _{1}\right) - \sum _{j\ne 1} \theta _{1j}\left( r+\pi _{j}\mu _{j}\right) - \frac{1}{2}\gamma _{1}\pi _{1}^{2}\sigma _{1}^{2} + \frac{1}{2}\sum _{j\ne 1} (\theta _{1j}^{2}(1-\gamma _{1}) + \theta _{1j})\ \pi _{j}^{2}\sigma _{j}^{2}\\&- \sum _{j\ne 1}\theta _{1j}(1-\gamma _{1})\pi _{1}\pi _{j}\sigma _{1}\sigma _{j}\rho _{1j} + \sum _{i=2}\sum _{j=i+1}\theta _{1i}\theta _{1j}(1-\gamma _{1})\pi _{i}\pi _{j}\sigma _{i}\sigma _{j}\rho _{ij} \end{aligned} \end{aligned}$$

Using the boundary condition of $f(T) = 1$ we solve the differential Eq. (A7), yielding $f(t) = e^{\xi _{1}(1-\gamma _{1})(T-t)}$. Thus, wealth is given as:

$$\begin{aligned} V(X_{1},\ldots ,t) = \frac{e^{\xi _{1}(1-\gamma _{1})(T-t)}}{1-\gamma _{1}}\left( X_{1t}^{{\hat{\theta }}_{1}}\prod _{j \ne 1}R_{1jt}^{\theta _{1j}}\right) ^{(1-\gamma _{1})} \end{aligned}$$

A.2 Equation (9) derivation

From Eq. (8) we have

$$\begin{aligned} \begin{aligned} \pi _{i}^{*}&= \frac{\mu + \rho \theta \sigma ^{2}(\gamma _{i} - 1)\xi }{\sigma ^{2}\gamma _{i} + \rho \theta \sigma ^{2}(\gamma _{i} - 1)} = \frac{\mu }{\sigma ^{2}(\gamma _{i}(1+\rho \theta ) - \rho \theta )} + \frac{\rho \theta (\gamma _{i} -1)\xi }{(\gamma _{i}(1+\rho \theta ) - \rho \theta )} \end{aligned} \end{aligned}$$

We sum up these equations over all players i as follows:

$$\begin{aligned} \begin{aligned} \sum _{i} \pi _{i}^{*}&= \sum _{i} \frac{\mu }{\sigma ^{2}(\gamma _{i}(1+\rho \theta ) - \rho \theta )} + \sum _{i} \frac{\rho \theta (\gamma _{i} -1)\xi }{(\gamma _{i}(1+\rho \theta ) - \rho \theta )}\\&= \left( \frac{\mu }{\sigma ^{2}}\right) \sum _{i} \frac{1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )} + \rho \theta \xi \sum _{i} \frac{(\gamma _{i} -1)}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )} \end{aligned} \end{aligned}$$

Recall that $\xi = \sum _{i}\pi _{i}^{*}$. This allows us to obtain

$$\begin{aligned} \begin{aligned} \xi \left( 1 - \rho \theta \sum _{i} \frac{\gamma _{i} -1}{\gamma _{i}(1+\rho \theta ) - \rho \theta }\right)&= \left( \frac{\mu }{\sigma ^{2}}\right) \sum _{i} \frac{1}{\gamma _{i}(1+\rho \theta ) - \rho \theta } \end{aligned} \end{aligned}$$

A.3 Lemma 1 proof

Evidently, it suffices to show that $\xi $ is strictly locally decreasing in $\gamma _{i}$ for all i. Recall that $\xi = A^{-1}B$, where A and B are as defined in Eq. (9). Then, using the chain rule,

$$\begin{aligned} \frac{\partial {\xi }}{\partial {\gamma _{i}}} = -A^{-2}B\frac{\partial {A}}{\partial {\gamma _{i}}} + A^{-1}\frac{\partial {B}}{\partial {\gamma _{i}}} \end{aligned}$$

Define $x_{i}$ as $x_{i} :=\gamma _{i}(1 + \rho \theta ) - \rho \theta $ and note that $\frac{\partial {A}}{\partial {\gamma _{i}}} = -\frac{\rho \theta }{x_{i}^{2}}$. Observe that $\frac{\partial {\xi }}{\partial {\gamma _{i}}} \le 0$ if and only if

$$\begin{aligned}&A\frac{\partial {B}}{\partial {\gamma _{i}}} - B\frac{\partial {A}}{\partial {\gamma _{i}}} \le 0\nonumber \\&\quad -\left( \frac{\mu }{\sigma ^{2}}\right) \left( \frac{1+\rho \theta }{x_{i}^{2}}\right) \left( 1 - \rho \theta \sum _{j} \frac{\gamma _{j} -1}{x_{j}}\right) + \left( \frac{\mu }{\sigma ^{2}}\right) \sum _{j} \frac{1}{x_{j}} \left( \frac{\rho \theta }{x_{i}^{2}} \right) \le 0\nonumber \\&\rho \theta \sum _{j} \frac{\gamma _{j}}{x_{j}} - \left( \frac{(\rho \theta )^{2}}{1 + \rho \theta }\right) \sum _{j}\frac{1}{x_{j}} - 1\le 0 \end{aligned}$$

(A8)

Define $\iota (\pmb {\gamma })$ as

$$\begin{aligned} \iota (\pmb {\gamma }) :=\rho \theta \sum _{j} \frac{\gamma _{j}}{x_{j}} - \left( \frac{(\rho \theta )^{2}}{1 + \rho \theta }\right) \sum _{j}\frac{1}{x_{j}} - 1 \end{aligned}$$

Then for arbitrary i,

$$\begin{aligned} \begin{aligned} \frac{\partial {\iota }}{\partial {\gamma _{i}}}&= \rho \theta \left( \frac{1}{x_{i}} - \frac{(1+\rho \theta )\gamma _{i}}{x_{i}^{2}}\right) + \left( \frac{(\rho \theta )^{2}}{1 + \rho \theta }\right) \left( \frac{(1 + \rho \theta )}{x_{i}^{2}}\right) = 0 \end{aligned} \end{aligned}$$

Thus, for all i, $\iota $ is constant in $\gamma _{i}$. Accordingly, we may simply substitute in any value of $\gamma _{i}$. For convenience, substitute in $\gamma _{i} = 1$ for all i. From (A8),

$$\begin{aligned} \begin{aligned} 0&\ge \rho \theta \sum _{i}1 - \left( \frac{(\rho \theta )^{2}}{1 + \rho \theta }\right) \sum _{i} 1 - 1 \quad \Leftrightarrow \quad \frac{1}{n-1} \ge \rho \theta \end{aligned} \end{aligned}$$

which clearly must hold. $\square $

A.4 Lemma 2 proof

We prove this result by contradiction. Let $\gamma _{i} > \frac{\rho \theta }{1 + \rho \theta }$ for all i and suppose that $g(\pmb {\gamma }) \le 0$. That is, $g(\pmb {\gamma }) \le 0$

$$\begin{aligned} \begin{aligned}&\Leftrightarrow \quad \qquad \qquad \prod _{i}(\gamma _{i}(1+\rho \theta ) - \rho \theta ) - \rho \theta \sum _{i}(\gamma _{i}-1)\prod _{j \ne i}(\gamma _{j}(1+\rho \theta ) - \rho \theta ) \le 0\\&\Leftrightarrow \quad \qquad \qquad 1 \le \rho \theta \sum _{i}\frac{\gamma _{i} - 1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )}\\ \end{aligned} \end{aligned}$$

Define $\psi (\gamma _{i})$ as

$$\begin{aligned} \psi (\gamma _{i}) :=\frac{\gamma _{i} - 1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )} \end{aligned}$$

Then,

$$\begin{aligned} \begin{aligned} \frac{\partial {\psi }}{\partial {\gamma _{i}}}&= \frac{1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )} - \frac{(\gamma _{i} - 1)(1 + \rho \theta )}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )^{2}}\\&= \frac{(\gamma _{i}(1+\rho \theta ) - \rho \theta ) - (\gamma _{i} - 1)(1 + \rho \theta )}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )^{2}}\\&= \frac{1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )^{2}}\\ \end{aligned} \end{aligned}$$

which is continuous and strictly greater than 0 for $\gamma _{i} \in \left( \frac{\rho \theta }{1 + \rho \theta }, \infty \right) $. Now we examine $\psi (\gamma _{i})$ in the limit:

$$\begin{aligned} \lim _{\gamma _{i} \rightarrow \infty } \psi (\gamma _{i}) = \frac{1}{(1+\rho \theta )} \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned} 1 \le \rho \theta \sum _{i}\frac{\gamma _{i} - 1}{(\gamma _{i}(1+\rho \theta ) - \rho \theta )}&< \frac{n \rho \theta }{(1+\rho \theta )} \end{aligned} \end{aligned}$$

Or, $1 < \rho \theta (n-1)$, which is clearly false. $\square $

A.5 Proposition 2 proof

If the parameters are completely symmetric across all players, then $\pi ^{*} = \pi _{i}^{*}$ for all i and

$$\begin{aligned} \pi ^{*} = \left( \frac{\mu }{\sigma ^{2}} \right) \left( \frac{1}{\gamma - \rho \theta (n-1)(\gamma -1)} \right) \end{aligned}$$

First, note that the denominator, $\gamma - \rho \theta (n-1)(\gamma -1)$ is strictly greater than 0 for permissible values of the parameters. Then, $\frac{\partial {\pi ^{*}}}{\partial {\gamma }} < 0$ if and only if $\rho \theta (n-1) -1 < 0$, which clearly must hold. Similarly, $\frac{\partial {\pi ^{*}}}{\partial {{\mathfrak {s}}}} > 0$ if and only if $\sigma > 0$. Likewise, $\frac{\partial {\pi ^{*}}}{\partial {\theta }}>$$(<)$ 0 if and only if $\rho (n-1)(\gamma -1)>$$(<)$ 0 if and only if $(\gamma -1)>$$(<)$ 0; and $\frac{\partial {\pi ^{*}}}{\partial {\rho }}>$$(<)$ 0 if and only if $\theta (n-1)(\gamma -1)>$$(<)$ 0 if and only if $(\gamma -1)>$$(<)$ 0. $\square $

A.6 Lemma 3 proof

Consider the situation with n players, indexed by i. Without loss of generality, suppose $\gamma _{n} < \gamma _{i}$ for all $i \ne n$. We prove by contradiction: suppose that in equilibrium there is some j such that $\pi _{j}^{*} \ge \pi _{n}^{*}$.

Since we are at equilibrium, $A \varvec{\pi }^{*} = {\mathbf {b}}$. For convenience, we divide both sides by the scalars $\rho $, $\theta $ and $\sigma ^{2}$ and so we may rewrite the relationship as:

$$\begin{aligned} \begin{pmatrix} \frac{\gamma _{1}}{\rho \theta } &{}\quad (1 - \gamma _{1}) &{}\quad \cdots &{}\quad \cdots &{}\quad (1 - \gamma _{1}) \\ (1 - \gamma _{2}) &{}\quad \frac{\gamma _{2}}{\rho \theta } &{}\quad (1 - \gamma _{2}) &{}\quad \ddots &{}\quad (1 - \gamma _{2})\\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots &{}\quad \vdots \\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots &{}\quad \vdots \\ (1 - \gamma _{n}) &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \frac{\gamma _{n}}{\rho \theta } \end{pmatrix} \begin{pmatrix} \pi _{1}^{*} \\ \pi _{2}^{*} \\ \vdots \\ \vdots \\ \pi ^{*} \end{pmatrix} = \begin{pmatrix} \frac{\mu }{\rho \theta \sigma ^{2}} \\ \frac{\mu }{\rho \theta \sigma ^{2}} \\ \vdots \\ \vdots \\ \frac{\mu }{\rho \theta \sigma ^{2}} \end{pmatrix} \end{aligned}$$

Thus, we have

$$\begin{aligned} \begin{aligned} \frac{1}{\rho \theta }\gamma _{j}\pi _{j}^{*} + (1 - \gamma _{j})\sum _{i \ne j}\pi _{i}^{*}&= \frac{\mu }{\rho \theta \sigma ^{2}}, \quad \frac{1}{\rho \theta }\gamma _{n}\pi _{n}^{*} \\&\quad + (1 - \gamma _{n})\sum _{i \ne n}\pi _{i}^{*} = \frac{\mu }{\rho \theta \sigma ^{2}}\\ \end{aligned} \end{aligned}$$

(A9)

Combining,

$$\begin{aligned} \pi _{j}^{*}\left( \frac{1}{\rho \theta }\gamma _{j} + \gamma _{n} -1\right) = \pi _{n}^{*}\left( \frac{1}{\rho \theta }\gamma _{n} + \gamma _{j} -1\right) + (\gamma _{j} - \gamma _{n})\sum _{i \ne j, n}\pi _{i}^{*} \end{aligned}$$

(A10)

The set $\left\{ \pi ^{*}_{i}\right\} _{i \ne j,n}$ is discrete and finite and thus compact. Therefore, it must have a set of maximal elements, one of whose members is WLOG $\pi _{m}^{*}$. Then, from our supposition above and (A10),

$$\begin{aligned} \pi _{j}\left( \frac{1}{\rho \theta }\gamma _{j} + \gamma _{n} -1\right) \le \pi _{j}\left( \frac{1}{\rho \theta }\gamma _{n} + \gamma _{j} -1\right) + (\gamma _{j} - \gamma _{n})(n-2)\pi _{m} \end{aligned}$$

Rearranging, we obtain $\pi _{j}^{*}\frac{1 - \rho \theta }{\rho \theta (n-2)} \le \pi _{m}^{*}$, whence we obtain $\pi _{m}^{*} > \pi _{j}^{*} \ge \pi _{n}^{*}$. Just as we have Eq. (A10), we can also obtain in analogous fashion:

$$\begin{aligned} \pi _{m}^{*}\left( \frac{1}{\rho \theta }\gamma _{m} + \gamma _{n} -1\right) = \pi _{n}^{*}\left( \frac{1}{\rho \theta }\gamma _{n} + \gamma _{m} -1\right) + (\gamma _{m} - \gamma _{n})\sum _{i \ne m, n}\pi _{i}^{*} \end{aligned}$$

Suppose $\pi _{j}^{*} \in \max _{i \ne m,n} \left\{ \pi ^{*}_{i}\right\} $. We must have

$$\begin{aligned} \pi _{m}^{*}\left( \frac{1}{\rho \theta }\gamma _{m} + \gamma _{n} -1\right) \le \pi _{m}^{*}\left( \frac{1}{\rho \theta }\gamma _{n} + \gamma _{m} -1\right) + (\gamma _{m} - \gamma _{n})(n-2)\pi _{j}^{*} \end{aligned}$$

Rearranging, we obtain $\pi _{m}^{*}\frac{1 - \rho \theta }{\rho \theta (n-2)} \le \pi _{j}^{*}$, whence we obtain $\pi _{j}^{*}> \pi _{m}^{*} > \pi _{j}^{*}$, which is obviously a contradiction. Thus $\pi _{j}^{*} \notin \max _{i \ne m,n} \left\{ \pi ^{*}_{i}\right\} $ and so we must pick another element of that set, say $\pi _{q}^{*}$. We must have

$$\begin{aligned} \pi _{m}^{*}\left( \frac{1}{\rho \theta }\gamma _{m} + \gamma _{n} -1\right) \le \pi _{m}^{*}\left( \frac{1}{\rho \theta }\gamma _{n} + \gamma _{m} -1\right) + (\gamma _{m} - \gamma _{n})(n-2)\pi _{q}^{*} \end{aligned}$$

Rearranging, we obtain $\pi _{m}^{*}\frac{1 - \rho \theta }{\rho \theta (n-2)} \le \pi _{q}^{*}$, whence we obtain $\pi _{q}^{*} > \pi _{m}^{*}$.

But this contradicts our our statement above, that $\pi _{m}^{*}$ is a maximal element of $\left\{ \pi ^{*}_{i}\right\} _{i \ne j,n}$. We have generated a contradiction, and since j indexed an arbitrary player, we conclude that if $\gamma _{n} < \gamma _{i}$$\forall i \ne n$, then $\pi _{n}^{*} > \pi _{i}^{*}$$\forall i \ne n$.

The final part of this lemma, the case where $\gamma _{n}$ is not uniquely minimal, and there exists some $\gamma _{k} = \gamma _{n}$ holds trivially: clearly $\pi _{k} = \pi _{n}$. $\square $

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Mark Whitmeyer
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1.Department of EconomicsUniversity of Texas at AustinAustinUSA

Relative performance concerns among investment managers

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