Nash equilibrium strategies and survival portfolio rules in evolutionary models of asset markets

Abstract

We consider a stochastic model of a financial market with one-period assets and endogenous asset prices. The model was initially developed and analyzed in the context of Evolutionary Finance with the main focus on questions of “survival and extinction” of investment strategies (portfolio rules). In this paper we view the model from a different, game-theoretic, perspective and analyze Nash equilibrium properties of survival portfolio rules.

Appendix

Proof of Lemma 1

Let us first assume that at least one of the coordinates of $\lambda$ is zero, i.e. $\mathbf {K}:=\{k:\lambda _{k} =0\}\ne \emptyset$. Denote by $S^{\prime }$ the set of those s for which $\sum _{k=1}^{K}R_{k}(s)\lambda _{k}=0$. If $s\in S^{\prime }$, then $R_{k}(s)\lambda _{k}=0$ for all k. Consequently,

$$\begin{aligned} B=\sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}=0,\ s\in S^{\prime }, \end{aligned}$$

and so $F(\lambda ,\kappa ;s)=+\infty$, $s\in S^{\prime }$. Thus for any $H>1$, we have

$$\begin{aligned}&E\min [H,\ln F(\lambda ,\kappa ;s)]\nonumber \\&=E\min [H,\ln F(\lambda ,\kappa ;s)]\mathbf {1}_{S^{\prime }}+E\min [H,\ln F(\lambda ,\kappa ;s)]\mathbf {1}_{S\backslash S^{\prime }}\ge \nonumber \\&HP(S^{\prime })+2(\ln c)(1-P(S^{\prime })), \end{aligned}$$

(29)

where $\mathbf {1}_{\Gamma }$ is the indicator function of the set $\Gamma$.

Assume that $P(S^{\prime })>0$. Then for all H large enough, the expression in (29) is greater than 1. The assertion of Lemma 1 will be true for any H for which (29) is greater than 1 and $\delta =1$.

Now let us assume that $P(S^{\prime })=0$, i.e. $\sum _{k=1}^{K}R_{k} (s)\lambda _{k}>0$ (a.s.). In this case, we will conduct the proof of the lemma in three steps.

Step 1: Let us prove that

$$\begin{aligned} E\ln F(\lambda ,\kappa ;s)>0\ \mathrm{for all}\ \kappa \in (0,1]. \end{aligned}$$

(30)

By applying Jensen’s inequality, we find

$$\begin{aligned} E\ln \sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}^{*}}{\lambda _{k}^{*} \kappa +\lambda _{k}(1-\kappa )}\ge E\sum _{k=1}^{K}R_{k}(s)\ln \frac{\lambda _{k}^{*}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}\end{aligned}$$

(31)

$$\begin{aligned} =\sum _{k=1}^{K}\lambda _{k}^{*}\ln \frac{\lambda _{k}^{*}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )} \end{aligned}$$

(32)

and

$$\begin{aligned} E\ln \sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}<\ln E\sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}\end{aligned}$$

(33)

$$\begin{aligned} =\ln \sum _{k=1}^{K}\lambda _{k}^{*}\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}. \end{aligned}$$

(34)

The inequality in (33) is strict because the right-hand side of (33) is finite (all $\lambda _{k}^{*}$ and some $\lambda _{k}$ are strictly positive) and there is no constant $\gamma$ such that

$$\begin{aligned} \sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}=\gamma \ \mathrm{(a.s.)}. \end{aligned}$$

(35)

Indeed, if (35) holds, then

$$\begin{aligned} \sum _{k=1}^{K}R_{k}(s)\nu _{k}=0\text {\ }\mathrm{(a.s.)}, \end{aligned}$$

(36)

where $\nu _{k}:=\lambda _{k}[\lambda _{k}^{*}\kappa +\lambda _{k} (1-\kappa )]^{-1}-\gamma$. Observe that at least one of the numbers $\nu _{k}$ is not equal to zero. Otherwise, $\lambda _{k}=[\lambda _{k}^{*} \kappa +\lambda _{k}(1-\kappa )]\gamma$ for all k, and by summing up these equalities over k, we get $\gamma =1$, which yields $\lambda _{k}=[\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )]$. Thus $\lambda _{k}=\lambda _{k} ^{*}$ for all k (recall that $\kappa \ne 0$). This is a contradiction because $\lambda \ne \lambda ^{*}$. From this we conclude that at least one of the numbers $v_{k}$ is not equal to zero, which shows that (36) cannot hold since the functions $R_{k}(s)$ are linearly independent with respect to the given distribution on S.

By combining (31)–(34), we get

$$\begin{aligned} E\ln F(\lambda ,\kappa ;s)>\Phi _{\kappa }(\lambda ), \end{aligned}$$

(37)

where

$$\begin{aligned} \Phi _{\kappa }(\lambda ):=\sum _{k=1}^{K}\lambda _{k}^{*}\ln \frac{\lambda _{k}^{*}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}-\ln \sum _{k=1} ^{K}\lambda _{k}^{*}\frac{\lambda _{k}}{\lambda _{k}^{*}\kappa +\lambda _{k}(1-\kappa )}. \end{aligned}$$

In [9], pp. 337–338, it was proved that $\Phi _{\kappa }(\lambda )>0$ for all $\lambda =(\lambda _{1},\ldots ,\lambda _{K})>0$. Therefore $\Phi _{\kappa }(\lambda (1-\varepsilon )+\varepsilon \lambda ^{*})>0$ for each $\varepsilon >0$. The function $\Phi _{\kappa }(\lambda )$ is finite and continuous on $\Delta ^{K}$. Consequently, $\Phi _{\kappa }(\lambda )=\lim _{\varepsilon \downarrow 0}\Phi _{\kappa }(\lambda (1-\varepsilon )+\varepsilon \lambda ^{*}))\ge 0$. This inequality combined with (37) yields (30).

Step 2: Let us show that for all $\kappa \in (0,1]$,

$$\begin{aligned} F(\lambda ,\kappa ,s)\ge F(\lambda ,1,s)=[\sum _{k=1}^{K}R_{k}(s)\frac{\lambda _{k}}{\lambda _{k}^{*}}]^{-1} \end{aligned}$$

(38)

($0^{-1}:=$$+\infty$). Fix any $s\in S$. If $R_{k}(s)\lambda _{k}=0$ for all k, then $F(\lambda ,\kappa ,s)=F(\lambda ,1,s)=+\infty$, and so inequality (38) holds. Let us assume that $R_{k}(s)\lambda _{k}\ne 0$ for some k. Then the denominator in (20) is strictly positive, and so $c^{2}\le F(\lambda ,\kappa ,s)<\infty$. For $\kappa \in (0,1)$, we have $F(\lambda ,\kappa ;s)=G(\lambda ,u,s)$, where

$$\begin{aligned} G(\lambda ,u,s):=\frac{\sum _{k=1}^{K}R_{k}(s)\dfrac{\lambda _{k}^{*}}{\lambda _{k}^{*}u+\lambda _{k}}}{\sum _{k=1}^{K}R_{k}(s)\dfrac{\lambda _{k} }{\lambda _{k}^{*}u+\lambda _{k}}}=\frac{\sum _{k=1}^{K}R_{k}(s)\dfrac{\lambda _{k}^{*}}{\lambda _{k}^{*}+\lambda _{k}/u}}{\sum _{k=1}^{K} R_{k}(s)\dfrac{\lambda _{k}}{\lambda _{k}^{*}+\lambda _{k}/u}}, \end{aligned}$$

and $u:=\kappa (1-\kappa )^{-1}$. When the variable $\kappa$ ranges through (0, 1), the variable u ranges through $(0,+\infty )$, and $G(\lambda ,u,s)\rightarrow [\sum _{k=1}^{K}R_{k}(s)(\lambda _{k}/\lambda _{k}^{*})^{-1}]^{-1}$ as $u\rightarrow \infty$. Thus, to verify (38), it is sufficient to show that the derivative $G(\lambda ,u;s)^{\prime }$ of the function $G(\lambda ,u;s)$ with respect to u is non-positive for all $u>0$. Assuming that the parameter s is fixed, we omit “s” in the notation and write

$$\begin{aligned}&G(\lambda ,u)^{\prime }\nonumber \\&=\frac{\left( \sum R_{k}\dfrac{\lambda _{k}^{*}}{\lambda _{k}^{*}u+\lambda _{k}}\right) ^{\prime }\left( \sum R_{k}\dfrac{\lambda _{k}}{\lambda _{k}^{*}u+\lambda _{k}}\right) -\left( \sum R_{k}\dfrac{\lambda _{k}^{*}}{\lambda _{k}^{*}u+\lambda _{k}}\right) \left( \sum R_{k}\dfrac{\lambda _{k}}{\lambda _{k}^{*}u+\lambda _{k}}\right) ^{\prime }}{\left( \sum R_{k}\dfrac{\lambda _{k}}{\lambda _{k}^{*}u+\lambda _{k}}\right) ^{2}}, \end{aligned}$$

where $\sum =\sum _{k=1}^{K}$. The sign of the above fraction is the same as the sign of its nominator

$$\begin{aligned} J:=-\,\sum R_{k}\frac{(\lambda _{k}^{*})^{2}}{(\lambda _{k}^{*}u+\lambda _{k})^{2}}\sum R_{k}\frac{\lambda _{k}}{\lambda _{k}^{*}u+\lambda _{k}}\nonumber \\ +\sum R_{k}\frac{\lambda _{k}^{*}}{\lambda _{k}^{*}u+\lambda _{k}}\sum R_{k} \frac{\lambda _{k}\lambda _{k}^{*}}{(\lambda _{k}^{*}u+\lambda _{k})^{2}} \end{aligned}$$

By setting $w_{k}:=\lambda _{k}^{*}u+\lambda _{k}$ and $v_{k}:=\lambda _{k}^{*}w_{k}^{-1}$, we obtain $\lambda _{k}=w_{k}-\lambda _{k}^{*}u$ and

$$\begin{aligned} J= & {} -\,\sum R_{k}(\lambda _{k}^{*})^{2}w_{k}^{-2}\sum R_{k}\lambda _{k}w_{k} ^{-1}+\sum R_{k}\lambda _{k}^{*}w_{k}^{-1}\sum R_{k}\lambda _{k}\lambda _{k}^{*}w_{k}^{-2}\nonumber \\= & {} -\,\sum R_{k}(\lambda _{k}^{*})^{2}w_{k}^{-2}\sum R_{k}(w_{k}-\lambda _{k}^{*}u)w_{k}^{-1}\nonumber \\&+\,\sum R_{k}\lambda _{k}^{*}w_{k}^{-1}\sum R_{k} (w_{k}-\lambda _{k}^{*}u)\lambda _{k}^{*}w_{k}^{-2}\nonumber \\= & {} -\,\sum R_{k}v_{k}^{2}\sum (R_{k}-R_{k}uv_{k})+\sum R_{k}v_{k}\sum R_{k} (v_{k}-v_{k}^{2}u)\nonumber \\&-\,\sum R_{k}v_{k}^{2}(1-\sum R_{k}uv_{k})+\sum R_{k}v_{k}\left( \sum R_{k}v_{k}-\sum R_{k}v_{k}^{2}u\right) \nonumber \\= & {} -\,\sum R_{k}v_{k}^{2}+\left( \sum R_{k}v_{k}^{2}\right) \sum R_{k}v_{k}u+\left( \sum R_{k} v_{k}\right) ^{2}-\left( \sum R_{k}v_{k}\right) \sum R_{k}v_{k}^{2}u\nonumber \\= & {} \left( \sum R_{k}v_{k}\right) ^{2}-\sum R_{k}v_{k}^{2}\le 0. \end{aligned}$$

The last expression is non-positive by virtue of the Schwartz inequality (we use here the fact that $R_{k}\ge 0$ and $\sum R_{k}=1$). This completes the proof of inequality (38).

Step 3. Let us show that there exists a natural number m such that $E\min \{m,\ln F(\lambda ,1,s)\}>0$. The sequence of random variables

$$\begin{aligned} \phi _{m}:=\min \{m,\ln F(\lambda ,1,s)\}, m=1,2,\ldots , \end{aligned}$$

is bounded below by $2\ln c$ [see (21)], is nondecreasing and tends to $\ln F(\lambda ,1,s)$ for each s. Consequently, we have $E\phi _{m}\rightarrow E\ln F(\lambda ,1,s)>0$ [see (30)], and so $E\phi _{m}>0$ for some $m=m_{0}$. By setting $H:=m_{0}$, $\delta :=E\min \{H,\ln F(\lambda ,1;s)\}>0$ and using (38), we find

$$\begin{aligned} E\min \{H,\ln F(\lambda ,\kappa ;s)\}\ge E\min \{H,\ln F(\lambda ,1;s)\}=\delta , \end{aligned}$$

which proves Lemma 1 in the case when at least one of the coordinates of $\lambda$ is zero.

Now assume that $\lambda$ has no zero coordinates: $\lambda _{k}>0$ for each k. Then the function $\ln F(\lambda ,\kappa ;s)$, $\kappa \in [0,1]$, is uniformly bounded:

$$\begin{aligned} 2\ln c\le \ln F(\lambda ,\kappa ;s)\le \ln (\min _{k}\lambda _{k})^{-2}, \end{aligned}$$

and so $E\ln F(\lambda ,\kappa ;s)$ is continuous in $\kappa \in [0,1]$. It suffices to show that the infimum of $E\ln F(\lambda ,\kappa ;s)$ with $\kappa \in [0,1]$ is strictly positive (then $\delta$ can be defined as this infimum and H as $2|\ln c|+2|\ln \min \lambda _{k}|)$. In view of the continuity of $E\ln F(\lambda ,\kappa ;s)$ this will be proved if we establish the inequality $E\ln F(\lambda ,\kappa ;s)>0$ for each $\kappa \in [0,1]$. For $\kappa \in (0,1]$ this was proved above [see (30)] under the condition that $\sum _{k=1}^{K}R_{k}(s)\lambda _{k}>0$ (a.s.), which holds if $\lambda >0$. In the case of $\kappa =0$, we have

$$\begin{aligned} E\ln F(\lambda ,0,s)=E\ln \sum _{k=1}^{K}R_{k}(s)\dfrac{\lambda _{k}^{*} }{\lambda _{k}}\ge \ln \sum _{k=1}^{K}\lambda _{k}^{*}\dfrac{\lambda _{k}^{*}}{\lambda _{k}}\ge \sum _{k=1}^{K}\lambda _{k}^{*}\ln \dfrac{\lambda _{k} ^{*}}{\lambda _{k}}>0 \end{aligned}$$

as long as $\lambda ^{*}$,$\lambda >0$ and $\lambda \ne \lambda ^{*}$. $\square$