A paradox in time-consistency in the mean–variance problem?

Abstract

We establish new conditions under which a constrained (no short-selling) time-consistent equilibrium strategy, starting at a certain time, will beat the unconstrained counterpart, as measured by the magnitude of their corresponding equilibrium mean–variance value functions. We further show that the pure strategy of solely investing in a risk-free bond can sometimes simultaneously dominate both constrained and unconstrained equilibrium strategies. With numerical experiments, we also illustrate that the constrained strategy can dominate the unconstrained one for most of the commencement dates (even more than 90%) of a prescribed planning horizon. Under a precommitment approach, the value function of an investor increases with the size of the admissible sets of strategies. However, this may fail to be true under the game-theoretic paradigm, as the constraint of time-consistency itself affects the value function differently when short-selling is and is not prohibited.

Appendix: Technical proofs in Sect. 4

1.1 A.1 Proof of Theorem 4.1

Recall (1.7) and (1.9). Then (4.1) implies $c^{(U)}_{T}=c^{(C)}_{T}<1$. Because we have $G(x)=x$ when $x\in(0,1)$ (see Theorem 1.6), we get for any $t$ with $c^{(C)}_{t} \in(0,1)$ that $c^{(C)}_{t}=G(\frac{\alpha_{t}}{\sigma^{2}_{t}} d^{(C)}_{t})=\frac{\alpha_{t}}{\sigma^{2}_{t}} d^{(C)}_{t}$ by (1.9). Thus both $c^{(C)}_{\cdot}$ and $c^{(U)}_{\cdot}$ satisfy the same form of integral equations (see (1.7) and (1.9)) over $[t,T]$. Therefore, $c^{(U)}_{t}=c^{(C)}_{t} \in(0,1)$ for any $t>\sup\{t< T: c^{(C)}_{t} \notin(0,1) \}$. By Lemmas 3.1 (i) and 3.5 (i), (4.2) implies that both $c^{(U)}$ and $c^{(C)}$ are increasing in $t$ whenever $c^{(U)}_{t}=c^{(C)}_{t}>0$. Define

$$ \tau^{*}_{0}:=\sup\{t< T : c^{(U)}_{t}=c^{(C)}_{t}=0 \}\in[-\infty, T). $$

We consider two separate cases:

(i) If $\tau^{*}_{0}=-\infty$, then $c^{(U)}_{t}=c^{(C)}_{t} \in(0,1)$ for all $t\leq T$ and it is clear that $V^{(U)}(t,x)= V^{(C)}(t,x)$ for all $t \leq T$ and $x>0$.

(ii) If $\tau^{*}_{0}>-\infty$, we have $c^{(U)}_{t}=c^{(C)}_{t} \in(0,1)$ for all $t \in(\tau^{*}_{0},T)$ and $c^{(U)}_{t} \leq c^{(C)}_{t}=0$ for all $t <\tau^{*}_{0}$ because of Lemmas 3.1 (i) and 3.5 (i). Therefore, we obtain $V^{(U)}(t,x)= V^{(C)}(t,x)$ for all $t \in[\tau^{*}_{0},T]$ and $x>0$. Recall that the equilibrium value function in terms of the investment-to-wealth ratio $c$ is

$$ V(t,x)= e^{\int^{T}_{t} (r_{s}+\alpha_{s} c_{s}) ds}x-\frac{\gamma_{t}}{2} \Big( e^{\int^{T}_{t} 2(r_{s}+\alpha_{s} c_{s})+\sigma_{s}^{2} (c_{s})^{2} ds}-e^{\int^{T}_{t} 2(r_{s}+\alpha_{s} c_{s})ds}\Big)x. $$

Differentiating the product $e^{-\int^{T}_{t} r_{s} ds} V(t,x)$ gives

$$\begin{aligned} \frac{d}{dt}\big(e^{-\int^{T}_{t} r_{s} ds} V(t,x)\big) =&\frac{r_{t} \gamma_{t}-\gamma'_{t}}{2}e^{\int^{T}_{t} r_{s}+2\alpha_{s} c_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c_{s})^{2} ds}-1\big)x \\ & -\alpha_{t} c_{t} \Big(e^{\int^{T}_{t} \alpha_{s} c_{s} ds}-\gamma_{t} e^{\int^{T}_{t} r_{s}+2\alpha_{s} c_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c_{s})^{2} ds}-1\big) \Big)x \\ & +\frac{\gamma_{t} \sigma_{t}^{2} (c_{t})^{2}}{2} e^{\int^{T}_{t} r_{s}+2\alpha_{s} c_{s}+\sigma_{s}^{2} (c_{s})^{2} ds}x. \end{aligned}$$

(A.1)

We can then deduce that for any $t < \tau^{*}_{0}$ and $x>0$,

$$\begin{aligned} &\frac{d}{dt}\Big(e^{-\int^{T}_{t} r_{s} ds} \big(V^{(U)}(t,x)-V^{(C)}(t,x)\big)\Big) \\ &= \frac{r_{t} \gamma_{t}-\gamma'_{t}}{2}e^{\int^{T}_{t} r_{s} ds}\Big(e^{2\int ^{T}_{t} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big) \\ &\phantom{:=:\frac{r_{t} \gamma_{t}-\gamma'_{t}}{2}e^{\int^{T}_{t} r_{s} ds}}-e^{2\int^{T}_{\tau^{*}_{0}} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{\tau ^{*}_{0}} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big)\Big)x \\ &\phantom{=:} -\frac{\gamma_{t} \sigma_{t}^{2} (c^{(U)}_{t})^{2}}{2} e^{\int^{T}_{t} r_{s}+2 \alpha _{s} c^{(U)}_{s}+\sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}x \\ &< \frac{r_{t} \gamma_{t}-\gamma'_{t}}{2} e^{\int^{T}_{t} r_{s} ds}\Big(e^{2\int ^{T}_{t} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big) \\ &\phantom{=: \frac{r_{t} \gamma_{t}-\gamma'_{t}}{2} e^{\int^{T}_{t} r_{s} ds}\Big(}-e^{2\int^{T}_{\tau^{*}_{0}} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{\tau ^{*}_{0}} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big)\Big)x, \end{aligned}$$

(A.2)

where the first equality follows because $c^{(C)}_{t}=0$ for $t <\tau^{*}_{0}$, $c^{(C)}_{t}=c^{(U)}_{t}$ for $t \in[\tau^{*}_{0},T]$ and $c^{(U)}_{t}$ satisfies (1.7). Next we compute

$$\begin{aligned} & \frac{d}{dt}\Big(e^{2\int^{T}_{t} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int ^{T}_{t} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big)\Big) \\ & = -2\alpha_{t} c^{(U)}_{t} e^{2\int^{T}_{t} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big) \\ & \phantom{=:} -\sigma^{2} (c^{(U)}_{s})^{2} e^{\int^{T}_{t} (2\alpha_{s} c^{(U)}_{s}+\sigma_{s}^{2} (c^{(U)}_{s})^{2}) ds} \\ & = \sigma^{2} (c^{(U)}_{s})^{2} e^{\int^{T}_{t} (2\alpha_{s} c^{(U)}_{s}+\sigma_{s}^{2} (c^{(U)}_{s})^{2}) ds}-\frac{2\alpha_{t}}{\gamma_{t}}c^{(U)}_{t} e^{-\int^{T}_{t} (r_{s}-\alpha_{s} c^{(U)}_{s} ) ds}\geq0 \end{aligned}$$

for $t<\tau^{*}_{0}$, where the second equality follows from (1.7), and the positivity from the fact that $c^{(U)}_{t} \leq0$ for $t<\tau^{*}_{0}$. Hence we have

$$ e^{2\int^{T}_{t} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big)\leq e^{2\int^{T}_{\tau^{*}_{0}} \alpha_{s} c^{(U)}_{s} ds} \big(e^{\int^{T}_{\tau^{*}_{0}} \sigma_{s}^{2} (c^{(U)}_{s})^{2} ds}-1\big) $$

for $t<\tau^{*}_{0}$. Together with the result in (4.2), the inequality (A.2) thus becomes

$$ \frac{d}{dt}\Big(e^{-\int^{T}_{t} r_{s} ds} \big(V^{(U)}(t,x)-V^{(C)}(t,x)\big)\Big) < 0 \qquad\text{ for $t< \tau^{*}_{0}$ and $x>0$.} $$

Since $V^{(U)}(\tau^{*}_{0},x)=V^{(C)}(\tau^{*}_{0},x)$, we have for $t<\tau ^{*}_{0}$ and $x>0$ that

$$ e^{-\int^{T}_{t} r_{s} ds} \big(V^{(U)}(t,x)-V^{(C)}(t,x)\big)\leq e^{-\int ^{T}_{\tau^{*}_{0}} r_{s} ds} \big(V^{(U)}(\tau^{*}_{0},x)-V^{(C)}(\tau^{*}_{0},x)\big)=0, $$

and so the claim follows. □

1.2 A.2 Proof of Theorem 4.3

We only prove the constrained case and omit the unconstrained one as its argument is analogous. Using the differentiation in (A.1), we have for $t \leq T$ and $x>0$ that

$$\begin{aligned} \frac{d}{dt}\frac{ V^{(C)}(t,x)}{V^{(Rf)}(t,x)}&=\frac{d}{dt}\frac{ V^{(C)}(t,x)}{e^{\int^{T}_{t} r_{s} ds} x} \\ &=\left(\frac{r_{t} \gamma_{t}-\gamma'_{t}}{2}\right)e^{\int^{T}_{t} r_{s}+2\alpha _{s} c^{(C)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(C)}_{s})^{2} ds}-1\big) \\ &\phantom{=:} -\alpha_{t} c^{(C)}_{t} \Big(e^{\int^{T}_{t} \alpha_{s} c^{(C)}_{s} ds}-\gamma_{t} e^{\int^{T}_{t} r_{s}+2\alpha_{s} c^{(C)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(C)}_{s})^{2} ds}-1\big) \Big) \\ &\phantom{=:} +\frac{\gamma_{t} \sigma_{t}^{2} (c^{(C)}_{t})^{2}}{2} e^{\int^{T}_{t} r_{s}+2 \alpha _{s} c^{(C)}_{s}+\sigma_{s}^{2} (c^{(C)}_{s})^{2} ds} \\ &\leq -\alpha_{t} c^{(C)}_{t} \Big(e^{\int^{T}_{t} \alpha_{s} c^{(C)}_{s} ds}-\gamma_{t} e^{\int^{T}_{t} r_{s}+2\alpha_{s} c^{(C)}_{s} ds} \big(e^{\int^{T}_{t} \sigma_{s}^{2} (c^{(C)}_{s})^{2} ds}-1\big) \Big) \\ &\phantom{=:} +\frac{\gamma_{t} \sigma_{t}^{2} (c^{(C)}_{t})^{2}}{2} e^{\int^{T}_{t} r_{s}+2\alpha_{s} c^{(C)}_{s}+\sigma_{s}^{2} (c^{(C)}_{s})^{2} ds} , \end{aligned}$$

(A.3)

where the first inequality follows from (4.3). On the other hand, (1.9) gives

$$ \frac{\alpha_{t}}{\sigma_{t}^{2}}\bigg(\frac{1}{\gamma_{t}}e^{-\int^{T}_{t} (r_{s}+\alpha_{s} c^{(C)}_{s}+\sigma_{s}^{2} (c^{(C)}_{s})^{2}) ds}+e^{-\int^{T}_{t} \sigma_{s}^{2} (c^{(C)}_{s})^{2} ds}-1\bigg) \geq c^{(C)}_{t} $$

whenever $c^{(C)}_{t} \geq0$. Thus (A.3) now becomes

$$\begin{aligned} &\frac{d}{dt}\frac{ V^{(C)}(t,x)}{V^{(Rf)}(t,x)} \\& \textstyle\begin{cases}\leq -\frac{\gamma_{t} \sigma_{t}^{2} (c^{(C)}_{t})^{2}}{2} e^{\int ^{T}_{t} r_{s}+2\alpha_{s} c^{(C)}_{s}+\sigma_{s}^{2} (c^{(C)}_{s})^{2} ds} \leq0 &\qquad \text{whenever } c^{(C)} \in(0,1],\\ \leq0 &\qquad \text{whenever } c^{(C)}=0. \end{cases}\displaystyle \end{aligned}$$

Hence

$$ \frac{V^{(C)}(t,x)}{V^{(Rf)}(t,x)} \geq\frac {V^{(C)}(T,x)}{V^{(Rf)}(T,x)}=1 \qquad\text{for } t \leq T, $$

and our claim is justified for the constrained case.  □