Portfolio selections under mean-variance preference with multiple priors for means and variances

Abstract

We study portfolio selections under mean-variance preference with multiple priors for means and variances. We introduce two types of multiple priors, the priors for means and the priors for variances of risky asset returns. As our framework, in the absence of a risk-free asset, the global minimum-variance portfolio is optimal when the investor is extremely ambiguity averse with respect to means, and the equally weighted portfolio is optimal when the investor is extremely ambiguity averse with respect to variances.

Appendix: Derivation of (12)

Let us denote by $v^{i,j}$ the $i\times j$ th element of V. The first derivative of the Lagrange function $L^{\theta ,V}$ with respect to $v^{i,j}$ is

$$\begin{aligned} \frac{\partial L^{\theta ,V}}{\partial v^{i,j}} = - \frac{\gamma }{2} \frac{\partial }{\partial v^{i,j}}u^\prime V u + \lambda _\theta \frac{\partial }{\partial v^{i,j}} \theta ^\prime (\varSigma + V)^{-1} \theta + \lambda _V\frac{\partial }{\partial v^{i,j}}\Vert V\Vert ^2. \end{aligned}$$

Recall that V is a symmetric matrix. We have

$$\begin{aligned} \frac{\partial }{\partial v^{i,j}}u^\prime V u = \left\{ \begin{array}{cc} 2 u^i u^j, &{}\quad \text{ if } i\ne j, \\ (u^i)^2, &{}\quad \text{ if } i= j, \end{array}\right. \quad \frac{\partial }{\partial v^{i,j}}\Vert V\Vert ^2 = \left\{ \begin{array}{cc} 4 v^{i,j}, &{}\quad \text{ if } i\ne j, \\ 2 v^{i,i}, &{}\quad \text{ if } i= j. \end{array}\right. \end{aligned}$$

(34)

We need to compute $\partial \theta ^\prime (\varSigma + V)^{-1} \theta / \partial v^{i,j}$. To do this, we need the following lemma.

Lemma 3

Let us denote $F:{\mathbb {R}}\rightarrow {\mathbb {R}}^{N\times N}$ by a matrix-valued function.

• 1. Suppose that F(x) is invertible for all x. Then, the following equality holds.

  $$\begin{aligned} \frac{\partial (F(x))^{-1}}{\partial x} = - (F(x))^{-1}\frac{\partial F(x)}{\partial x}(F(x))^{-1}. \end{aligned}$$

  (35)

• 2. Let us denote $g:{\mathbb {R}}^{N\times N}\rightarrow {\mathbb {R}}$ by a real-valued function. Then, the following equality holds.

  $$\begin{aligned} \frac{\partial g(F(x))}{\partial x} = \mathrm {tr}\left\{ \left( \frac{\partial g(F(x))}{\partial F(x)}\right) ^\prime \frac{\partial F(x)}{\partial x}\right\} . \end{aligned}$$

  (36)

Proof

We first prove (35). For all x, F(x) satisfies the following identity.

$$\begin{aligned} F(x)(F(x))^{-1} = I_N. \end{aligned}$$

Differentiating the above identity with respect to x, we have

$$\begin{aligned} \frac{\partial F(x)}{\partial x}(F(x))^{-1} + F(x)\frac{\partial (F(x))^{-1}}{\partial x} = {\mathbf {O}}_N, \end{aligned}$$

where ${\mathbf {O}}_N$ is an N-dimensional zero matrix, and we have used the chain rule of derivatives. Hence, we obtain (35).

Next, we prove (36). By the chain rule, we have

$$\begin{aligned} \frac{\partial g(F(x))}{\partial x} = \sum _{i = 1}^N\sum _{j = 1}^N \frac{\partial g(F(x))}{\partial F^{i,j}(x)}\frac{\partial F^{i,j}(x)}{\partial x} = \mathrm {tr}\left\{ \left( \frac{\partial g(F(x))}{\partial F(x)}\right) ^\prime \frac{\partial F(x)}{\partial x}\right\} , \end{aligned}$$

where $F^{i,j}(x)$ is the $i\times j$ th element of F(x).$\square$

Let $g(F) := \theta ^\prime F^{-1}\theta$ and $F := \varSigma + V$. Then, by (36), we have

$$\begin{aligned} \frac{\partial }{\partial v^{i,j}} \theta ^\prime (\varSigma + V)^{-1} \theta = \mathrm {tr}\left\{ \left( \frac{\partial g(F)}{\partial F}\right) ^\prime \frac{\partial F}{\partial v^{i,j}}\right\} . \end{aligned}$$

By the symmetry of V,

$$\begin{aligned} \frac{\partial F}{\partial v^{i,j}} = \frac{\partial (\varSigma + V)}{\partial v^{i,j}} = M^{i,j}, \end{aligned}$$

where $M^{i,j}$ has been defined in the proof of Lemma 1. Since g(F) is linear in all elements of $F^{-1}$, we have

$$\begin{aligned} \frac{\partial g(F)}{\partial F^{i,j}} = \theta ^\prime \frac{\partial F^{-1}}{\partial F^{i,j}} \theta = - \theta ^\prime F^{-1} \frac{\partial F}{\partial F^{i,j}}F^{-1}\theta , \end{aligned}$$

where we have used (35). It can be easily seen that $\theta ^\prime F^{-1} (\partial F/\partial F^{i,j})F^{-1}\theta$ is equal to the $i\times j$ th element of $F^{-1}\theta \theta ^\prime F^{-1}$. Hence,

$$\begin{aligned} \frac{\partial g(F)}{\partial F} = -F^{-1}\theta \theta ^\prime F^{-1} = - (\varSigma + V)^{-1}\theta \theta ^\prime (\varSigma + V)^{-1}. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\partial }{\partial v^{i,j}} \theta ^\prime (\varSigma + V)^{-1} \theta = -\mathrm {tr}\left\{ (\varSigma + V)^{-1}\theta \theta ^\prime (\varSigma + V)^{-1}M^{i,j}\right\} . \end{aligned}$$

(37)

Substituting (34) and (37) into $\partial L^{\theta , V}/\partial v^{i,j} = 0$, we can derive the first order condition (12).