Sharpe Portfolio Using a Cross-Efficiency Evaluation

Abstract

The Sharpe ratio is a way to compare the excess returns (over the risk-free asset) of portfolios for each unit of volatility that is generated by a portfolio. In this paper, we introduce a robust Sharpe ratio portfolio under the assumption that the risk-free asset is unknown. We propose a robust portfolio that maximizes the Sharpe ratio when the risk-free asset is unknown, but is within a given interval. To compute the best Sharpe ratio portfolio, all the Sharpe ratios for any risk-free asset are considered and compared by using the so-called cross-efficiency evaluation. An explicit expression of the Cross-Efficiency Sharpe Ratio portfolio is presented when short selling is allowed.

A previous version of this manuscript is available on arXiv.org.

15.6 Appendix

Proof of Proposition 2. The efficient portfolio $i=(\sigma ^*_{MSR_i}, r^*_{MSR_i})$   that maximizes the cross-efficiency  $CE_i$,  in the interval  $[r_{1},r_{2}]$,  is reached when

$$\begin{aligned} r_i^* = r_{GMV}^*+\sigma _{GMV}^* \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{\sigma ^*_{MSR_{2}}} - \frac{r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} }{ \ln \left( \displaystyle \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{ \displaystyle \sigma ^*_{MSR_{2}}} - \frac{ r^*_{GMV} - r_{2}}{\sigma ^*_{GMV} }}{ \displaystyle \frac{ \displaystyle r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} -\frac{ \displaystyle r^*_{GMV} - r_{{1}}}{\sigma ^*_{GMV} } } \right) }. \end{aligned}$$

The cross-efficiency of portfolio i, $CE_i$, depends of the risk-free rate, $r_i$, associated with the portfolio i. We can considered $CE_i$ as a function of $r_i$, for $r_i\in [r_{\min },r_{\max }]$. We can write $CE_i(r_i)$ as follows

$$ CE_i(r_i)=\displaystyle \frac{\displaystyle r^*_{MSR_i}}{\sigma ^*_{MSR_i}} \int _{r_{\min }}^{r_{\max }}\frac{ \displaystyle \sigma ^*_{MSR}}{r^*_{MSR}-r_f} \, \text {d}r_{f} - \frac{1}{\displaystyle \sigma ^*_{MSR_i}} \int _{r_{\min }}^{r_{\max }}\frac{ \displaystyle \sigma ^*_{MSR} \, \, r_f}{ \displaystyle r^*_{MSR}-r} \, \text {d}r_{f}. $$

From expressions (15.4), (15.9) and (15.10), using notation of (15.5), we can derivate the following identities for the expected return and variance of GMV and MSR portfolios:

$$\begin{aligned}&r_{GMV}^*=\frac{c}{b}, \quad \sigma _{GMV}^*=\frac{1}{\sqrt{b}},\quad \frac{r_{GMV}^*-r_f}{\sigma ^{*2}_{GMV}}=c-b\,\, r_f, \quad \frac{r_{MSR}^*}{\sigma _{MSR}^*}= \frac{a-c \,\, r_f}{\sqrt{a-2c \,\, r_f+b \,\, r_f^2 }}, \nonumber \\&\frac{1}{\sigma _{MSR}^*}= \frac{c-b\,\, r_f}{\sqrt{a-2c\,\, r_f+b\,\, r_f^2 }}, \quad \text {and} \quad \frac{\sigma _{MSR}^*}{r_{MSR}^*-r_f}= \frac{1}{\sqrt{a-2c\,\, r_f+b\,\, r_f^2 }}, \end{aligned}$$

(15.27)

and write the cross-efficiency, $CE_i(r_i)$, in terms of variable $r_i$.

$$ CE_i(r_i) = \; \displaystyle \frac{\displaystyle a-c\,\, r_i}{\sqrt{a-2c\,\, r_i+b\,\, r_i^2}} I_1 - \frac{c-b\,\, r_i}{\sqrt{a-2c\,\, r_i+b \,\, r_i^2}} I_2$$

where

$$ I_1=\frac{1}{r_{\max }-r_{\min }}\int _{r_{\min }}^{r_{\max }}\frac{ \displaystyle \text {d}r_{f} }{\sqrt{a-2c \,\, r_f+b \,\, r_f^2}} \quad \text {and} \quad I_2=\frac{1}{r_{\max }-r_{\min }} \int _{r_{\min }}^{r_{\max }}\frac{r_f\,\, \, \text {d}r_{f} }{\sqrt{a-2c\,\, r_f+b \,\, r_f^2}}. $$

The function $CE_i(R_i)$ has first derivate

$$\begin{aligned} CE'_i(r_i)=&\frac{-c \sqrt{a-2cr_i+br_i^2} -(a-cr_i)(br-c)/\sqrt{a-2cr_i+br_i^2}}{\left( \sqrt{a-2cr_i+br_i^2}\right) ^2}I_1- \\&-\frac{-b\sqrt{a-2cr_i+br_i^2}-(c-br_i)(br_i-c)/\sqrt{a-2cr_i+br_i^2}}{\left( \sqrt{a-2cr_i+br_i^2}\right) ^2}I_2 \\ =&\frac{(c^2r_i-abr_i)I_1+(ba-c^2)I_2}{\left( \sqrt{a-2cr_i+br_i^2}\right) ^3}. \end{aligned}$$

It is left to show that $CE'_i(r_i)=0$ for $r_i=I_2/I_1$, therefore, $r_i=I_2/I_1$ is a point with slope zero, and it is a candidate to a maximum in the interval $[r_{\min },r_{\max }]$. The second derivate of the function $CE_i(r_i)$ is given by the following expression

$$\begin{aligned} CE''_i(r_i)=&\frac{(c^2-ab)I_1\left( \sqrt{a-2cr_i+br_i^2}\right) ^3}{\left( \sqrt{a-2cr_i+br_i^2}\right) ^6} - \frac{3\left( (ba-c^2)I_2+(c^2-ab)I_1r_i \right) (br_i-c)}{\left( \sqrt{a-2cr_i+br_i^2}\right) ^5} \end{aligned}$$

(15.28)

and the second term of (15.28) is zero at $r_i=I_2/I_1$, and

$$ CE''(I_2/I_1)= \frac{(c^2-ab)I_1}{(a-2c \, I_2/I_1+b \, I_2^2/I_1^2)}. $$

Since $\Sigma ^{-1}$ is positive-definite matrix, then $(\mu -r)^T\Sigma ^{-1}(\mu -r)= a-2c r +b r^2 > 0$, with discriminant $4(c^2-ab) < 0$, then the second derivate at $r_i=I_2/I_1$, $CE''(I_2/I_1)$, is less to 0.

Next, we show the expression of $I_2/I_1$.

$$\begin{aligned} (r_{\max }-r_{\min }) I_1&=\int _{r_{\min }}^{r_{\max }}\frac{ \displaystyle \text {d}r_{f} }{\sqrt{a-2cr_f+br_f^2}} =\left[ \frac{1}{\sqrt{b}} \ln \left( \sqrt{b} \sqrt{a-2cr_f+br_f^2} + b r_f -c \right) \right] _{r_{\min }}^{r_{\max }} \\ (r_{\max }-r_{\min }) I_2&= \int _{r_{\min }}^{r_{\max }}\frac{r_f\, \text {d}r_{f} }{\sqrt{a-2cr_f+br_f^2}}=\nonumber \\&= \left[ \frac{c}{\sqrt{b}^3} \ln \left( \sqrt{b} \sqrt{a-2cr_f+br_f^2} + b r_f -c \right) + \frac{1}{b}\sqrt{a-2cr_f+br_f^2} \right] _{r_{\min }}^{r_{\max }}. \end{aligned}$$

Now, we can write the above expression using the identities of (15.27) as follows:

$$\begin{aligned} (r_{\max }-r_{\min }) I_1&= \sigma _{GMV}^* \ln \left( \displaystyle \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{ \displaystyle \sigma ^*_{MSR_{2}}} - \frac{ r^*_{GMV} - r_{2}}{\sigma ^*_{GMV} }}{ \displaystyle \frac{ \displaystyle r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} -\frac{ \displaystyle r^*_{GMV} - r_{{1}}}{\sigma ^*_{GMV} } } \right) \\ (r_{\max }-r_{\min }) I_2&= r_{GMV}^* \sigma _{GMV}^* \ln \left( \displaystyle \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{ \displaystyle \sigma ^*_{MSR_{2}}} - \frac{ r^*_{GMV} - r_{2}}{\sigma ^*_{GMV} }}{ \displaystyle \frac{ \displaystyle r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} -\frac{ \displaystyle r^*_{GMV} - r_{{1}}}{\sigma ^*_{GMV} } } \right) + \\&+ \sigma _{GMV}^{*2} \left( \frac{r_{MSR_{2}}^*-r_{2}}{\sigma ^*_{{MSR}_2}} - \frac{r_{MSR_{1}}^*-r_{1}}{\sigma ^*_{{MSR}_1}} \right) \end{aligned}$$

and, finally we can write the maximum $r_i$ as

$$\begin{aligned} r_i^* = r_{GMV}^*+\sigma _{GMV}^* \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{\sigma ^*_{MSR_{2}}} - \frac{r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} }{ \ln \left( \displaystyle \frac{ \displaystyle \frac{ \displaystyle r^*_{MSR_{2}} - r_{2}}{ \displaystyle \sigma ^*_{MSR_{2}}} - \frac{ r^*_{GMV} - r_{2}}{\sigma ^*_{GMV} }}{ \displaystyle \frac{ \displaystyle r^*_{MSR_{1}} - r_{1}}{\sigma ^*_{MSR_{1}}} -\frac{ \displaystyle r^*_{GMV} - r_{{1}}}{\sigma ^*_{GMV} } } \right) }. \end{aligned}$$

(15.29)

   $\square$

Proof of Proposition 3. There exists a Pythagorean relationship between the slopes of the Tangent and Global Minimum portfolios and the slope of the asymptote of $W_{\rho }$.

$$\begin{aligned} m_{TP}^2= m_{ah}^2+m_{GMV}^2. \end{aligned}$$

(15.30)

From expressions (15.5), we can derivate the following identities for the $m_{TP}$, $m_{ah}$ and $m_{GMV}$ slopes:

$$\begin{aligned}&m_{TP}=\sqrt{a}, \qquad m_{ah}= \sqrt{\frac{ab-c^2}{b}} \quad \text {and} \quad m_{GMV}=\frac{c}{\sqrt{b}}. \end{aligned}$$

(15.31)

and now, we can derivate the relationship $m_{TP}^2= m_{ah}^2+m_{GMV}^2$,

$$\begin{aligned} m_{ah}^2+m_{GMV}^2= \frac{ab-c^2}{b} + \frac{c^2}{b} = a= m_{TP}^2 \end{aligned}$$

   $\square$

Proof of Corollary 2. The maximum cross-efficiency (MCESR) portfolio in  $[0,r_{GMV}^*]$  depends only of Minimal Global Variance and Tangent portfolios.

$$\begin{aligned} r_i^* = r^*_{GMV}\left( 1- \frac{\displaystyle \sqrt{\frac{r^*_{TP}}{r^*_{GMV} }}- \sqrt{\frac{r^*_{TP}}{r^*_{GMV} }-1} }{ \ln \left( \displaystyle \sqrt{\frac{r^*_{TP}}{r^*_{GMV} } -1} \right) - \ln \left( \displaystyle \sqrt{\frac{r^*_{TP}}{r^*_{GMV} }} -1 \right) } \right) . \end{aligned}$$

(15.32)

From (15.27) and (15.31), we can derivate the following expressions:

$$\begin{aligned}&\frac{m_{TP}^2}{m_{GMV}^2}=\frac{a}{c^2/\sqrt{b}^2}= \frac{a/c}{c/b}= \frac{r^*_{TP}}{r^*_{GMV}}, \, \text {then} \quad \frac{m_{TP}}{m_{GMV}}=\sqrt{\frac{r^*_{TP}}{r^*_{GMV}}} \\&\frac{m_{ah}^2}{m_{GMV}^2}= \frac{ \frac{ab-c^2}{b}}{c^2/b} = \frac{ab-c^2}{c^2}=\frac{ab}{c^2}-1= \frac{r^*_{TP}}{r^*_{GMV}} -1, \, \text {then} \quad \frac{m_{ah}}{m_{GMV}}=\sqrt{\frac{r^*_{TP}}{r^*_{GMV}}-1} \\&\frac{m_{TP}^2}{m_{ah}^2}=\frac{a}{\frac{ab-c^2}{b}}=\frac{a/c}{a/c-c/b} =\frac{r^*_{TP}}{r^*_{TP}-r^*_{GMV}}, \, \text {then} \quad \frac{m_{TP}}{m_{ah}}= \sqrt{\frac{r^*_{TP}}{r^*_{TP}-r^*_{GMV}}}. \end{aligned}$$

From the expression (15.21), it is left to show that (15.32) is true.    $\square$