A theoretical foundation of portfolio resampling

Abstract

A portfolio-resampling procedure invented by Richard and Robert Michaud is a subject of highly controversial discussion and big scientific dispute. It has been evaluated in many empirical studies and Monte Carlo experiments. Apart from the contradictory findings, the Michaud approach still lacks a theoretical foundation. I prove that portfolio resampling has a strong foundation in the classic theory of rational behavior. Every noise trader could do better by applying the Michaud procedure. By contrast, a signal trader who has enough prediction power and risk-management skills should refrain from portfolio resampling. The key note is that in most simulation studies, investors are considered as noise traders. This explains why portfolio resampling performs well in simulation studies, but could be mediocre in real life.

Proofs

In the following, it is implicitly assumed that the given expectations, variances, and covariances exist and are finite. Moreover, it is assumed that $\Sigma$ is positive definite so that the inverse $\Sigma ^{-1}$ exists.

Proposition 1

Let $f$ be an increasing and strictly concave function from $\mathbb {R}$ to $\mathbb {R}\,$. Further, let $g$ be a concave function from $\mathbb {R}^N$ to $\mathbb {R}\,$. Then $f\circ g$ is strictly concave, too.

Proof

Since $f$ is strictly concave, it holds that

$$\begin{aligned} \pi f\big (g(x_1)\big ) + (1-\pi ) f\big (g(x_2)\big ) < f\big (\pi g(x_1) + (1-\pi ) g(x_2)\big ) \end{aligned}$$

for every $\pi \in ]0,1[\,$. Further, since $g$ is concave, we have that

$$\begin{aligned} \pi g(x_1) + (1-\pi ) g(x_2) \le g\big (\pi x_1+(1-\pi )x_2\big ). \end{aligned}$$

Since $f$ is increasing, it turns out that

$$\begin{aligned} \pi f\big (g(x_1)\big ) + (1-\pi ) f\big (g(x_2)\big ) < f\big (g(\pi x_1+(1-\pi )x_2)\big ) \end{aligned}$$

for every $\pi \in ]0,1[\,$, which means that $f\circ g$ is strictly concave.

1.1 Theorem 1

We have that

$$\begin{aligned} E \Bigl (u\bigl (W\,(1+r+\tilde{w}'R)\bigr )\Bigr ) = \int \int u\bigl (W\,(1+r+w'r)\bigr )\,p(w,r)\,\mathrm{d}r\,\mathrm{d}w \end{aligned}$$

and, since $\tilde{w}$ is independent of $R\,$, it follows that $p(w,r)=p(w)p(r)\,$. This means

$$\begin{aligned} E \Bigl (u\bigl (W\,(1+r+\tilde{w}'R)\bigr )\Bigr )&= \int \int u\bigl (W\,(1+r+w'r)\bigr )\,p(w)p(r)\mathrm{d}r\mathrm{d}w \\&= \int \left[ \int u\bigl (W\,(1+r+w'r)\bigr )\,p(w)\,\mathrm{d}w\right] p(r)\,\mathrm{d}r. \end{aligned}$$

Since $u$ is increasing and strictly concave in $W\,(1+r+w'r)$, which is linear and thus concave in $w$, Proposition 1 implies that $u\big (W\,(1+r+w'r)\big )$ is strictly concave in $w\,$. Further, since $\tilde{w}$ is mixed, its probability distribution is not degenerated and thus

$$\begin{aligned} \int u\bigl (W\,(1+r+w'r)\bigr )\,p(w)\,\mathrm{d}w < u\bigl (W(1+r+\bar{w}'r)\bigr ) \end{aligned}$$

by Jensen’s inequality. Hence, it follows that

$$\begin{aligned} E \Big (u\bigl (W\,(1+r+\tilde{w}'R)\bigr )\Big )&< \int u\bigl (W(1+r+\bar{w}'r)\bigr )\,p(r)\,\mathrm{d}r\\&= E \Bigl (u\bigl (W\,(1+r+\bar{w}'R)\bigr )\Bigr ). \end{aligned}$$

Since $\mathcal {C}$ is convex and $\tilde{w}\in \mathcal {C}$, we have that $\bar{w}=E (\tilde{w})\in \mathcal {C}\,$.

1.2 Theorem 2

The out-of-sample performance of $w^*=\Sigma ^{-1}\mu /\alpha$ amounts to

$$\begin{aligned} \varphi _\alpha (w^*) = \frac{\mu '\Sigma ^{-1}\mu }{2\alpha }. \end{aligned}$$

Further, we have that

$$\begin{aligned} (\tilde{w}-w^*)'\Sigma \,(\tilde{w}-w^*)&= \tilde{w}'\Sigma \,\tilde{w}-2\tilde{w}'\Sigma \,w^*+w^{*\prime }\Sigma \,w^*\\&= \tilde{w}'\Sigma \,\tilde{w}-\frac{2\tilde{w}'\mu }{\alpha }+\frac{\mu '\Sigma ^{-1}\mu }{\alpha ^2}, \end{aligned}$$

so that

$$\begin{aligned} \frac{\alpha }{2}\,(\tilde{w}-w^*)'\Sigma \,(\tilde{w}-w^*) = \frac{\alpha }{2}\,\tilde{w}'\Sigma \,\tilde{w} - \tilde{w}'\mu + \frac{\mu '\Sigma ^{-1}\mu }{2\alpha } = \varphi _\alpha (w^*) - \varphi _\alpha (\tilde{w}). \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\alpha }{2}\,E \Bigl ((\tilde{w}-w^*)'\Sigma \,(\tilde{w}-w^*)\Bigr ) = E \Bigl (\varphi _\alpha (w^*)-\varphi _\alpha (\tilde{w})\Bigr ) = \mathcal {L}_\alpha (\tilde{w}). \end{aligned}$$

The decomposition of $\mathcal {L}_\alpha (\tilde{w})$ into $\mathcal {B}_\alpha (\tilde{w})$ and $\mathcal {V}_\alpha (\tilde{w})$ follows immediately from

$$\begin{aligned} \mathcal {L}_\alpha (\tilde{w})&= \frac{\alpha }{2}\,E \Bigl (\bigl [(\tilde{w}-\bar{w}) +(\bar{w}-w^*)\bigr ]'\Sigma \,\bigl [(\tilde{w}-\bar{w}) +(\bar{w}-w^*)\bigr ]\Bigr )\\&= \frac{\alpha }{2}\,E \Bigl ((\tilde{w}-\bar{w})'\Sigma \,(\tilde{w}-\bar{w})\Bigr ) +\frac{\alpha }{2}\,(\bar{w}-w^*)'\Sigma \,(\bar{w}-w^*) = \mathcal {V}_\alpha (\tilde{w}) + \mathcal {B}_\alpha (\tilde{w}). \end{aligned}$$

Since $\Sigma$ is positive definite we have that

$$\begin{aligned} \mathcal {V}_\alpha (\tilde{w}) = \frac{\alpha }{2}\,E \Bigl ((\tilde{w}-\bar{w})'\Sigma \,(\tilde{w}-\bar{w})\Bigr ) \ge 0 \end{aligned}$$

and from

$$\begin{aligned} \mathcal {B}_\alpha (\tilde{w}) = \frac{\alpha }{2}\,(\bar{w}-w^*)'\Sigma \,(\bar{w}-w^*) = \mathcal {L}_\alpha (\bar{w}) \end{aligned}$$

it follows that $\mathcal {L}_\alpha (\tilde{w})\ge \mathcal {L}_\alpha (\bar{w})\,$. Moreover, since $\tilde{w}\in \mathcal {C}$ and $\mathcal {C}$ is a convex set, it holds that $\bar{w}=E (\tilde{w})\in \mathcal {C}$. If $\tilde{w}$ is a mixed strategy, Jensen’s inequality implies that $\mathcal {V}_\alpha (\tilde{w})>0\,$, i.e., the inequality is strict. Vice versa, if the inequality is strict, it must hold that $\mathcal {V}_\alpha (\tilde{w})>0\,$. This is only possible if $\tilde{w}$ is mixed.

1.3 Theorem 3

First of all, we have that

$$\begin{aligned} E \big (\phi _\alpha (\tilde{w})\big ) = E (\tilde{w}'R) - \frac{\alpha }{2}\,E \Big ((\tilde{w}'R)^2\Big ), \end{aligned}$$

where

$$\begin{aligned} E (\tilde{w}'R) = \sum _{i=1}^NE (\tilde{w}_iR_i) = \sum _{i=1}^N\bar{w}_i\mu _i + \sum _{i=1}^NCov (\tilde{w}_i,R_i) = \bar{w}'\mu + \sum _{i=1}^NCov (\tilde{w}_i,R_i). \end{aligned}$$

Moreover, it follows that

$$\begin{aligned} E \Big ((\tilde{w}'R)^2\Big ) = E \big (\tilde{w}'RR'\tilde{w}\big ) = E \big (\tilde{w}'\Sigma \,\tilde{w}\big ) + E \Big (\tilde{w}'\big (RR'-\Sigma \big )\tilde{w}\Big ), \end{aligned}$$

where

$$\begin{aligned} E \Big (\tilde{w}'\big (RR'-\Sigma \big )\tilde{w}\Big ) = \sum _{i,j=1}^N E \Big (\tilde{w}_i\tilde{w}_j\big (R_iR_j-E (R_iR_j)\big )\Big ) = \sum _{i,j=1}^N Cov \big (\tilde{w}_i\tilde{w}_j,R_iR_j\big ). \end{aligned}$$

This means

$$\begin{aligned} E \big (\phi _\alpha (\tilde{w})\big ) = \bar{w}'\mu + \sum _{i=1}^NCov (\tilde{w}_i,R_i) - \frac{\alpha }{2}\left( E \big (\tilde{w}'\Sigma \,\tilde{w}\big )+\sum _{i,j=1}^NCov \big (\tilde{w}_i\tilde{w}_j,R_iR_j\big )\right) . \end{aligned}$$

1.4 Theorem 4

Since $\tilde{w}$ and its copies $\tilde{w}_1,\tilde{w}_2,\ldots ,\tilde{w}_n$ are irrelevant, it follows that

$$\begin{aligned} p(w_t,r) = p(w_t)p(r) = p(w)p(r) = p(w,r) \end{aligned}$$

and thus

$$\begin{aligned} E \Big (u\big (W\,(1+r+\tilde{w}'_tR)\big )\Big )&= \int \int u\big (W\,(1+r+w'_tr)\big )\,p(w_t,r)\,\mathrm{d}r\mathrm{d}w_t \\&= \int \int u\big (W\,(1+r+w'r)\big )\,p(w,r)\,\mathrm{d}r\mathrm{d}w \\&= E \Big (u\big (W\,(1+r+\tilde{w}'R)\big )\Big ),\qquad t = 1,2,\ldots ,n, \end{aligned}$$

i.e.,

$$\begin{aligned} E \Big (u\big (W\,(1+r+\tilde{w}'R)\big )\Big ) = E \left( \frac{1}{n}\sum _{t=1}^{n} u\big (W\,(1+r+\tilde{w}'_tR)\big )\right) . \end{aligned}$$

Since $u\big (W\,(1+r+x)\big )$ is strictly concave in $x$, we have that

$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n} u\big (W\,(1+r+\tilde{w}'_tR)\big )&\le u\left( W\bigg (1+r+\frac{1}{n}\sum _{t=1}^{n}\tilde{w}'_tR\bigg )\right) \\&= u\Big (W\,(1+r+\hat{\bar{w}}'R)\Big ) \end{aligned}$$

with probability one, but since $\tilde{w}'_1R,\tilde{w}'_2R,\ldots ,\tilde{w}'_nR$ are not almost surely identical, it follows that

$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n} u\big (W\,(1+r+\tilde{w}'_tR)\big ) < u\Big (W\,(1+r+\hat{\bar{w}}'R)\Big ) \end{aligned}$$

with positive probability. This means

$$\begin{aligned} E \Big (u\big (W\,(1+r+\tilde{w}'R)\big )\Big ) < E \Big (u\big (W\,(1+r+\hat{\bar{w}}'R)\big )\Big ). \end{aligned}$$

Finally, since $\mathcal {C}$ is convex and $\tilde{w}_t\in \mathcal {C}$ for $t=1,2,\ldots ,n$, we have that $\hat{\bar{w}}=\frac{1}{n}\sum _{t=1}^{n}\tilde{w}_t\in \mathcal {C}$.