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A theoretical foundation of portfolio resampling


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.

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Fig. 1


  1. 1.

    The Michaud approach is a US patented procedure with patent number 6,003,018.

  2. 2.

    In the following, I use both terms “investor” and “trader” synonymously.

  3. 3.

    Here it is implicitly assumed that the search costs for the information \(\mathcal {I}\) are negligible.

  4. 4.

    Here \(\bar{w}\) denotes the theoretical expectation of \(\tilde{w}\) but not some corresponding approximation or estimate.

  5. 5.

    Here the mean portfolio \(\bar{w}_t\) is a function of time, since the distribution of \(\tilde{w}_t\) might depend on \(t\in \mathbb {Z}\,\).

  6. 6.

    In Sect. 4.2 I discuss the apparent link between the random-walk and the efficient-market hypothesis.

  7. 7.

    In most publications, \(\Sigma \) typically denotes the covariance matrix \(Var (R)\) but not the uncentered second-moment matrix of \(R\). For short-term asset returns, the difference between \(Var (R)\) and \(E (RR')\) is negligible.

  8. 8.

    Here \(\Sigma \) refers to the covariance matrix of \(R\) (Michaud 1998).

  9. 9.

    Hence, \(\mathcal {C}\) belongs to the \(N\)-dimensional Euclidean simplex.

  10. 10.

    This point has been suggested by Christoph Memmel in a personal communication.

  11. 11.

    In the random-walk model (see Example 4), \(\theta \) is the parameter of \(F\), i.e., the joint cumulative distribution function of the asset returns.

  12. 12.

    For example, it is typically assumed that the random-walk model is true, \(\Sigma \) is known, and \(\mathcal {C}=\mathbb {R}^N\). Then the desired portfolio corresponds to \(w^*=\Sigma ^{-1}\mu /\alpha \), which is a linear function of \(\mu \,\).

  13. 13.

    I would like to thank Winfried Pohlmeier for this hint.

  14. 14.

    For a precise definition of \(\lambda _{t,T}\) see Back (2010) and Frahm (2013).

  15. 15.

    More details on this subject can be found in Frahm (2013).

  16. 16.

    This means \(R\) is stochastically independent of \(\tilde{w}_t\sim \tilde{w}\) for \(t=1,2,\ldots ,n\,\).

  17. 17.

    Here I do not distinguish between theories where a rational individual act as if he maximizes the expected utility based on his subjective probabilities and other theories implying that he actually does.

  18. 18.

    This is also reflected by Markowitz’ two-stage portfolio optimization approach discussed in Sect. 2.


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Author information

Correspondence to Gabriel Frahm.

Additional information

Many thanks belong to Uwe Jaekel, Bob Korkie, Christoph Memmel, Gert Mittring, Winfried Pohlmeier, Rainer Schüssler, and Niklas Wagner for their fruitful comments and suggestions. Moreover, I thank all participants of the workshop on portfolio optimization in Konstanz (2012) for their contributions.



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.


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.

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 \(\bar{w}=E (\tilde{w})\in \mathcal {C}\,\).

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

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


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

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.

Theorem 3

First of all, we have that

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


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

Moreover, it follows that

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


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

This means

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

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

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


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

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 \(\hat{\bar{w}}=\frac{1}{n}\sum _{t=1}^{n}\tilde{w}_t\in \mathcal {C}\).

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Frahm, G. A theoretical foundation of portfolio resampling. Theory Decis 79, 107–132 (2015). https://doi.org/10.1007/s11238-014-9453-0

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  • Asset allocation
  • Mean–variance analysis
  • Noise trader
  • Out-of-sample performance
  • Portfolio resampling
  • Resampled efficiency
  • Signal trader

JEL Classification

  • Primary G11
  • Secondary D81