Diversified minimum-variance portfolios

Abstract

We build on a one parameter family of weighting schemes arising from $L^2$-constrained portfolio optimization problems. The parameter allows to fine tune the trade-off between the volatility and the diversification of the portfolio. We propose two criteria in order to determine two unique portfolios: the first criterion requires that no weights be negative while the second one imposes a target diversification which is median between full concentration and full diversification. Both portfolios are empirically compared to classical benchmarks. The first one behaves very much like other popular Long-Only weighting schemes while the second displays a more aggressive profile, while generating moderate turnover. We also discuss implementation issues, as well as estimation related problems.

Appendices

Appendix A: The formal proof of why $\gamma ^{LO}$ exists and is unique

The main difficulty in (3) is the matrix inversion but it can be handled in the following way. We factorize $\varvec{\varSigma }$ using its eigendecomposition: $\varvec{\varSigma =P\varLambda P'}$. Basic algebra then implies

$$\begin{aligned} \varvec{Q}=(\varvec{\varSigma }+\gamma \varvec{I_N})^{-1}=\varvec{P}(\varvec{\varLambda }+\gamma \varvec{I_N})^{-1}\varvec{P'}, \end{aligned}$$

(6)

where the inverse matrix is diagonal with values $(\varLambda _{i,i}+\gamma )^{-1}$ for $i=1,\ldots ,N$. We use the standard notation $M_{i,j}$ for the elements of the matrix $\varvec{M}$. Further computations lead to:

$$\begin{aligned} Q_{i,j} = \sum _{k=1}^N P_{i,k} (\varLambda _{k,k}+\gamma )^{-1} P_{j,k}. \end{aligned}$$

Therefore, summing the lines of $\varvec{Q}$, we see that the weights in (3) are proportional to

$$\begin{aligned} \sum _{j=1}^N Q_{i,j}=\sum _{j=1}^N \sum _{k=1}^N P_{i,k} (\varLambda _{k,k}+\gamma )^{-1} P_{j,k}, \quad i=1,\ldots ,N. \end{aligned}$$

Because we are interested in Long-Only portfolios, we require that

$$\begin{aligned} \sum _{j=1}^N \sum _{k=1}^N P_{i,k} (\varLambda _{k,k}+\gamma )^{-1} P_{j,k} \ge 0, \quad i=1,\ldots ,N. \end{aligned}$$

Getting rid of the denominators (the inverse parts), this is equivalent to

$$\begin{aligned} \sum _{j=1}^N \sum _{k=1}^N P_{i,k}P_{j,k} \prod _{l \ne k} (\varLambda _{l,l}+\gamma ) \ge 0, \quad i=1,\ldots ,N. \end{aligned}$$

(7)

The function on the l.h.s. of (7) is polynomial in $\gamma$ with degree $N-1$. The leading term (in $\gamma ^{N-1}$) has coefficient

$$\begin{aligned} \sum _{j=1}^N \sum _{k=1}^N P_{i,k}P_{j,k}=\sum _{j=1}^N(\varvec{PP'})_{i,j}=1, \quad i=1,\ldots ,N, \end{aligned}$$

because $\varvec{P}$ is orthogonal. This latter expression ensures that as $\gamma$ increases, all of the weights will progressively become positive. Some of them are already positive, even for $\gamma =0$, because they correspond to the long positions of $\varvec{w}_{MV}$.

Appendix B: Proof of Proposition 1

For notational ease, we will write $\varvec{w}$ instead $\varvec{w}_\gamma$ during the proof.

Using the eigendecomposition $\varvec{\varSigma =P\varLambda P'}$ and the fact that $\varvec{P'P=PP'=I_N}$,

$$\begin{aligned} \varvec{w' \varSigma w}&=\frac{\varvec{1'(\varSigma +\gamma I_N)^{-1}\varSigma (\varSigma +\gamma I_N)^{-1} 1}}{(\varvec{1'(\varSigma +\gamma I_N)^{-1}1})^2} = \frac{\varvec{1'P(\varLambda +\gamma I_N)^{-1} \varLambda (\varLambda +\gamma I_N)^{-1} P'1}}{(\varvec{1'P(\varLambda +\gamma I_N)^{-1}P'1})^2}\\&= \frac{\displaystyle \sum \nolimits _{i=1}^N\sum _{j=1}^N\sum _{k=1}^N P_{i,k}P_{j,k} \varLambda _{k,k}( \varLambda _{k,k}+\gamma )^{-2}}{\displaystyle \left( \sum \nolimits _{i=1}^N\sum \nolimits _{j=1}^N\sum \nolimits _{k=1}^N P_{i,k}P_{j,k}(\varLambda _{k,k}+\gamma )^{-1}\right) ^2} = \frac{\displaystyle \sum \nolimits _{k=1}^N \varLambda _{k,k}(\varLambda _{k,k}+\gamma )^{-2}\left( \sum \nolimits _{l=1}^N P_{l,k}\right) ^2}{\displaystyle \left( \sum \nolimits _{k=1}^N(\varLambda _{k,k}+\gamma )^{-1} \left( \sum \nolimits _{l=1}^N P_{l,k}\right) ^2\right) ^2} \end{aligned}$$

We now introduce the following quantity, which will be ubiquitous in the remainder of the proofs:

$$\begin{aligned} L_k=\left( \sum _{l=1}^N P_{l,k}\right) ^2. \end{aligned}$$

Straightforward differentiation gives

$$\begin{aligned}&\frac{\partial }{\partial \gamma }\varvec{w' \varSigma w}\nonumber \\&\quad = \frac{\displaystyle -2\sum \nolimits _{k=1}^N\frac{ \varLambda _{k,k}}{(\varLambda _{k,k}\!+\!\!\gamma )^{3}}L_k \!\times \! \sum \nolimits _{k=1}^N(\varLambda _{k,k}\!\!+\!\gamma )^{-1}L_k\!\!+\!2 \sum \nolimits _{k=1}^N\frac{ \varLambda _{k,k}}{(\varLambda _{k,k}\!\!+\!\gamma )^{2}}L_k \!\times \! \sum \nolimits _{k=1}^N(\varLambda _{k,k}\!+\!\gamma )^{-2}L_k }{\displaystyle \left( \sum \nolimits _{k=1}^N(\varLambda _{k,k}\!\!+\!\gamma )^{-1} L_k\right) ^3}. \end{aligned}$$

The denominator is always positive, therefore, the sign of $\frac{\partial }{\partial \gamma }\varvec{w' \varSigma w}$ is equal to that of

$$\begin{aligned}&\sum _{k=1}^N \sum _{i=k+1}^NL_kL_i \left[ \frac{\varLambda _{k,k}}{( \varLambda _{k,k}+\gamma )^2} (\varLambda _{i,i}+\gamma )^{-2}-\frac{\varLambda _{k,k}}{(\varLambda _{k,k}+\gamma )^3}(\varLambda _{i,i}+\gamma )^{-1} \right. \\&\qquad \left. + \frac{\varLambda _{i,i}}{(\varLambda _{i,i}+\gamma )^2}(\varLambda _{k,k}+\gamma )^{-2} - \frac{\varLambda _{i,i}}{(\varLambda _{i,i}+\gamma )^3}(\varLambda _{k,k}+\gamma )^{-1}\right] \\&= \sum _{k=1}^N \sum _{i=k+1}^NL_kL_i (\varLambda _{i,i}+\gamma )^{-1}(\varLambda _{k,k}+\gamma )^{-1} f_\gamma (\varLambda _{i,i},\varLambda _{k,k}) \end{aligned}$$

where

$$\begin{aligned} f_{\gamma }(x,y)=(x+\gamma )^{-1}(y+\gamma )^{-1}(x+y)-x(x+\gamma )^{-2}-y(y+\gamma )^{-2}. \end{aligned}$$

Further, we have

$$\begin{aligned} \frac{\partial }{\partial x}f_{\gamma }(x,y)=\frac{2\gamma (x-y)}{(x+\gamma )^3(y+\gamma )}, \quad \frac{\partial ^2}{\partial ^2 x}f_{\gamma }(x,y)=\frac{2\gamma (\gamma -2x+3y)}{(x+\gamma )^2(y+\gamma )}, \end{aligned}$$

and consequently the mapping $x \mapsto f_{\gamma }(x,y)$ reaches its minimum value (which is zero) at $x=y$. Therefore, $\frac{\partial }{\partial \gamma }\varvec{w' \varSigma w} \ge 0$. The derivative can only reach zero if all eigenvalues are equal, which is excluded because $\varvec{\varSigma }$ is not a multiple if $\varvec{I_N}$. The derivative of $\varvec{w' w}$ with respect to $\gamma$ is dealt with in the same fashion¹⁰ except that $f_{\gamma }(x,y)$ must be replaced by

$$\begin{aligned} g_{\gamma }(x,y)=2(x+\gamma )^{-1}(y+\gamma )^{-1}-(x+\gamma )^{-2}-(y+\gamma )^{-2}=\frac{-(x-y)^2}{(x+\gamma )^2(y+\gamma )^2}, \end{aligned}$$

which is always strictly negative for $x,y,\gamma > 0$ and $x \ne y$ (which occurs because $\varvec{\varSigma }\ne \alpha \varvec{I_N}$). Moreover, it is obvious that $\frac{\partial }{\partial \gamma }\varvec{w' w}$ bounded for $x,y>0$ and $\gamma \ge 0$.

Appendix C: Proof of the lemmas

We start with Lemma 1.

Proof

For the first assertion, if we consider $\gamma >U$, then we can write $\varvec{\varSigma }+\gamma \varvec{I_N}=\gamma (\varvec{\varSigma }/\gamma + \varvec{I_N})$. By the definition of $U$, the largest eigenvalue of $\varvec{\varSigma }/\gamma$ is strictly smaller than one and hence, the Neumann series of the inverse is given by

$$\begin{aligned} (\varvec{\varSigma }+\gamma \varvec{I_N})^{-1}&=\gamma ^{-1} \sum _{k=0}^\infty (-\varvec{\varSigma }/\gamma )^k \\&=\gamma ^{-1} \sum _{k=0}^\infty \left[ (\varvec{\varSigma }/\gamma )^{2k}-(\varvec{\varSigma }/\gamma )^{2k+1} \right] \end{aligned}$$

We now write $s_{i,j}^{(2k)}$ for the elements of $(\varvec{\varSigma }/\gamma )^{2k}$. The $i$th row sum of $(\varvec{\varSigma }/\gamma )^{2k+1}$ is then given by

$$\begin{aligned} \sum _{j=1}^N s_{i,j}^{(2k+1)}&=\sum _{j=1}^N \sum _{l=1}^N s_{i,l}^{(2k)} \times \varSigma _{l,j}/\gamma \\&= \sum _{l=1}^N s_{i,l}^{(2k)} \sum _{j=1}^N \varSigma _{l,j}/\gamma \\&\le \sum _{l=1}^N s_{i,l}^{(2k)}. \end{aligned}$$

In the last inequality, we have used the definition of $U$ (as well as the symmetry of $\varvec{\varSigma }$) and the fact that by the assumption of the lemma, the row sums of even powers of $\varvec{\varSigma }$ are positive. The inequality shows that all of the row sums of $(\varvec{\varSigma }+\gamma \varvec{I_N})^{-1}$ will be positive, thereby fulfilling the condition in the definition of $\gamma ^{LO}$ in (4).

We now turn to Lemma 2.

Proof

We use the notations of “Appendix B”, and by replacing $\varvec{\varSigma }$ by $\varvec{I_N}$ in the computation of the ex-ante variance, we get

$$\begin{aligned} D(\varvec{w}_\gamma )=\frac{\displaystyle \sum \nolimits _{k=1}^N (\varLambda _{k,k}+\gamma )^{-2}L_k}{\displaystyle \left( \sum \nolimits _{k=1}^N(\varLambda _{k,k}+\gamma )^{-1} L_k\right) ^2}. \end{aligned}$$

First, we have

$$\begin{aligned} \displaystyle \sum _{k=1}^N (\varLambda _{k,k}+\delta \varLambda _{1,1})^{-2}L_k \le (\delta \varLambda _{1,1})^{-2}\sum _{k=1}^N L_k, \end{aligned}$$

while at the same time

$$\begin{aligned} \displaystyle \left( \sum _{k=1}^N(\varLambda _{k,k}+\delta \varLambda _{1,1} )^{-1} L_k\right) ^2 \ge \displaystyle \left( \sum _{k=1}^N((1+\delta )\varLambda _{1,1})^{-1} L_k\right) ^2. \end{aligned}$$

Therefore,

$$\begin{aligned} D(\varvec{w}_\gamma )^{-1} \ge \frac{\delta ^2}{(\delta +1)^2}\sum _{k=1}^N L_k=\frac{\delta ^2}{(\delta +1)^2}N, \end{aligned}$$

where the equality stems from the definition of $L_k$ and the fact that $\varvec{P}$ is orthogonal (the sum of all elements of $\varvec{PP'}$ is equal to $N$).

Appendix D: Proof of Proposition 2

We start by looking at the $L^2$-norm difference of the two portfolios (3), with common $\gamma$, but different $\varvec{\varSigma }^{(i)}$. With the notations of the former proof:

$$\begin{aligned} \epsilon&=|(\varvec{w}_\gamma ^{(1)})'\varvec{w}_\gamma ^{(1)}-(\varvec{w}_\gamma ^{(2)})'\varvec{w}_\gamma ^{(2)}| \nonumber \\&=\left| \frac{a(\varLambda ^{(1)},\gamma )}{b(\varLambda ^{(1)},\gamma )^2}-\frac{a(\varLambda ^{(2)},\gamma )}{b(\varLambda ^{(2)},\gamma )^2} \right| , \nonumber \\&=\frac{a(\varLambda ^{(1)},\gamma )[b(\varLambda ^{(2)},\gamma )^2\!-\!b(\varLambda ^{(1)},\gamma )^2]+b(\varLambda ^{(1)},\gamma )^2[a(\varLambda ^{(1)},\gamma )-a(\varLambda ^{(2)},\gamma )]}{b(\varLambda ^{(1)},\gamma )^2b(\varLambda ^{(2)},\gamma )^2}, \end{aligned}$$

(8)

with

$$\begin{aligned} a(\varLambda ^{(i)},\gamma )=\sum _{k=1}^N (\varLambda ^{(i)}_{k,k}+\gamma )^{-2}L_k, \quad \quad b(\varLambda ^{(i)},\gamma )=\sum _{k=1}^N (\varLambda ^{(i)}_{k,k}+\gamma )^{-1}L_k. \end{aligned}$$

(9)

Basic computations further yield

$$\begin{aligned} |a(\varLambda ^{1},\gamma )-a(\varLambda ^{2},\gamma )|&= \sum _{k=1}^N \frac{(\varLambda ^{(2)}_{k,k}+\gamma )^2-(\varLambda ^{(1)}_{k,k}+\gamma )^2}{(\varLambda ^{(1)}_{k,k}+\gamma )^2(\varLambda ^{(2)}_{k,k}+\gamma )^2} L_k \nonumber \\&= \sum _{k=1}^N \frac{(\varLambda ^{(2)}_{k,k}-\varLambda ^{(1)}_{k,k})((\varLambda ^{(2)}_{k,k}+\varLambda ^{(1)}_{k,k}+2\gamma )}{(\varLambda ^{(1)}_{k,k}+\gamma )^2(\varLambda ^{(2)}_{k,k}+\gamma )^2} L_k \nonumber \\&\le \sum _{k=1}^N \frac{2|\varLambda ^{(2)}_{k,k}-\varLambda ^{(1)}_{k,k}|}{\left( \varLambda ^{(1)}_{k,k}+\gamma \right) \left( \varLambda ^{(2)}_{k,k}+\gamma \right) \left( \min \left( \varLambda ^{(1)}_{k,k},\varLambda ^{(2)}_{k,k}\right) +\gamma \right) } L_k \nonumber \\&\le \frac{2\eta N}{\left( \varLambda ^{(1)}_{N,N}+\gamma \right) \left( \varLambda ^{(2)}_{N,N}+\gamma \right) \left( \min \left( \varLambda ^{(1)}_{N,N},\varLambda ^{(2)}_{N,N}\right) +\gamma \right) } \nonumber \\&:= \kappa \end{aligned}$$

(10)

where the first inequality stems from the fact that $x+y+2z \le 2(\max (x,y)+z)$ for $x,y,z \ge 0$. In the second inequality, we have set $\eta = {\text {max}}_{i=1,\ldots ,N} \left| \varLambda ^{(2)}_{i,i}-\varLambda ^{(1)}_{i,i}\right|$ and used the identity $\sum _{k=1}^N L_k=N$.

Moreover,

$$\begin{aligned}&|b(\varLambda ^{(1)},\gamma )^2-b(\varLambda ^{(2)},\gamma )^2|\nonumber \\&\quad =(b(\varLambda ^{(1)},\gamma )+b(\varLambda ^{(2)},\gamma ))|b(\varLambda ^{(1)},\gamma )-b(\varLambda ^{(2)},\gamma )| \nonumber \\&\quad = \left( \sum _{k=1}^N \frac{L_k}{\varLambda ^{(1)}_{k,k}+\gamma }+\sum _{k=1}^N \frac{L_k}{\varLambda ^{(2)}_{k,k}+\gamma }\right) \left| \sum _{k=1}^N \frac{L_k}{\varLambda ^{(1)}_{k,k}+\gamma }-\sum _{k=1}^N \frac{L_k}{\varLambda ^{(2)}_{k,k}+\gamma }\right| \nonumber \\&\quad =\left( \sum _{k=1}^N\frac{\varLambda ^{(1)}_{k,k}+\varLambda ^{(2)}_{k,k}+2\gamma }{\left( \varLambda ^{(1)}_{N,N}+\gamma \right) \left( \varLambda ^{(2)}_{N,N}+\gamma \right) }L_k \right) \left| \sum _{k=1}^N\frac{\varLambda ^{(2)}_{k,k}-\varLambda ^{(1)}_{k,k}}{\left( \varLambda ^{(1)}_{N,N}+\gamma \right) \left( \varLambda ^{(2)}_{N,N}+\gamma \right) }L_k \right| \nonumber \\&\quad \le \frac{2N}{\min \left( \varLambda ^{(1)}_{N,N},\varLambda ^{(2)}_{N,N}\right) +\gamma } \!\times \! \frac{\eta N}{\left( \varLambda ^{(1)}_{N,N}\!+\!\gamma \right) \left( \varLambda ^{(2)}_{N,N}\!+\!\gamma \right) } \nonumber \\&\quad = \kappa N \end{aligned}$$

(11)

Combining (8), (10) and (11), we get

$$\begin{aligned} \epsilon \le \frac{a(\varLambda ^{(1)},\gamma )\kappa N+b(\varLambda ^{(1)},\gamma )^2 \kappa }{b(\varLambda ^{(1)},\gamma )^2b(\varLambda ^{(2)},\gamma )^2}. \end{aligned}$$

Lastly, the simple bounds

$$\begin{aligned} \frac{N}{(\varLambda ^{(i)}_{1,1}+\gamma )^2} \le a(\varLambda ^{(i)},\gamma )\le \frac{N}{(\varLambda ^{(i)}_{N,N}+\gamma )^2} \end{aligned}$$

and

$$\begin{aligned} \frac{N}{\varLambda ^{(i)}_{1,1}+\gamma } \le b(\varLambda ^{(i)},\gamma )\le \frac{N}{\varLambda ^{(i)}_{N,N}+\gamma }, \end{aligned}$$

(12)

lead to

$$\begin{aligned} \epsilon \le \frac{2 \kappa }{N^2} \frac{(\varLambda ^{(1)}_{1,1}+\gamma )^2(\varLambda ^{(2)}_{1,1}+\gamma ) ^2}{(\varLambda ^{(1)}_{N,N}+\gamma )^2} :=c \eta /N, \end{aligned}$$

(13)

for some constant $c>0$ (which is all the more large that the eigenvalues are dispersed).

We now consider the functions $h_{i}(\gamma )=(\varvec{w}^{(i)}_\gamma )'\varvec{w}^{(i)}_\gamma$ for $i=1,2$. Recall from the proof of Proposition 1, that the derivatives $h_i'$ are strictly negative. Therefore, the inverse functions $h^{-1}_i$ are also strictly decreasing and Lipschitz continuous on the interval $(1/N+\varepsilon ,h_i(0))$ for any strictly positive $\varepsilon$, as is depicted in the right side of Fig. 2.

Fig. 2

Graphical illustration of the functions $h_i$ and their inverses $h_i^{-1}$

Since $\varvec{w}^{(i)}_{\gamma _i}$ both satisfy (2), then

$$\begin{aligned} 0&=(\varvec{w}^{(1)}_{\gamma ^{(1)}})'\varvec{w}^{(1)}_{\gamma ^{(1)}}-(\varvec{w}^{(2)}_{\gamma ^{(2)}})'\varvec{w}^{(2)}_{\gamma ^{(2)}}\\&=\left[ (\varvec{w}^{(1)}_{\gamma ^{(1)}})'\varvec{w}^{(1)}_{\gamma ^{(1)}}-(\varvec{w}^{(1)}_{\gamma ^{(2)}})'\varvec{w}^{(1)}_{\gamma ^{(2)}}\right] +\left[ (\varvec{w}^{(1)}_{\gamma ^{(2)}})'\varvec{w}^{(1)}_{\gamma ^{(2)}}-(\varvec{w}^{(2)}_{\gamma ^{(2)}})'\varvec{w}^{(2)}_{\gamma ^{(2)}}\right] \end{aligned}$$

Now, because the $h^{-1}_i$ are Lipschitz on $(\delta ,h_i(0))$, the first bracket satisfies

$$\begin{aligned} |\gamma ^{(1)}-\gamma ^{(2)}| \le c \left| \varvec{w}^{(1)}_{\gamma ^{(1)}})'\varvec{w}^{(1)}_{\gamma ^{(1)}}-(\varvec{w}^{(1)}_{\gamma ^{(2)}})'\varvec{w}^{(1)}_{\gamma ^{(2)}} \right| , \end{aligned}$$

while the second, as we have shown in (13) is $O(\eta /N)$, therefore it must also hold that

$$\begin{aligned} |\gamma ^{(1)}-\gamma ^{(2)}| \le C \eta /N, \end{aligned}$$

for some other constant $C$ which depends on the maximum absolute values of $(h_i^{-1})'$ on the interval $[\delta ,h_i(0)]$ - note that it is not a priori obvious (or true) that the $h_i$ are convex. Overall, we therefore expect $|\gamma ^{(1)}-\gamma ^{(2)}|$ to be smaller for larger values of $\delta$. When the norm constraint is loose, the $\gamma ^{(i)}$ will usually be very small and so will their difference.

Appendix E: Proof of Theorem 1

Simplifying the notations of the preceding proof, we will henceforth write $b_i= b(\varLambda ^{(i)},\gamma ^{(i)})$ (defined in (9)).

$$\begin{aligned} \epsilon&= (\varvec{w}^{(1)}-\varvec{w}^{(2)})'(\varvec{w}^{(1)}-\varvec{w}^{(2)}) \nonumber \\&=(\varvec{w}^{(1)})'\varvec{w}^{(1)}+(\varvec{w}^{(2)})'\varvec{w}^{(2)}-2(\varvec{w}^{(1)})'\varvec{w}^{(2)} \nonumber \\&= \sum _{k=1}^N X_k \times L_k \end{aligned}$$

(14)

where

$$\begin{aligned} X_k =&\frac{1}{\left( \varLambda _{k,k}^{(1)}\!+\!\gamma ^{(1)}\right) ^2b_1^2} \!+\! \frac{1}{\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) ^2b_2^2}-2\frac{1}{\left( \varLambda _{k,k}^{(1)}\!+\!\gamma ^{(1)}\right) \left( \varLambda _{k,k}^{(2)}\!+\!\gamma ^{(2)}\right) b_1b_2} \nonumber \\ =&\frac{\left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) ^2b_1^2\!+\!\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) ^2b_2^2\!-\!2\left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) \left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) b_1b_2}{\left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) ^2\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) ^2b_1^2b_2^2} \nonumber \\ =&\frac{\left( \left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) b_1-\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) b_2\right) ^2}{\left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) ^2\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) ^2b_1^2b_2^2} \nonumber \\ =&\frac{\left( b_1(\varLambda _{k,k}^{(1)}-\varLambda _{k,k}^{(2)})+b_1(\gamma ^{(1)}-\gamma ^{(2)})+ \varLambda _{k,k}^{(2)}(b_1-b_2)+\gamma ^{(2)}(b_1-b_2)\right) ^2 }{\left( \varLambda _{k,k}^{(1)}+\gamma ^{(1)}\right) ^2\left( \varLambda _{k,k}^{(2)}+\gamma ^{(2)}\right) ^2b_1^2b_2^2} \end{aligned}$$

(15)

The denominator of $X_k$ is bounded from below by

$$\begin{aligned} d=N^4\frac{\left( \varLambda _{N,N}^{(1)}+\gamma ^{(1)}\right) ^2\left( \varLambda _{N,N}^{(2)}+\gamma ^{(2)}\right) ^2}{\left( \varLambda _{1,1}^{(1)}+\gamma ^{(1)}\right) ^2\left( \varLambda _{1,1}^{(2)}+\gamma ^{(2)}\right) ^2}, \end{aligned}$$

and this constant, given Lemma 2, can be increased so as to be unrelated to the $\gamma ^{(i)}$.

Moreover, if we recall that $\eta = {\text {max}}_{i=1,\ldots ,N} \left| \varLambda ^{(2)}_{i,i}-\varLambda ^{(1)}_{i,i}\right|$, the following bound also holds

$$\begin{aligned} |b_1-b_2|\!\le \!\sum _{k=1}^N\frac{|\varLambda ^{(2)}_{k,k}\!-\!\varLambda ^{(1)}_{k,k}|}{\left( \varLambda ^{(1)}_{k,k}\!+\!\gamma ^{(1)}\right) \left( \varLambda ^{(2)}_{k,k}\!+\!\gamma ^{(2)} \right) }L_k\!\le \! \frac{\eta N}{\left( \varLambda ^{(1)}_{N,N}+\gamma ^{(1)}\right) \left( \varLambda ^{(2)}_{N,N}+\gamma ^{(2)} \right) }. \nonumber \\ \end{aligned}$$

(16)

We now reconnect the pieces. There are 4 terms in the squared numerator in (15):

• the first one is bounded by a multiple of $\eta N$

• the second one, by Proposition 2 and (12) is $O(\eta )$

• the third one, by (16) is $O(\eta N)$ like the first one

• lastly, by (16) and Lemma 2, the fourth one is also $O(\eta N)$

Consequently, going back to (14) and plugging $\sum _{k=1}^N L_k=N$, we get

$$\begin{aligned} \epsilon = (\varvec{w}^{(1)}-\varvec{w}^{(2)})'(\varvec{w}^{(1)}-\varvec{w}^{(2)})=O(\eta ^2/N). \end{aligned}$$

Remark Instead of looking at the parameters $\eta$ and $N$, one may wonder what is the impact of $\delta$ on $\epsilon$. It can essentially be measured through the magnitude of the $\gamma ^{(i)}$. First, by the last remark of the proof of Proposition 2, we expect $|\gamma ^{(1)}-\gamma ^{(2)}|$ to be a decreasing function of $\delta$. But this term in (15) is associated with the only term which does not increase with $N$. Moreover, the terms in $b_i$ and $|b_1-b_2|$ are all bounded above and below by functions which decrease $\gamma ^{(i)}$ and hence increase with $\delta$ (by Proposition 1). In the end, it is very likely that $(\varvec{w}^{(1)}-\varvec{w}^{(2)})'(\varvec{w}^{(1)}-\varvec{w}^{(2)})$ will be most of the time an increasing function of $\delta$. This seems logical because as $\delta$ decreases to $1/N$, both $\varvec{w}^{(1)}$ and $\varvec{w}^{(2)}$ will converge to the equally weighted portfolio. When the constraint is tight, there is not much room for a large differences between $\varvec{w}^{(1)}$ and $\varvec{w}^{(2)}$, even if $\varvec{\varSigma }^{(1)}$ and $\varvec{\varSigma }^{(2)}$ are very different.