The "diversification return" of a portfolio is the difference between the geometric average (compound) return of a rebalanced portfolio and the weighted sum of the geometric average returns of the assets. My understanding is this diversification return is always nonnegative in a portfolio with positive asset weights (only long positions)

I have not found a rigorous proof that diversification return is never negative for the most general case involving $N$ assets. I have found references that show diversification return is approximately the difference between the weighted average of asset variances and the portfolio variance which must be nonnegative. But I would like to see a proof that does not rely on any approximation for an arbitrary number of assets.

To formalize the problem, suppose we have a portfolio with N assets that is always rebalanced to maintain constant asset weights $w_1, w_2, \ldots, w_N$. Let $r_{ij}$ denote the return of asset $i$ in holding period $j$ where $i=1,2,\ldots,N$ and $j = 1,2,\ldots, T$. The return of the portfolio in period $j$ is $r_{Pj} = w_1r_{1j} + \ldots w_N r_{Nj}$ and so the geometric average return of the portfolio over the $T$ periods is

$$g_P = [(1+r_{P1})(1+r_{P2})\cdots (1+r_{PT})]^{1/T} -1$$

The geometric average return of asset $i$ is $$g_i = [(1+r_{i1})(1+r_{i2})\cdots (1+r_{iT})]^{1/T} -1$$

Can it be proved rigorously that $g_P \geq w_1g_1 + w_2 g_2 + \ldots + w_N g_N$?

  • $\begingroup$ That is true and is sometimes referred to as volatility pumping. For two assets it is relatively easy to show, but a little trickier for general $N$. $\endgroup$
    – RRL
    Commented Sep 12, 2022 at 20:59

1 Answer 1


Assuming portfolio weights that sum to $1$, the geometric average portfolio return is given by

$$(1+g_P)^T= \prod_{j=1}^T\left[1+ \sum_{i=1}^N w_i r_{ij} \right] =\prod_{j=1}^T \sum_{i=1}^N w_i(1+ r_{ij}) $$

To proceed, we can apply a form of the Cauchy-Schwarz inequality. The most familiar form is

$$\left(\sum_{i=1}^N X_iY_i\right)^2 \leqslant \left(\sum_{i=1}^NX_i^2 \right)\left(\sum_{i=1}^NY_i^2 \right)$$

and this can be generalized for $T$-fold products to

$$\tag{1}\left(\sum_{i=1}^N\prod_{j=1}^TX_{ij}\right)^T \leqslant \prod_{j=1}^T\sum_{i=1}^NX_{ij}^T$$

Since the weights are nonnegative and assuming $r_{ij} \geqslant -1$, we can define $X_{ij} = [w_i(1+ r_{ij})]^{1/T}$ and substitute into (1) to obtain

$$\tag{2}\prod_{j=1}^T\sum_{i=1}^Nw_i(1+r_{ij}) \geqslant \left(\sum_{i=1}^N\prod_{j=1}^T[w_i(1+r_{ij})]^{1/T}\right)^T = \left(\sum_{i=1}^Nw_i\left[\prod_{j=1}^T(1+r_{ij})\right]^{1/T}\right)^T $$

The geometric average return of asset $i$ satisfies $1+g_i = \left[\prod_{j=1}^T(1+r_{ij})\right]^{1/T}$, and substituting into the RHS of (2) we get

$$\prod_{j=1}^T\sum_{i=1}^Nw_i(1+r_{ij}) \geqslant \left(\sum_{i=1}^N w_i(1+g_i)\right)^T = \left(1+\sum_{i=1}^N w_ig_i\right)^T $$

Since the LHS is equal to $(1+g_P)^T$, it follows that

$$g_P \geqslant \sum_{i=1}^N w_ig_i,$$ as was to be shown.

Proof of generalized inequality (1).

We can derive the generalized Cauchy-Schwarz inequality from the inequality of arithmetic and geometric means (AM-GM inequality) which for nonnegative $x_1,x_2,\ldots, x_n$ is written as $$\left(\prod_{j=1}^n x_j\right)^{1/n} =\sqrt[n]{x_1x_2\cdots x_n}\leqslant \frac{x_1+x_2+ \ldots + x_n}{n} =\frac{1}{n} \sum_{j=1}^n x_j$$

Taking $S_j = \sum_{i=1}^N X_{ij}^T$, we have by the AM-GM inequality,

$$\left(\prod_{j=1}^T\frac{X_{ij}^T}{S_j}\right)^{1/T}\leqslant \frac{1}{T}\sum_{j=1}^T \frac{X_{ij}^T}{S_j},$$

and it follows that

$$\frac{\sum_{i=1}^N \prod_{j=1}^T X_{ij}}{\left(\prod_{j=1}^T S_j \right)^{1/T}}= \sum_{i=1}^N\left(\prod_{j=1}^T\frac{X_{ij}^T}{S_j}\right)^{1/T}\leqslant \sum_{i=1}^N\frac{1}{T}\sum_{j=1}^T \frac{X_{ij}^T}{S_j}\\= \frac{1}{T}\sum_{j=1}^T\sum_{i=1}^N\frac{X_{ij}^T}{S_j } = \frac{1}{T}\sum_{j=1}^T\frac{\sum_{i=1}^NX_{ij}^T}{\sum_{i=1}^NX_{ij}^T} =1$$

Multiplying both sides by the denominator on the LHS we get

$$\sum_{i=1}^N \prod_{j=1}^T X_{ij}\leqslant\left(\prod_{j=1}^T S_j \right)^{1/T} = \left(\prod_{j=1}^T \sum_{i=1}^N X_{ij}^T \right)^{1/T}, $$

and, thus,

$$\left(\sum_{i=1}^N \prod_{j=1}^T X_{ij}\right)^{T}\leqslant\prod_{j=1}^T \sum_{i=1}^N X_{ij}^T $$

  • 1
    $\begingroup$ Thanks. I was hoping to see something like this. I know the basic Cauchy-Schwarz inequality $ \left(\sum x_ix_i\right)^2 \leq \left(\sum x_i^2 \right)\left(\sum y_i^2 \right)$ but have not seen this generalization for $N > 2$. Can you provide a reference or explain briefly where that comes from? $\endgroup$
    – AlRacoon
    Commented Sep 13, 2022 at 0:05
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    $\begingroup$ @AlRacoon: Having used this result many times over the years, I no longer recall any references. I can derive it from the arithmetic-mean (AM) / geometric-mean (GM) inequality and add it above if you like. The AM-GM inequality is a very well-known and useful tool which is worth learning if you don't already know it. For instance it comes up in showing that geometric-mean return is always dominated by arithmetic-mean return. $\endgroup$
    – RRL
    Commented Sep 13, 2022 at 0:51
  • 1
    $\begingroup$ @AlRacoon You can also use a proof by reccurrence on T. $\endgroup$
    – Xman
    Commented Sep 13, 2022 at 12:24
  • 1
    $\begingroup$ @AlRacoon: I added a derivation. $\endgroup$
    – RRL
    Commented Sep 13, 2022 at 16:17

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