What is the intuition behind the "spanning" result in the following statement?

For a fixed pair of distinct frontier portfolios $\phi_p$ and $\phi_q$, any frontier portfolio $\phi$ can be obtained as a linear combination of $\phi_p$ and $\phi_q$. Thus, we say the portfolio frontier is spanned by two frontier portfolios.


2 Answers 2


I prefer to interpret the mean-variance frontier as a consequence of linear algebra as developed in Hansen and Richard (1987) and discussed in Cochrane (2005). In brief:

  • The space of returns is a hyperplane in the vector space of payoffs.
  • The set of returns on the mean-variance frontier is a line in the space of returns.
    • Any two distinct points on a line define the line. (This is basically what the two-fund separation theorem boils down to.)

Why a line? Speaking loosely, there's only one direction in return space for up (higher expected return), and moving in any perpendicular direction doesn't change the expected return (but does change variance). Move up the line for a higher expected return, move down the line for a lower expected the return, and move perpendicular off the line for more variance and the same expected return.

I'll briefly sketch some of the arguments but read Cochrane for a more in depth discussion.


Let $X$ and $Y$ denote random variables. Observe that $\operatorname{E}[XY]$ is an inner product. $X$ and $Y$ are called orthogonal if their inner product is zero, i.e., $\operatorname{E}[XY] = 0$.

Let $R^*$ denote the projection of the stochastic discount factor onto the space of payoffs and scaled so that it's a return. The hyperplane of excess returns is orthogonal to the discount factor and $R^*$. Let $R^{e*}$ denote the projection of a constant $1$ onto the space of excess returns. In return space, moving in the direction $R^{e*}$ will give a higher expected return.

Hansen Richard orthogonal decomposition

The important point is that any return $R_i$ can be written using the following orthogonal decomposition:

$$ R_i = R^* + w_i R^{e*} + \eta_i$$

Different returns $R_i$ will have different $w_i$ and $\eta_i$. (Note $w_i$ is a scalar and $\eta_i$ is a random variable.) Observe that by construction, $\eta_i$ is orthogonal to 1, hence $\operatorname{E}[\eta_i]= 0$. Therefore:

$$ \operatorname{E}[R_i] = \operatorname{E}[R^*] + w_i \operatorname{E}[R^{e*}]$$

Using $R^*$, $R^{e*}$, and $\eta_i$ are all orthogonal to eachother, you can further show that:

$$ \operatorname{Var}(R_i) = \operatorname{Var}\left( R^* + w_i R^{e*} \right) + \operatorname{Var}(\eta_i) $$

The set of returns on the mean-variance frontier is the line $ R^{mv} = \left\{ R^* + \alpha R^{e*} \mid \alpha \in \mathbb{R} \right\}$. Any return along the mean-variance frontier has $\eta_i = 0$. A non-zero $\eta_i$ gives you variance but no change in expected return (because the space in which $\eta_i$ lies is orthogonal to $R^{e*}$, the projection of $1$ onto the space of excess returns). Why is $1$ special? The inner product of a random variable with $1$ gives its mean.

Hansen Richard in the space of security weights (instead of return space)

  • Let $ \mathbf{R} = \begin{bmatrix} R_1 \\ \ldots \\ R_k \end{bmatrix} $ be a random vector denoting the returns of $k$ securities.
  • Let covariance matrix $\Sigma = \operatorname{Cov}(\mathbf{R})$ and mean return vector $\boldsymbol{\mu} = \operatorname{E}[\mathbf{R}]$.
  • For convenience, let $A = \Sigma + \boldsymbol{\mu} \boldsymbol{\mu'}$. (Hence $A = \operatorname{E}[\mathbf{R}\mathbf{R}']$.)
  • Define inner product $\langle \mathbf{x}, \mathbf{y} \rangle_A \equiv \mathbf{x}' A \mathbf{y}$.
  • Let $\mathbf{1}$ denote a vector of 1s.

Security weights $ \mathbf{w}^*$ (a vector) and corresponding return $R^*$ (a random variable) are given by: $$ \mathbf{w}^* = \frac{A^{-1}\mathbf{1}}{\mathbf{1}'A^{-1}\mathbf{1}} \quad \quad R^* = \mathbf{w}^* \cdot \mathbf{R} $$

Security weights $ \mathbf{w}^{e*}$ and excess return $R^{e*}$ are given by: $$ \mathbf{w}^{e*} = A^{-1}\boldsymbol{\mu} - \left( \frac{\mathbf{1}' A^{-1} \boldsymbol{\mu}}{\mathbf{1}'A^{-1}\mathbf{1}}\right) A^{-1}\mathbf{1} \quad \quad R^{e*} = \mathbf{w}^{e*} \cdot \mathbf{R}$$

Security weights for portfolios on the mean-variance frontier are:

$$ \left\{ \mathbf{w}^{*} + \alpha \mathbf{w}^{e*} \mid \alpha \in \mathbb{R} \right\} $$

$$ \left\{ \frac{A^{-1}\mathbf{1}}{\mathbf{1}'A^{-1}\mathbf{1}} + \alpha \left[ A^{-1}\boldsymbol{\mu} - \left( \frac{\mathbf{1}' A^{-1} \boldsymbol{\mu}}{\mathbf{1}'A^{-1}\mathbf{1}}\right) A^{-1}\mathbf{1} \right] \,\middle|\, \alpha \in \mathbb{R} \right\} $$

Or equivalently:

$$ \left\{ \left( 1 - \beta \right) \frac{A^{-1}\mathbf{1}}{\mathbf{1}'A^{-1}\mathbf{1}} + \beta \frac{ A^{-1}\boldsymbol{\mu}}{\mathbf{1}' A^{-1} \boldsymbol{\mu}} \; \middle| \ \beta \in \mathbb{R} \right\} $$

This is the same mean-variance frontier traced out by combinations of the minimum variance portfolio $\mathbf{w}_{\mathrm{mv}} = \frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}'\Sigma^{-1}\mathbf{1}}$ and tangency portfolio $\mathbf{w}_{\mathrm{tan}} = \frac{\Sigma^{-1}\boldsymbol{\mu}}{\mathbf{1}\Sigma^{-1}\boldsymbol{\mu}}$ but the algebra to see that is a bit horrible.


Cochrane, John, Asset Pricing, 2005

Hansen, Lars Peter and Scott F. Richard, "The Role of Conditioning Information in Deducing Testable Restrictions Implied by Dynamic Asset Pricing Models," Econometrica, 1987 link


This is easier to explain with an example.

Let's say there is no risk free rate as your question seems to imply.

Also assume there are two portfolios $\phi_p$ and $\phi_q$ with the following characteristics:

  1. Both lie on the frontier;
  2. The first one has an expected return of 0.04 and standard deviation of 0.2
  3. The second one has an expected return of 0.10 and standard deviation of 0.3

Then let's have a weight of $w_1$ on the first portfolio and of $w_2 = 1 - w_1$ on the second portfolio. If portfolio 1 and 2 are indeed in the frontier then any combination of these portfolios will be in the frontier. I have plotted such frontier in excel (see below):

enter image description here

This is called the two fund separation theorem. A good reference is Modern Portfolio Theory and Investment Analysis.

One important implication is that all investors can obtain their desired portfolio by mixing two funds, on the frontier (just superimpose indifference curves to the graph above).

  • $\begingroup$ Maybe I should restate the question: what is the intuition behind the two fund separation theorem? $\endgroup$
    – Ola
    May 28, 2018 at 21:57
  • $\begingroup$ I am not sure what you mean by intuition. It is a mathematical fact that if you have 2 assets or combination of assets that lie on the frontier then that is all you need to span the entire frontier. There is no other combination of assets that will yield a higher return per unit of risk ... $\endgroup$
    – phdstudent
    May 28, 2018 at 22:00

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