# Efficient frontier portfolio's analytical solution for a given expected return $r$

$$$$\boldsymbol{w}(r) = \frac{r\mathbf\Sigma^{-1} \boldsymbol{\mu}}{\boldsymbol{\mu}^{\top} \mathbf{\Sigma}^{-1}\boldsymbol{\mu}}$$$$ is the closed-form analytical solution for the portfolio weight vector of any portfolio along the efficient frontier, whose expected return level is some scalar value $$r$$. It appears in a 2010s article without citation so it must've been introduced much earlier.

Which source first derived this solution, as well as the analytical solutions for its mean and variance, $$\mu_p[\boldsymbol{w}(r)], \sigma_p[\boldsymbol{w}(r)]$$? (not to be confused with the minimum-variance and maximum Sharpe portfolio solutions)

• Hi, could you please explain more in detail what you define $w$ to be? It is a vector of weight? It does not sum to 1, then, no? Also: have you already read the papers by Markowitz anf others? Where have you already looked for information? Where did you get that result from? That would be a helpful starting point for us - for me at least :) – Kermittfrog Nov 16 '20 at 7:11

I agree with @Kermittfrog's comment, that this only works if you do not impose any budget constraint (in the sense that your weights sum up to one). Other than that, I am sorry that I can not precisely answer your question where it was first derived (tbh: I am not even sure if it was explicity derived at all somewhere because it simply follows from the very definition of a target-return constraint optimziation problem). In brief:

Let $$\mu$$ be a $$n$$-dimensional vector that contains the single-asset returns, and $$\Sigma$$ the corresponding $$n \times n$$ covariance matrix. Moreover, let $$\mathbf{w}$$ be a $$n$$-dimensional vector that assigns the portfolio weights to the $$n$$ individual assets. (Here we do not impose summation to unity, see comment above!) Furthermore, express by $$r$$ the target return an investor expects to achieve on her portfolio (thus, $$r$$ is a scalar). In other words, the optimal solution must in addition satisfy $$r = \mu^{T} \mathbf{w}$$, where superscript $$T$$ denotes a vector/matrix transpose.

Now, we seek to minimize (for mathematical convenience: half of) the portfolio variance, which is given by $$\mathbf{w}^{T}\Sigma\mathbf{w}$$. And we will minimize it under the target return restriction presented above. As we have formulated it as an equality constraint, we can use the Lagrangian method to solve this problem. Defining the Lagrangian multiplier $$\lambda$$, we can write the first-order condition of the optimization problem (with respect to the weight vector) as follows:

$$\Sigma \mathbf{w} - \lambda \mu = \mathbf{0}$$

Note that $$\mathbf{0}$$ describes the all-zero vector of dimension $$n$$. Solving this for $$\mathbf{w}$$ is easy, and defining by $$\Sigma^{-1}$$ the inverse of our covariance matrix, we obtain that:

$$\mathbf{w} = \Sigma^{-1} \lambda \mu$$

Moreover, recall our optimization constraint, which will help us to find $$\lambda$$; it reads $$r = \mu^{T} \mathbf{w}$$. If you plug in the solution for $$\mathbf{w}$$ and rearrange, you will get that $$\lambda = \dfrac{r}{\mu^{T} \Sigma^{-1} \mu}$$.

Clearly, you see that this Lagrangian "slack" parameter depends on your choice of target return, $$r$$. Plugging this Lambda back into our weight solution, you obtain the proposed result:

$$\mathbf{w}(r) = \dfrac{r \Sigma^{-1} \mu}{\mu^{T} \Sigma^{-1} \mu}$$.

Finally, in a similar fashion proposed in the answer to your other post (Closed-form analytical solution for the variance of the minimum-variance portfolio?), once you have your ($$r$$-dependent) weights, the ($$r$$-dependent) mean and variance of the portfolio follow therefrom.

• So the analytical moments are the minimum-variance portfolio's moment solutions except multiplied by $r$? – develarist Nov 18 '20 at 12:05
• I don't think that this is generally true – KevinT Nov 24 '20 at 7:53
• why not, given that $$$\boldsymbol{w}_{MV}= \frac{\mathbf\Sigma^{-1} \boldsymbol{\mu}}{\boldsymbol{\mu}^{\top} \mathbf{\Sigma}^{-1}\boldsymbol{\mu}}$$$, whereas $\mathbf{w}(r) = \dfrac{r \Sigma^{-1} \mu}{\mu^{T} \Sigma^{-1} \mu}$? – develarist Nov 24 '20 at 7:55
• It should be $\mathbf{w}_{MVP} = \dfrac{\Sigma^{-1} \mathbf{1}}{\mathbf{1}^{T} \Sigma^{-1} \mathbf{1}}$. – KevinT Nov 24 '20 at 8:28
• Hi @develarist, if your question is solved by this answer, please consider accepting it – KevinT Mar 31 at 6:47