# Setup for proving equation 3.4 from Grinold

I'm studying from Grinold's Active Portfolio Management right now, and used the below equation to answer one of the exercises:

.. let us assume that the correlation between the returns of all pairs of stocks is equal to $$\rho$$. Then the risk of an equally weighted portfolio is

$$\sigma_P = \sigma \cdot \sqrt{\frac{1 + \rho \cdot (N - 1)}{N}}$$

Using this equation though, I realized I don't really know why it's true. Maybe an obvious question, but how would you go about proving this equation is true? I realized I don't quite know how you'd conveniently write out the total risk equation here and reduce it.

In order to derive the simplified portfolio volatility, there is also an assumption of equal variance $$\sigma_i = \sigma$$ for all $$i$$.

Assume we have an equally weighted portfolio of $$n$$ assets with common correlation and variance. Hence, $$w_i = \frac{1}{n}$$, $$\rho_{ij} = \rho$$ and $$\sigma_i = \sigma$$ for all $$i,j=1,\ldots,n$$.

Then, the portfolio variance can be simplified algebraically:

\begin{align*} \sigma^2_p &= \sum_{i=1}^n \sum_{j=1}^n w_i w_j\sigma_i\sigma_j \rho_{ij} \\ &= \sum_{i=1}^n \sum_{j=1}^n \frac{1}{n}\frac{1}{n} \sigma \cdot \sigma \cdot \rho\\ &= \sum_{i=1}^n \sum_{j=1}^n \frac{1}{n^2} \sigma^2 \cdot \rho\\ &= \frac{\sigma^2 }{n^2}\sum_{i=1}^n \sum_{j=1}^n \rho \end{align*} Now, for $$i=j$$ we know that the correlation with itself is $$\rho_{ii} = 1$$. There are $$n$$ diagonal elements meaning that we have $$\sum_{i=1}^n \rho_{ii} = n$$. Moreover there exists $$n^2-n$$ off-diagonal elements in a $$n \times n$$ matrix. Using these results we further get: \begin{align*} \ldots &= \frac{\sigma^2 }{n^2}\left(\sum_{i=1}^n \rho_{ii} + \sum_{i=1}^n \sum_{j\neq i} \rho_{ij} \right)\\ &= \frac{\sigma^2 }{n^2}\left(n+ \rho \cdot (n^2 - n) \right)\\ &=\frac{\sigma^2 }{n^2} \cdot n \cdot \left(1 + \rho \cdot (n - 1) \right)\\ &=\sigma^2 \cdot \frac{1 + \rho \cdot (n - 1)}{n} \end{align*}

Now, the formula for portfolio volatility follows directly by taking the square-root:

$$\sigma_p = \sigma \cdot \sqrt{\frac{1 + \rho \cdot (n - 1)}{n}}$$

Note that the portfolio variance can also be derived using matrix algebra on $$\sigma_p^2 = w^T \Sigma w$$.

• Ahhh, the equal variance assumption is what I was missing to get anywhere from first step - thank you! Commented Jun 19 at 20:29

For sake of completeness, let me add the approach using linear algebra. Let the covariance matrix

\begin{align} \Sigma&=\sigma^2\begin{pmatrix} 1&\rho&\rho&\ldots&\rho \\ \rho&1&\rho&\ldots&\rho \\ \ldots&\ldots&\ldots&\ldots&\ldots \\ \rho&\rho&\rho&\ldots&1 \end{pmatrix}\\ &=\sigma^2\left((1-\rho)\mathbf{I}+\rho\mathbf{1}\mathbf{1}^T\right) \end{align}

Given the weight vector $$w=\frac{\mathbf{1}}{\mathbf{1}^T\mathbf{1}}$$, i.e. $$w_i=1/n$$, the portfolio variance is

\begin{align} \mathrm{Var}_w&=w^T\Sigma w\\ &=\sigma^2\frac{\mathbf{1}^T}{\mathbf{1}^T\mathbf{1}}\left((1-\rho)\mathbf{I}+\rho\mathbf{1}\mathbf{1}^T\right)\frac{\mathbf{1}}{\mathbf{1}^T\mathbf{1}}\\ &=\sigma^2\left((1-\rho)\frac{1}{\mathbf{1}^T\mathbf{1}}+\rho\right)\\ &=\sigma^2\frac{1+\rho(n-1)}{n} \end{align}