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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.

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2 Answers 2

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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 $\sigma_p^2 = w^T \Sigma w$.

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

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

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