# Correlation of assets to portfolio of assets

How do you calculate the correlation of an asset to a portfolio, when for all assets in the portfolio you know there: correlation to each other, volatility and weight in portfolio.

For example: Assets 1,2,3&4 all have volatility of 15%. Assets 1&2 have a correlation of 1 and all other pairs of assets correlation = 0.

With a portfolio of 16.7% in Assets 1 & 2 and 33.3% in 3 & 4, What I am reading states that all assets (1,2,3,4) have a correlation of 0.577 with the portfolio.

How is this calculated? Is there a formula that can be applied to broader examples with more varied asset volatilities and correlations?

This is to basic for this website, but I will answer it anyway as I think it is interesting.

You have a correlation matrix of 4 assets (1, 2, 3, 4). This is how it looks:

$Correl = \begin{bmatrix} 1 &1& 0& 0 \\ 1 &1& 0& 0 \\ 0 &0& 1& 0 \\ 0 &0& 0& 1 \\ \end{bmatrix}$

Thus the covariance matrix is (show this yourself):

$Cov= \begin{bmatrix} 0.0225 &0.0225& 0& 0 \\ 0.0225 &0.0225& 0& 0 \\ 0 &0& 0.0225& 0 \\ 0 &0& 0& 0.0225 \\ \end{bmatrix}$

The weights matrix is:

$w= \begin{bmatrix} 0.167 \\ 0.167 \\ 0.333 \\ 0.333 \\ \end{bmatrix}$

Therefore the standard deviation of the whole portfolio is: $Std(P) = \sqrt{w' Cov \text{ }w} = 0.0866$.

Now what is the correlation of the first asset ($A_1$) with the portfolio?

Well it is given by:

$\rho_{P,A_1} = \frac{Cov(A_1,P)}{std(A) std(P)} = \frac{Cov(A_1,P)}{std(A) std(P)} = \frac{w_1 cov(A_1, A_1) + w_2 cov(A_1, A_2) + w_3 cov(A_1, A_3) +w_4 cov(A_1, A_4)}{std(A) std(P)} = 0.57$

Then just repeat the last step for the other assets (2, 3 and 4).

• As you can see from this example, the general rule for the correlation between the portfolio and the i-th security is: take the inner product between the weight vector and the i-th row of the covariance matrix, then divide by the std dev of the i-th security and the std dev of the portfolio. (The former is the sqrt of the [i,i] entry in the covariance matrix, the latter is $\sqrt{w^{'} \Sigma w}$). Commented May 24, 2018 at 11:41