New answers tagged covariance-matrix
1
Yes, indeed. It's a simple Linear Algebra and Expectation result:
Given:
$Var(w'r) = \mathbb{E}[(w'r)^2] = \mathbb{E}[(w'rr'w)]$
With $w$ and $r$ the vectors of weights and returns. As $w$ is constant, it holds:
$\mathbb{E}[w'rr'w] = w'\mathbb{E}[rr']w$
The sample variance, $\hat{\Sigma}$, is a estimator of for $\mathbb{E}[rr']$. Therefore, it holds what you ...
4
This turns out to be a general drawback of the HRP algorithm, as pointed out by Pfitzinger, J., & Katzke, N. (2019) (my highlights):
As shown in Figure 2.3, the naive bisection rule can violate the intuitive character of the result, by placing similar assets into separate clusters for allocation purposes. While centered bisection yields a symmetric ...
1
Let $\mathbb{1}$ denote a vector of ones. With the definition of risk parity in the question, we have
$$
Sw=c\mathbb{1}
$$
with $c$ some constant, thus
$$
w=cS^{-1}\mathbb{1}
$$
As $\mathbb{1}^Tw=1$, we have
$$
c\mathbb{1}^TS^{-1}c\mathbb{1}=1 \Rightarrow c=\frac{1}{\mathbb{1}^TS^{-1}\mathbb{1}}
$$
and hence
$$
w=\frac{S^{-1}\mathbb{1}}{\mathbb{1}^TS^{-1}\...
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