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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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