# Tag Info

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}\...

Top 50 recent answers are included