How can I calculate $\frac{\partial \log |\Sigma|}{\partial \rho }$ where $\Sigma=(1-\rho)I+\rho\mathbf{1}\mathbf{1}^\top$?

Question

I need to calculate the $\dfrac{\partial \log |\Sigma|}{\partial \rho }$ when $\Sigma = (1-\rho) I + \rho \mathbf{1} \mathbf{1}^\top$ and $\Sigma$ has dimension $p \times p$.

I try to use the formula presented here and here but the result is not right.

Without trying anything fancy, you can simply calculate $|\Sigma|$ and proceed as usual. — StubbornAtom, Apr 25 '19 at 15:38

score 0 · Answer 1 · edited Apr 25 '19 at 15:39

0

The derivative wrt $\rho$ is $\operatorname{trace}((-I+11^T){\Sigma}^{-1})$.

edited Apr 25 '19 at 15:39

StubbornAtom

17,052

answered Apr 25 '19 at 09:50

score 0 · Answer 2 · answered Nov 24 '19 at 02:05

To obtain loup blanc's result $$\eqalign{ \Sigma &= I + \rho\,({\tt\large 11}^T-I) \\ d\Sigma &= ({\tt\large 11}^T-I)\,d\rho \\ \beta &= \det(\Sigma) \\ d\beta &= \beta\,\Sigma^{-1} \\ \lambda &= \log(\beta) \\ d\lambda &= \frac{d\beta}{\beta} \\ &= \Sigma^{-1}:d\Sigma \\ &= \Sigma^{-1}:({\tt\large 11}^T-I)\,d\rho \\ \frac{d\lambda}{d\rho} &= \Sigma^{-1}:({\tt\large 11}^T-I) \\ }$$ where a colon is used as a convenient product notation for the trace, i.e. $$\eqalign{A:B = {\rm tr}(A^TB)}$$ and Jacobi's formula is used to differentiate $\beta$.

How can I calculate $\frac{\partial \log |\Sigma|}{\partial \rho }$ where $\Sigma=(1-\rho)I+\rho\mathbf{1}\mathbf{1}^\top$?

2 Answers2