Let $ f(L) $ be a scalar function of the lower triangular Cholesky decomposition of the covariance matrix $ \Sigma $ such as $ \Sigma = LL' $.
Let's assume that we know the first and second derivative of $ f(L) $ wrt to $L{_i}{_j} $:
$$ dL=\frac{\text{d } f(L)}{\text{d vech}(L)} $$
$$ hL=\frac{\text{d}^2 f(L)}{\text{d vech}(L)^2} $$
where $ dL $ is a vector of the same size of $ vech(L) $ and $ hL $ is the corresponding hessian matrix.

To pass from the Cholesky factor to the (lower triangular) covariance matrix, Iain Murray has found an interesting formula for the first derivative:
$$ \text{tril}(d\Sigma) = \Phi(L^{-T}(P+P^T)L^{-1}) \text{ where } P=\Phi(L^TdL) $$
and $ \Phi()  $ takes the lower-triangular part of a matrix and halves its diagonal.

What I would like to find is the formula for $ h\Sigma $ as a function of $ dL $ and $ hL $.

A similar question has been asked [here][1] for the first derivative, and Iain Murray has written an [interesting paper][2] on the question of first derivatives.

Thanks!

  [1]: https://mathoverflow.net/questions/150427/the-derivative-of-the-cholesky-factor
  [2]: http://arxiv.org/abs/1602.07527