Consider a multivariate normal distribution with covariance matrix $\Sigma$ of size $n \times n$, which can be written in terms of its lower triangular Cholesky decomposition $L$ as $$\Sigma = L \cdot L^\intercal.$$
Consider the marginal joint normal distribution of $m$ of the original $n$ variables. The covariance matrix of the marginal distribution $\Sigma '$ can be written as: $$\Sigma '= M (L L^\intercal) M^\intercal$$ where $M$ is a matrix of size $m\times n$ consisting of zeros and ones, so that $M$ is dropping the rows and columns corresponding to the unwanted $n-m$ variables.
Let us write $\Sigma'$ in block matrix form as $$\Sigma' = \begin{bmatrix} \Sigma'_{11} & \Sigma'_{12} \\ \Sigma_{12}^{'\intercal} & \Sigma'_{22} \end{bmatrix}$$ where $\Sigma'_{11}$ is a $(m-1) \times (m-1)$ matrix, $\Sigma'_{12}$ is a $(m-1) \times 1$ vector and $\Sigma'_{22}$ is an single entry.
I want to derive a conditional normal distribution, therefore I need to compute the Schur complement of $\Sigma'_{22}$, i.e. $S = \Sigma'_{22}-\Sigma_{12}^{'\intercal}\Sigma_{11}^{'-1}\Sigma'_{12}$. I would like to express $S$ as a function of the entries of the Cholesky decomposition $L$ of the original matrix $\Sigma$ since the entries of $L$ have a statistical meaning in my model. Is it possible? A similar question was raised in this Stack topic, but it remained unanswered.
In this answer, it is shown that we can relate the Schur complement of a matrix to its Cholesky factorization. Therefore, I tried to derive the Cholesky factorization of $\Sigma'$ as a rank $(n-m)$ downdate of Cholesky factorization of $\Sigma$, but this way seems rather complicated to follow (see Julia's blog).
Any help would be much appreciated.