Suppose I have the following model:
$$\begin{aligned} \mathcal L &\sim \mathcal N \left(\vec {\mu_L}, \mathbf \Sigma_L \right) \\ \mathcal X &\sim \mathcal N \left(\vec 0, \mathbf \Sigma_X \right) \\ \mathcal P &= \mathcal L + \mathcal X \\ \\ \end{aligned}$$
Or, equivalently:
$$\begin{aligned} \mathcal L &\sim \mathcal N \left(\vec {\mu_L}, \mathbf \Sigma_L \right) \\ \mathcal P | \mathcal L = \vec l &\sim \mathcal N \left(\vec {l}, \mathbf \Sigma_X \right) \end{aligned}$$
What's $\mathcal L | \mathcal P = \vec p$ distributed as, in general? In the one-dimensional case, it is very easy to see through straightforward algebraic manipulation that:
$$\mathcal L | \mathcal P = p \sim \mathcal N \left( \frac{\frac 1 {\sigma_L^{2}} \mu_L + \frac 1 {\sigma_X^{2}} p}{\frac 1 {\sigma_L^{2}} + \frac 1 {\sigma_X^{2}}}, \left(\frac 1 {\sigma_L^{2}} + \frac 1 {\sigma_X^{2}}\right)^{-1} \right)$$
That is, the posterior mean is the average between $\mu_L$ and $p$ weighted by the precisions of each distribution, and the posterior precision is the sum of the precisions of each distribution (so the posterior variance is its inverse).
Algebraic manipulation seems like it shouldn't be the only way to get to this result, though. What other way is there? And how do I use it to generalise this result to higher dimensions? It seems to me that the result should be something like:
$$\mathcal L | \mathcal P = \vec p \sim \mathcal N \left( \frac{\mathbf\Sigma_L^{-1}\vec{\mu_L} + \mathbf\Sigma_X^{-1}\vec p}{|\mathbf\Sigma_L^{-1}| + |\mathbf\Sigma_X^{-1}|}, \left(\mathbf\Sigma_L^{-1} + \mathbf\Sigma_X^{-1}\right)^{-1} \right)$$
But I have no idea if that's actually the case or how to prove it if so.
I found the answer in a previously-answered question here. It turns out I was slightly wrong, the correct form is:
$$\mathcal L | \mathcal P = \vec p \sim \mathcal N \left(\left(\mathbf\Sigma_L^{-1} + \mathbf\Sigma_X^{-1} \right)^{-1} \left(\mathbf\Sigma_L^{-1} \vec{\mu_L} + \mathbf\Sigma_X^{-1} \vec p \right), \left(\mathbf\Sigma_L^{-1} + \mathbf\Sigma_X^{-1} \right)^{-1} \right)$$
Since it wasn't mentioned there, I will have to assume there isn't a non-algebraically-manipulative way of figuring this out.