# Probability density of conditional multivariate distribution [duplicate]

We have a multivariate normal vector $${\boldsymbol Y} \sim \mathcal{N}(\boldsymbol\mu, \Sigma)$$. Consider partitioning $${\boldsymbol Y}$$ into

$${\boldsymbol Y}=\begin{bmatrix}{\boldsymbol y}_1 \\ {\boldsymbol y}_2 \end{bmatrix}$$

Say that $$f_Y$$ is a probability density function for $$Y$$. What can we say about the probability density function $$f_{y_1|y_2 = a}$$ of conditional multivariate distribution $$(y_1|y_2 = a)$$?

Does this hold? :

$$f_{y_1|y_2 = a}(y_1) \stackrel{?}{=} {f_Y([y_1,y_2 = a]) \over \int_{y_1} f_Y([y_1,y_2 = a])}$$

Intuitively, I would assume that we could use $$f_Y$$ in the appropriate $$y_2 = a$$, and then normalize it to integrate to 1... Or is there some "gotcha" that the density scales differently in the space of $$y_1$$? I know that it the conditional distribution could be computed with the Schur complement but this could save computational time in cases when you don't actually need a density normalized to 1, which is my case.

PS: and then, probably, $$\int_{y_1} f_Y([y_1,y_2 = a]) \stackrel{?}{=} f_{y_2}(y_2 = a)$$, so perhaps the normalization would also be simple?

• – whuber
Apr 2 '20 at 16:14
• Has anybody from the closers actually read what I am asking? Did you notice that I am asking about PROBABILITY DENSITY FUNCTION and NONE of these questions is asking about it? Apr 2 '20 at 20:26
• ...and every one of them is giving you everything you need to know about the density function as an answer, regardless, because it is characteristic of a Normal distribution that its density is completely--and very simply--determined in a very well-documented way by the mean vector and covariance matrix. To specify them is to specify the density and vice versa.
– whuber
Apr 2 '20 at 20:37