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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?

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  • $\begingroup$ See stats.stackexchange.com/…. $\endgroup$
    – whuber
    Apr 2 '20 at 16:14
  • $\begingroup$ 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? $\endgroup$
    – Tomas
    Apr 2 '20 at 20:26
  • $\begingroup$ ...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. $\endgroup$
    – whuber
    Apr 2 '20 at 20:37