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How to compute the conditional variance of a sum of three normally distributed random variables given two other random variables? Assume pairwise correlations exist and the following joint distributions:

$X\sim \mathcal N(\mu_X,\sigma_X^2)$; $Y\sim \mathcal N(\mu_Y,\sigma_Y^2)$; $Z\sim \mathcal N(\mu_Z,\sigma_Z^2)$; $\theta_1\sim \mathcal N(\mu_{\theta_1},\sigma_{\theta_1}^2)$; $\theta_2\sim \mathcal N(\mu_{\theta_2},\sigma_{\theta_2}^2)$;

The multidimensional linear projection theorem can be directly applied to find the conditional mean and variance of random variable/partition of random variables A given random variable/partition of random variables B, but what if A is a combination of three random variables? I am new to the community, so please do not close the question without detailing what else is needed. Thanks.

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  • $\begingroup$ To ask a good question, do not miss specifying the distributions of r.v.s appeared in the expression of your interest. For example, what are the distribution of $\theta_i$? Are $(X, Y, Z, \theta_1, \theta_2)$ jointly normal? Also, if all "5 random variables have different $\mu$ and $\sigma^2$", it is better to use a different notation instead of "$\sim N(\mu, \sigma^2)$". Based on your second paragraph, I think your question can be better framed as "If $(X, Y, Z, \theta_1, \theta_2) \sim N_5(\mathbf{\mu}, \Sigma)$, what is $\operatorname{Var}(X + Y + Z | \theta_1, \theta_2)$?" $\endgroup$
    – Zhanxiong
    Commented Apr 6, 2023 at 12:36
  • $\begingroup$ Thank you for the edits. However, the updated question is still not clear enough to state the "joint distribution" of these 5 r.v.s. Do you know the well-known result that "marginal normality" does not imply "joint normality"? See this thread. The joint normality is essential to derive a closed-form of the conditional variance of your interest. Without it (or with the info you provided so far), it is undetermined. $\endgroup$
    – Zhanxiong
    Commented Apr 6, 2023 at 12:57
  • $\begingroup$ To clarify, the title looks good. It is the question body that needs to be further updated. $\endgroup$
    – Zhanxiong
    Commented Apr 6, 2023 at 13:09
  • $\begingroup$ stats.stackexchange.com/… $\endgroup$
    – whuber
    Commented Apr 6, 2023 at 13:45
  • $\begingroup$ I significantly edited the body of my question - sorry for any inconvenience caused. $\endgroup$ Commented Apr 6, 2023 at 14:01

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Denote $(X, Y, Z)$ and $(\theta_1, \theta_2)$ by $X_1$ and $X_2$, and partition the mean vector $\mu$ and the covariance matrix $\Sigma$ accordingly to as follows: \begin{align*} \mu = \begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \quad \Sigma = \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix}. \end{align*} Then by the conditional distribution property of the multivariate normal distribution, the conditional variance of $X_1$ given $X_2$ is $\bar{\Sigma} = \Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$ (this result is also cited in another question by yourself, though in a more verbose way). Now write $X + Y + Z = e'X_1$, where $e = (1, 1, 1)'$. It then follows that:
\begin{align} & \operatorname{Var}(X + Y + Z|\theta_1, \theta_2) \\ =& \operatorname{Var}(e'X_1|X_2) \\ =& e'\operatorname{Var}(X_1|X_2)e \\ =& e'\bar{\Sigma}e. \end{align} Of course, you can proceed to expand $e'\bar{\Sigma}e$ by finishing tedious matrix operations to express the final result in terms of $\sigma_X^2, \ldots, \sigma_{\theta_2}^2$ and their pairwise correlations $\rho_{ij}$. To me, I think $e'\bar{\Sigma}e$ is a more succinct and equally clear result, so I will be satisfied to stop here.

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  • $\begingroup$ Thank you for your answer. I significantly edited the body of my question - sorry for any inconvenience caused. $\endgroup$ Commented Apr 6, 2023 at 13:58

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