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For a simulation in a research project, I am trying to randomly "appropriate" (meaning subdivide into two components) known values of $Z$ into $X_1$ and $X_2$ such that $Z=X_1+X_2$ in a way that respects their known bivariate normal distribution ($X_1,X_2 \sim N(\mu,\Sigma)$). Is there a way to do this directly?

Not knowing a direct way, I imagined I could compute a distribution of a new variable $U$ that is the conditional distribution given $X_1+X_2=c$, where $c$ is some constant. That is $U\mid(X_1+X_2=c)$, then "appropriate" $Z$ into $X_1$,$X_2$ depending on the realization of $U\mid(Z)$. Perhaps this is equivalent to centering the axis at $\mu$ then rotating the axis 45 degrees?

Once I know the theory, I will need to implement this in R.

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  • $\begingroup$ It's unclear what you mean by "appropriate." Of what variable do you wish to compute the conditional distribution? BTW, we have many threads giving general formulas for conditional distributions in multivariate Normal settings: searching our site will likely answer whatever your question might be. $\endgroup$
    – whuber
    Commented Nov 21, 2019 at 16:22
  • $\begingroup$ @whuber tks. I edited the question to clarify the meaning of appropriate and so that it is not a duplicate of conditional distribution questions. $\endgroup$ Commented Nov 21, 2019 at 17:50
  • $\begingroup$ Thank you. Could you clarify how this question is not simply asking for the distribution of $(X_1,X_2)$ conditional on $X_1+X_2$? $\endgroup$
    – whuber
    Commented Nov 21, 2019 at 17:55
  • $\begingroup$ @whuber. I am happy with the current answer. But feel free to mark this a duplicate of a conditional distribution question if you think so. The current answer helped me to understand the setup and what I have to do better (although once you know this, the answer is more textbook-like) $\endgroup$ Commented Nov 21, 2019 at 22:18

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When $(X_1,X_2)$ enjoy a bivariate normal distribution, $Z = X_1+X_2$ is a normal random variable also, with mean $\mu_Z = \mu_1+\mu_2$ and variance $\sigma_Z^2 = \Sigma_{11} + \Sigma_{22} + 2\Sigma_{12}$. So, to simulate $Z$, all you need to ask R to do is generate samples of a $N(\mu_Z,\sigma_Z^2)$ random variable.

If you wish to determine the conditional distribution of $(X_1,X_2)$ given $Z$, then be aware that the conditional distribution is not a bivariate normal distribution in the usual sense of the bivariate normal joint density. $X_1$ and $X_2$ indeed are bivariate normal random variables but their covariance matrix $\Sigma$ is singular and they don't have a two-dimensional joint density. The distribution is degenerate in that given the value of $Z$ is $c$, the value of $X_2$ is necessarily $c-X_1$. So, to simulate the conditional distribution given $Z = c$, you can simulate the value of $X_1$ given $Z=c$ and then set $X_2$ equal to $c-X_1$. Note that $(X_1, X_1+X_2)$ is bivariate normal with mean $(\mu_1, \mu_1+\mu_2)$ and covariance matrix $$\begin{bmatrix} \sigma_1^2 & \sigma_1^2 + \Sigma_{12}\\ \sigma_1^2 + \Sigma_{12}& \sigma_Z^2\end{bmatrix}.$$ I will leave it as an exercise for you to determine the conditional distribution of $X_1$ given $X_1+X_2 =c$ from this information. Or, if you are feeling lazy, see this answer for ideas on how to proceed.

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  • $\begingroup$ This seems to answer a different question than the one in the title, which asks for some kind of "conditional" distribution. $\endgroup$
    – whuber
    Commented Nov 21, 2019 at 16:27
  • $\begingroup$ @whuber Please see revised answer. $\endgroup$ Commented Nov 21, 2019 at 16:31
  • $\begingroup$ Thank you. I don't think it's an effective answer, though, because it reduces the problem to finding the conditional distribution of $X_1,$ which you don't specify. BTW, many people would allow that a degenerate bivariate Normal distribution is still bivariate Normal. Maintaining that distinction would unnecessarily complicate the statements of many basic results about Normal distributions. (E.g., that linear transformations of multivariate Normal variables are multivariate Normal.) $\endgroup$
    – whuber
    Commented Nov 21, 2019 at 16:36
  • $\begingroup$ @DilipSarwate, OP here. Thanks for the insightful explanation. Indeed I was expecting the conditional distribution to be univariate, as X2 is now a transformation of X1 $\endgroup$ Commented Nov 21, 2019 at 17:31

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