I'm reading a paper where (simplifying):

$Y_{i} = \beta_\theta \theta_i + \beta_x X_i + \epsilon_i $,

$\epsilon_i \sim N(0, \sigma_\epsilon^2) \perp (\theta_i, X_i) $, and $\beta_\theta$ and $\beta_x$ are fixed constants, and $\theta_i$ and $X_i$ are jointly normally distributed such that:

$\begin{pmatrix} \theta_i \\ X_i \end{pmatrix} \sim N \left( \begin{pmatrix} \mu_{\theta} \\ \mu_{x} \end{pmatrix}, \begin{pmatrix} \sigma_{\theta}^2 & \rho \sigma_{\theta} \sigma_{x} \\ \rho \sigma_{\theta} \sigma_{x} & \sigma_{x}^2 \end{pmatrix} \right) $.

In a proof, the authors refer to the multivariate Gaussian of $(\theta_i, X_i, Y_i)$.

My question: How do the authors know that the joint distribution of $(\theta_i, X_i, Y_i)$ is Normal?

I (think) I understand why the distribution of $Y_i$ is univariate normal: it's the sum of jointly normal variables ($\theta_i$ and $X_i$) and an independent normally distributed variable ($\epsilon_i$). Right?

But what are the characteristics of this case which make $Y_i$ jointly normal with $\theta_i$ and $X_i$?

Context: I'm more of an applied econometrics person, so I know some basic probability stuff but have forgotten a lot of it (clearly).

  • $\begingroup$ Just thinking: is it because any linear combination of $Y_i$ and $\theta_i$ and $X_i$ reduces to a linear combination of $\theta_i$ and $X_i$ and $\epsilon_i$, and we know that this linear combination will be normally distributed for the reason I outline in the post? $\endgroup$ Commented Jun 26, 2022 at 22:43

1 Answer 1


It comes from the properties of multivariate normal vectors.

Since $(\theta, X)$ are jointly normal and $\epsilon$ is normal independent of them, wouldn't you say that $(\theta, X, \epsilon)$ is jointly normal as well. (Exercise: Find the mean and covariance matrix)

The (random) vector $$(\theta, X, Y)=(\theta, X, \beta_\theta \theta + \beta_X X + \epsilon) = \begin{pmatrix}1&0&0\\ 0&1&0\\\beta_\theta&\beta_X&1\end{pmatrix}\begin{pmatrix}\theta\\X\\\epsilon\end{pmatrix}$$

Call the matrix in front "A". From the properties of MVRNM's, we have $A\mu=\mathcal N(A\mu_0,A\Sigma_0A^T)$ if $\mu\sim \mathcal N(\mu_0, \Sigma_0)$. We can clearly see that the random vector a few lines prior is multivariate normal as well.

  • $\begingroup$ Thanks for this. Is the intuition that if $(\theta, X, \epsilon)$ is jointly normal, then by definition a linear combination of $(\theta, X, \epsilon)$ (here via the matrix $A$) must also be jointly normal? $\endgroup$ Commented Jun 27, 2022 at 9:04
  • $\begingroup$ @galelamanzi that is correct. $\endgroup$
    – user308286
    Commented Jun 27, 2022 at 18:24
  • $\begingroup$ thanks! very helpful. $\endgroup$ Commented Jun 27, 2022 at 20:09

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