Joint distribution of difference of normals Consider the following three independent random variables, $X_1 \sim N(\mu_1,\sigma^2_1)$, $X_2 \sim N(\mu_2,\sigma^2_2)$, and $X_3 \sim N(\mu_3,\sigma^2_3)$.
Now let me define the pairwise differences as $\delta_{12} = X_1-X_2$, $\delta_{13} = X_1-X_3$, and $\delta_{23} = X_2-X_3$.
I know that these pairwise differences are easy to calculate as normal random variables, e.g.,
$$\delta_{ij}\sim N(\mu_i - \mu_j, \sigma^2_i + \sigma^2_j)$$
but are we able to calculate the joint distribution of $\delta_{12},\delta_{13},\delta_{23}$ in closed form?
In other words, what is the joint distribution of $p(\delta_{12},\delta_{13},\delta_{23}| \mu_1, \mu_2,\mu_3,\sigma^2_1,\sigma_2^2,\sigma_3^2)$?
Or do we need to place more assumptions?

It's trivial. The marginal means and variances remain the same, i.e. exactly as you have computed them. All you need to know is to calculate the off-diagonal of the variance-covariance matrix. To do so, this simple (useful) formula tells you what you need to know:
$$ \text{var}(X_1 - X_2) = \text{var}(X_1) + \text{var}(X_2) - 2 \text{cov}(X_1, X_2) $$
