Deriving covariance from "embedded" random variable I'm trying to derive covariance from "embedded" random variable (r.v.). Say we have two Gaussian r.v.'s
$$ X_1 \sim N(\mu_1, \sigma^2_1)$$
$$ X_2 \sim N(\mu_2, \sigma^2_2)$$
where $\mu_1$ and $\mu_2$ are also Gaussian r.v,
$$\mu_1 \sim N(\theta_1, \delta^2_1)$$
$$\mu_2 \sim N(\theta_2, \delta^2_2)$$
with covariance $\delta_{1,2}$. Here we assume that $\sigma$, $\theta$, and $\delta$ are all known.
I'm trying to get $\mathrm{cov}(X_1, X_2)$ following the definition,
$$\mathrm{cov}(X_1, X_2) = E[X_1, X_2] - E[X_1]E[X_2] \>.$$
However, I'm not sure how to incorporate $\delta_{1,2}$ into $E[X_1, X_2]$ when marginalizing out $\mu_1$ and $\mu_2$.
Or, vice versa, given covariance $\sigma_{1,2}$ between $X_1$ and $X_2$, can we derive $\delta_{1,2}$?
Any hint would be appreciated!
 A: Firstly you should be more rigorous when defining the distribution of $X_1$ and $X_2$. I think you mean 
$$ (X_1 \mid \mu_1) \sim N(\mu_1, \sigma^2_1)$$
$$ (X_2 \mid \mu_2) \sim N(\mu_2, \sigma^2_2)$$
and even more precisely you should say that $X_1$ and $X_2$ are conditionally independent given $\mu_1, \mu_2$. Right ?
Then use the well-known formula
$$\boxed{\mathrm{cov}(X_1, X_2) = \mathbb{E}\left[\mathrm{cov}(X_1, X_2 \mid \mu_1, \mu_2)\right] + \mathrm{cov}\bigl(\mathbb{E}[X_1 \mid \mu_1, \mu_2], \mathbb{E}[X_2 \mid \mu_1, \mu_2]\bigr)}.$$
Now $\mathrm{cov}(X_1, X_2 \mid \mu_1, \mu_2)=0$ (conditional independance) and $\mathbb{E}[X_1 \mid \mu_1, \mu_2]=\mu_1$, $\mathbb{E}[X_2 \mid \mu_1, \mu_2]=\mu_2$, then it is easy to conclude.
A: Rewrite $X=(X_1,X_2)$ as
$$
X = \mu + \epsilon\,,\quad \mu \sim \mathcal{N}_2(\theta,\Delta)\,,\epsilon\sim \mathcal{N}_2(0,\Sigma)\,,
$$
with independence between $\mu$ and $\epsilon$. Then 
$$
X \sim \mathcal{N}_2(\theta,\Sigma+\Delta)
$$
gives the marginal distribution of the vector $X$.
A: Based on the answer to my comment, I would have thought you could write 


*

*$X_1 = \theta_1 + V_1 + W_1$ 

*$X_2 = \theta_2 + V_2 + W_2$
where 


*

*$V_1 =\mu_1 - \theta_1 \sim N(0, \delta_1^2)$, 

*$V_2 =\mu_2 - \theta_2 \sim N(0, \delta_2^2)$, 

*$W_1 = X_1 - \mu_1 \sim N(0, \sigma_1^2)$, and 

*$W_2 = X_2 - \mu_2 \sim N(0, \sigma_2^2)$, 


and with $W_1$ and $W_2$ independent of each other and of $V_1$ and $V_2$, but $\mathrm{cov}(V_1,V_2)=\delta_{1,2}$.
In that case it seems obvious $\mathrm{cov}(X_1,X_2)=\delta_{1,2}$.         
