Finding the marginal distribution If $p(y) =N(0,\Gamma) $ and if $p(x|y) =N(\Lambda y, \psi I) $where $I$ denotes the identity matrix. Also, we have the conditions:
$\Gamma_{ij} =0$ for $i \neq j$ (diagonal) 
$\Lambda \Lambda^T = I$
What will the distribution of $x$ be? It is clear to me it should be normal and it has been suggested that we can find the mean and variance using the iterated law of expectations.
For example,
$E_X(X) =E_Y[E_{X|Y} [X|Y]] = E_Y[\Lambda y] =0$
The variance is more complicated though:
$Var_X[X] = E_Y[Var_{X|Y} [X|Y]] + Var_Y[E_{X|Y} [X|Y]] = E_Y[\psi I] + Var_Y[\Lambda y] $
I'm not sure how to proceed..I guess the first term is just $\psi I$ but the second is not clear how to use the diagonal covariance $\Gamma$. 
 A: We will use moment generating functions (mgf's) for the proof. The mgf of a vector random variable $X$ is 
$$ \DeclareMathOperator{\E}{\mathbb{E}} 
  M_X(t) = \E e^{t^T X}
$$ and remember that if $X$ is multivariate normal $\mathcal{N}(\mu,\Sigma)$ then $M_X(t)= \exp\{\mu^T t +\frac12 t^T \Sigma t \} $. Now, we have given the conditional distribution of $X \mid Y=y$ is $\mathcal{N}(\Lambda y, \psi I)$ and the marginal distribution of $Y$ is $\mathcal{N}(0,\Gamma)$.
For the mgf we have in general
$$
   M_X(t) =\E e^{t^T X} = \E [\E e^{t^T X}\mid Y=y]= \E M_{X\mid Y=y}(t;y)
$$
This conditional mgf is 
$$
  M_{X\mid Y=y}(t;y) = \exp\{t^T\Lambda y +\frac12 t^T\psi I t\}
$$
and calculating 
$$\begin{gather}
 M_X(t)&=\E_Y M_{X\mid Y=y}(t;y)=\E \exp\{(\Lambda t)^T Y+\frac12 \psi t^T t\} \\ &= \exp\{\frac12 \psi t^T t\}\E \exp\{(\Lambda t)^T Y\} \\
&= \exp\{\frac12 \psi t^T t\} M_Y( \Lambda t) \\
&= \exp\{ \frac12 t^T [\psi I+\Lambda^T \Gamma \Lambda] t\}
\end{gather}
$$
so we can finally conclude that $X$ is marginally $\mathcal{N}(0,\psi I+\Lambda^T \Gamma \Lambda)$.
