Covariance knowning conditional distribution of the two variables

Question

For an excercise at my university I need to find Var(X) where $X=(Y^t,Z^t)^t$ knowing that $Z \sim N(0,I_2)$ and $Y|Z \sim N(BZ,I_4)$ given that $B \in M_{4,2}$ is a matrix and the $I_2$ and $I_4$ are the identitis. I've managed to find that $Var(Y)=I_4+B^tB$ and I also already know $Var(Z)$, I am stuck at the $Cov(Y,Z)=E[YZ^t]$. Does someone know how to compute it?

Answer 1

Sanity check: $B^t B$ is $2 \times 2$, so $I_4 + B^t B$ does not make sense.

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The law of total variance implies $$\text{Var}(Y) = E[\text{Var}(Y \mid Z)] + \text{Var}(E[Y \mid X]) = I_4 + \text{Var}(BZ) = I_4 + BE[ZZ^t] B^t = I_4 + BB^t.$$

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Similarly, the law of total covariance implies \begin{align} \text{Cov}(Y,Z) = E[\text{Cov}(Y, Z \mid Z)] + \text{Cov}(E[Y \mid Z], E[Z \mid Z]) = 0 + B = B. \end{align}

Alternatively, letting $Y = BZ + U$ where $U \sim N(0, I_4)$ is independent of $Z$, \begin{align} \text{Cov}(Y, Z) &= E[(Y-E[Y])(Z-E[Z])^t] \\ &= E[YZ^t] \\ &= E[E[YZ^t \mid Z]] \\ &= E[(BZ+U)Z^t] \\ &= BE[ZZ^t] + B E[UZ^t] \\ &= B. \end{align}