Covariance of two random vectors IF I have $X,Y_1,...,Y_n$ iid then how do I calculate:
cov  $\left [\begin{pmatrix}X\\.\\.\\.\\X \end{pmatrix}, \begin{pmatrix}Y_1\\.\\.\\.\\Y_n \end{pmatrix}\right]$?
 A: This is known as the cross-covariance between vectors, and is defined by
$$
\text{cov}[\boldsymbol{X},\boldsymbol{Y}] = \text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}]
$$
where 
$$
\boldsymbol{\mu_X} = \text{E}[\boldsymbol{X}]\\
\boldsymbol{\mu_Y} = \text{E}[\boldsymbol{Y}]
$$
In your case, because all the components of $\boldsymbol{X}$ are the same, things simplify greatly. 
$$
\boldsymbol{X} = X
\left[
\begin{array}{c}1\\1\\\vdots\\1\end{array}
\right], \;\;
\boldsymbol{\mu_X} = \mu_X
\left[
\begin{array}{c}1\\1\\\vdots\\1\end{array}
\right]
$$
Where $\mu_X=\text{E}[X]$. Then
$$
\boldsymbol{X}-\boldsymbol{\mu_X} = (X-\mu_X)
\left[
\begin{array}{c}1\\1\\\vdots\\1\end{array}
\right]
$$
Now
$$
(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T} = 
(X-\mu_X)
\left[
\begin{array}{c}1\\1\\\vdots\\1\end{array}
\right]\left[
\begin{array}{cccc}Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\end{array}
\right]
$$
where $\mu_m=\text{E}[Y_m]$ for $m\in[1,2,\cdots,n]$. Expanding out that matrix product we have
$$
(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T} =
(X-\mu_X)\left[
\begin{array}{cccc}
Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\\
Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n\\
\vdots&\vdots&\ddots&\vdots\\
Y_1-\mu_1&Y_2-\mu_2&\cdots&Y_n-\mu_n
\end{array}
\right]
$$
Taking that scalar inside the matrix, we see it multiplies each entry in the matrix. Then taking the expectation of the result finally gives
$$
\text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}] =
\left[
\begin{array}{cccc}
\text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)]\\
\text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)]\\
\vdots&\vdots&\ddots&\vdots\\
\text{E}[(X-\mu_X)(Y_1-\mu_1)]&\text{E}[(X-\mu_X)(Y_2-\mu_2)]&\cdots&\text{E}[(X-\mu_X)(Y_n-\mu_n)]
\end{array}
\right]
$$
$$
=
\left[
\begin{array}{cccc}
\text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n)\\
\text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n)\\
\vdots&\vdots&\ddots&\vdots\\
\text{cov}(X,Y_1)&\text{cov}(X,Y_2)&\cdots&\text{cov}(X,Y_n)
\end{array}
\right]
$$
Now we are at the answer: you specified all the variables to be identically distributed and independent. Independent variables have covariance $0$. SO, you get the all zeros matrix for your answer
$$
\text{cov}(\boldsymbol{X},\boldsymbol{Y})=\text{E}[(\boldsymbol{X}-\boldsymbol{\mu_X})(\boldsymbol{Y}-\boldsymbol{\mu_Y})^\text{T}] =
\left[
\begin{array}{cccc}
0&0&\cdots&0\\
0&0&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&0\\
\end{array}
\right]
$$
A: Covariance of $2$ vectors is basically what is called a variance-covariance matrix $(\Sigma)$ defined as
$$((\Sigma_{ij}))=Cov(X_i,Y_j)$$ where $Cov(A,B)=E(AB)-E(A)E(B)$
For more details, just Google Variance Covariance matrix. 
Specifically, because of the iid character of your variables, $Cov$ will be $0$ for all.
