# the concatenation of bivariate iid

suppose that $$X \sim N\left( {0,{\sigma ^2}{I_2}} \right)$$ is a bivariate white noise, and the samples $${X_1}, \cdots ,{X_N}$$ are drawn from it, if we define the new random variable $$Y$$ with its samples $${Y_t} = \left[ {\begin{array}{*{20}{c}} {{X_t}}\\ \vdots \\ {{X_{t + n - 1}}} \end{array}} \right]$$, $$t = 1, \cdots ,N - n + 1$$, what distribution would $$Y$$ follow and what is its covariance matrix

• Your notation is not clear. – kjetil b halvorsen Apr 15 '20 at 12:14
• how is that , what is not clear? – redha mahir Apr 15 '20 at 14:46
• You did not define $n$. Is the idea that you are stacking the (column) vectors $X_i$? – kjetil b halvorsen Apr 15 '20 at 16:34
• yes basically, n is finite ofcourse, lets assume it n=2 – redha mahir Apr 15 '20 at 18:23
• what would be the answer – redha mahir Apr 15 '20 at 18:23

You have some iid random vectors, each is bivariate $$\mathcal{N}_2(0,I)$$ which you then are stacking as one vector of some length (depending on some $$n$$ you did not explain.) Then you have a vector of iid standard normals, of some length. Mean vector is zero, covariance matrix identity matrix of some dimension. There is nothing I can see in the question but play on notation. (maybe I misunderstood ...)
• are you sure covariance matrix is identity because the sample covariance $R = \frac{1}{N-n+1}\sum\limits_{i = 1}^{N-n+1} {{Y_i}Y_i^T}$ does not seem to give me identity, thats why im confused – redha mahir Apr 16 '20 at 7:05