0
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ Your notation is not clear. $\endgroup$ – kjetil b halvorsen Apr 15 '20 at 12:14
  • $\begingroup$ how is that , what is not clear? $\endgroup$ – redha mahir Apr 15 '20 at 14:46
  • $\begingroup$ You did not define $n$. Is the idea that you are stacking the (column) vectors $X_i$? $\endgroup$ – kjetil b halvorsen Apr 15 '20 at 16:34
  • $\begingroup$ yes basically, n is finite ofcourse, lets assume it n=2 $\endgroup$ – redha mahir Apr 15 '20 at 18:23
  • $\begingroup$ what would be the answer $\endgroup$ – redha mahir Apr 15 '20 at 18:23
0
$\begingroup$

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 ...)

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$ – redha mahir Apr 16 '20 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.