# Expectation of a covariance matrix

I came across an equation that I don't understand how it was derived. The equation was calculating the expectation of the covariance matrix of an error.

The error is $\tilde\theta_N$: $$\tilde\theta_N=R(N)^{-1}\sum_{t=1}^N\varphi(t)e(t)$$ where $$\varphi(t) = \left[u(t-1)~~~u(t-2)~~~...~~~u(t-m)\right]^T$$ $u(t)$ is a time-series signal and the number $m$ is unknown at this time.

$e(t)$ is a white noise sequence with variance $\lambda$. According to the book,

$e(t)$ can be described as a sequence of independent random variables with zero mean values and variances $\lambda$.

$$R(N) = \sum_{t=1}^N\varphi(t)\varphi^T(t)$$

The expectation of its covariance matrix is $P_N$: $$P_N=E\tilde\theta_N\tilde\theta_N^T=ER(N)^{-1}\sum_{t,s=1}^N\varphi(t)e(t)e(s)\varphi^T(s)R(N)^{-1}$$ My question is: how did the equation for $P_N$ came along?

This is what I've got so far:

$$P_N = E\tilde\theta_N\tilde\theta_N^T = ER(N)^{-1}\sum_{t=1}^N\varphi(t)e(t)\left[\sum_{s=1}^N\varphi(s)e(s)\right]^T\left(R^{-1}(N)\right)^T$$

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what are $R(N),\,\varphi(t)$ and $e(t)$? Is $R(N)$ symmetric? Is $\phi(t)$ a vector function? Is $e(t)$ the $t$-th vector in the standard basis? Which ones of these entities are random? –  user1551 Mar 20 '13 at 11:56
Hi @user1551, thanks for your feedback. I've updated my post to the best of my understanding. –  CherryQu Mar 20 '13 at 13:15

## 1 Answer

As $R(N)$ is symmetric, so is $R(N)^{-1}$, i.e. the $\left(R^{-1}(N)\right)^T$ on the RHS of your derivation is equal to $R(N)^{-1}$. Since $e(s)$ is a scalar function, $\left(\varphi(s)e(s)\right)^T=e(s)\varphi(s)^T$. Now the summation sign $\sum\limits_{s,t=1}^n$ is simply a shorthand for $\sum\limits_{t=1}^n\sum\limits_{s=1}^n$. So, $$\sum_{t=1}^N\varphi(t)e(t)\left[\sum_{s=1}^N\varphi(s)e(s)\right]^T =\sum_{s,t=1}^N\varphi(t)e(t)\left(\varphi(s)e(s)\right)^T =\sum_{s,t=1}^N\varphi(t)e(t)e(s)\varphi(s)^T$$ and your derived equation is equivalent to the one in the book.

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