Mean and autocorrrelation function

[TomBombadil]

I'm trying to solve a random process problem:

"Let $x(n) = v(n) + 3v(n-1)$ where $v(n)$ is a sequence of independent random variables with mean µ and variance s². Whats its mean and autocorrelation function? Is this process stationary? Justifiy."

To discover the mean, I applied the expected value at x(n):

$E[x(n)] = E[ v(n) ] + 3 E[v(n-1)] = \mu + 3\mu = 4\mu$

To obtain the autocorrelation function:

$r(n_1,n_2) = E[x(n_1)x(n_2)] = E[v(n_1)v(n_2) +3v(n_1)v(n_2 -1) + 3v(n_1-1)v(n_2) + 9v(n_1 -1)v(n_2-1)]$

Since v(n) is an independent sequence..
$r(n_1,n_2) = E[v(n_1)] E[v(n_2)] + 3E[v(n_1)] E[v(n_2-1)] + 3 E[v(n_1-1)] E[v(n_2)] + 9 E[v(n_1 -1 ) ] E[v(n_2 -1)]$

But $E[v(n)] = \mu$, then

$r(n_1,n_2) = \mu^2 + 3\mu^2 + 3 \mu^2 + 9 \mu^2 = 16\mu^2$

I'm afraid that I made mistake evaluating the autocorrelation function..So, is there any error? If yes, could someone correct me? Thanks!

 

[marcusl]

Your mistake is here

To obtain the autocorrelation function:

$r(n_1,n_2) = E[x(n_1)x(n_2)] = E[v(n_1)v(n_2) +3v(n_1)v(n_2 -1) + 3v(n_1-1)v(n_2) + 9v(n_1 -1)v(n_2-1)]$

Cross-correlation is defined as

$\rho_j=\sum_{k=-\infty}^{\infty} x_k y_{k+j}$

so the factorization you tried to do is not allowed. (Note that definitions vary. Sometimes c-correlation is defined as

$\rho_j=\sum_{k=-\infty}^{\infty} (x_k - \mu_x)(y_{k+j} - \mu_y)$

instead. Use the definition from your text or class.)

Modify this to fit your problem and it should work out better.

 

[mathman]

If n1 and n2 differ by 1, then one of the terms of r(n1,n2) cannot be split up as a product of expectations, since the argument for both v's in the product would be the same.