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I have an equation given by $$ \phi(k)=\sqrt{1-\rho^{2}}\sum_{j=1}^{k-1}\rho^{k-j-1}e(j) $$ where $\rho$ has value between 0 to 1 and $e$ is modeled as $\mathcal{C}\mathcal{N}(0,\sigma^{2})$, i.e. circularly symmetric complex Gaussian.

How do I compute the variance of such an expression?

Richard Hardy
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charu
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    What is $\mathcal{CN}$? Looks like $\phi(k)$ is simply a linear combination of $e(j)$. The formula of variance of a linear combination is pretty simple, it involves the sums of variances and covariances of the constitutent random variables. The distribution of the random variables does not matter as long as variance is well defined for them. – Richard Hardy Sep 11 '20 at 11:24
  • @richard, $\mathcal{C}\mathcal{N}$ represents circularly symmetric complex Gaussian complex gaussian. – charu Sep 11 '20 at 11:34
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    Are the e(j) independent from each other? – Three Diag Sep 11 '20 at 11:52
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    The sum goes from $j=...$? – Richard Hardy Sep 11 '20 at 12:21
  • @three diag, yes they are independent. – charu Sep 11 '20 at 18:36
  • Then use the usual formula for the variance of a linear combination: it simplifies greatly by virtue of the independence, becoming the sum of variances of the individual terms. – whuber Sep 11 '20 at 18:41
  • Then, you should use the fact that $\mathbb{V}(\sum_i a_i X_i) =\sum_i a_i^2 \mathbb{V}( X_i) $ when the X are independent, which is a standard property of variance. – Three Diag Sep 11 '20 at 21:38
  • @richard, the sum goes from $j=1$ to $k-1$ – charu Sep 12 '20 at 00:50
  • @richard, @ three diag got the solution..... – charu Sep 17 '20 at 01:52

1 Answers1

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I assume the OP means a Complex normal distribution whose covariance and relation matrices are identity matrices. To get at the result we use a few properties; first $\mathbb{V}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$, second that diagonal covariance and relation matrices imply independence of random variables, third is the geometric series.

Start with:

$$\mathbb{E}(\phi(k)) = \sqrt{1-\rho^2}\sum_{j=1}^{k-1}\mathbb{E}(\rho^{k-j-1}e(j)) $$

and notice that the $\rho$ is constant with respect to the expectation so that each term in the summand is multiplied by $0$. Therefore, $\mathbb{E}(\phi(k)) = 0$.

Next look at the second moment:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\left(\sum_{j=1}^{k-1}\rho^{k-j-1}e(j)\right)^2\right) $$ which, expanding the squares, simplifies to:

$$\mathbb{E}(\phi(k)^2) = (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2 + \sum_{j=1, j\neq i}^{k-1}\sum_{i=1}^{k-1} \rho^{2k-i-j-2} e(i)e(j)\right) $$

Now putting the expectation operator through to the two terms of the square shows that the second term in the sum is $0$, because all terms in the double sum are multiplied by a $0$ term. The first term in the sum has:

$$\mathbb{E}(e(i)^2) = \sigma^2 $$

So that the formula simplifies:

$$ (1-\rho^2)\mathbb{E}\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}e(j)^2\right) = (1-\rho^2)\left(\sum_{j=1}^{k-1}\rho^{2(k-j-1)}\sigma^2 \right)$$

Next factor out the common $\sigma^2$ term and identify the sum as (part of) a geometric series (missing the $j=0$ term. Using the formula

$$\sum_{j=0}^{k-1} (\rho^2)^j = \frac{1-\rho^{2k}}{1-\rho^2}$$

we simplify to

$$ (1 - \rho^2) \sigma^2 \left( \sum_{j=1}^{k-1}\rho^{2(k-j-1)} \right) = (1-\rho^2)\sigma^2 \rho^{2(k-1)} \left( \frac{1-\rho^{2k}}{1-\rho^2} - 1 \right). $$

This last term on the right hand side is what you'd likely use (or some variation). For example, you could turn the $-1$ term into a common denominator and factor out a $\rho^2$ if you like too but this general approach.

Also, here we are assuming $k \geq 1$.

Lucas Roberts
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