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From some process I got a series of values. I want to compute the variance of the mean from this series. The series is built with contiguous sub-series. In each sub-series the values are correlated. All sub-series follow the same pattern but they have different total lengths.

I read on Wikipedia that for correlated variables holds

$$\operatorname{Var}\left(\overline{X}\right) = \frac{\sigma^2}{n} + \frac{n - 1}{n}\rho\sigma^2 $$

where $\rho$ is the average correlation.

Q1: What does average correlation means or how it is computed?

Intuitively I would expect it to be something like the lag or the sum of all lags of the autocorrelation function of the series, but I don't know.

Q2: Is this approach reasonable?


EDIT: I checked these questions:

According to comments, I should use:

$$\operatorname{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \operatorname{Var}\left(X_i\right) + 2\sum_{1\le i<j\le n}\operatorname{Cov}\left(X_i, X_j\right)$$

but I am not sure on how to compute the $\operatorname{Cov}\left(X_i, X_j\right)$.

If it helps, the data arises from an stationary process (although it always starts with a complete sub-series).

Variance of a sum of identically distributed random variables that are not independent

Variance of sum of dependent random variables

and this article.

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  • $\begingroup$ Use the formula just stated above in wikipedia. $\endgroup$ Commented Sep 14, 2022 at 15:49
  • $\begingroup$ @User1865345, Thank you! I am not sure how to do that. I edited the question according, could you check it out? $\endgroup$ Commented Sep 14, 2022 at 17:11
  • $\begingroup$ What is covariance in terms of correlation? Use that. All the variances are identical. The covariance between each pair is same. Proceed then to calculate the variance of the mean (divide both sides by $n^2$). $\endgroup$ Commented Sep 14, 2022 at 17:42
  • $\begingroup$ @User1865345, Thank you! Does not the covariance depends on $j-i$ ? Something like $cov(x_i,x_j)=\rho[j-i]\sigma^2$. ? $\endgroup$ Commented Sep 14, 2022 at 17:45
  • $\begingroup$ Check Wikipedia. $\endgroup$ Commented Sep 14, 2022 at 17:47

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The general formula is

$$\text{Var}\left(\sum_{k=1}^n a_k X_k\right) = \sum_{k=1}^n a_k^2 \text{Var}\left(X_k\right) + \sum_{k=1,l=1 \\k\neq l}^n a_k a_l \text{Cov}\left(X_k,X_l\right) $$

And when all the variances are equal to $\sigma^2$ and all $a_k = \frac{1}{n}$ then

$$\sum_{k=1}^n a_k^2 \text{Var}\left(X_k\right) + \sum_{k=1,l=1 \\k\neq l}^n a_k a_l \text{Cov}\left(X_k,X_l\right) = \sum_{k=1}^n \frac{1}{n^2} \sigma^2 + \sum_{k=1,l=1 \\k\neq l}^n \frac{1}{n^2} \rho_{k,l}\sigma^2 = \frac{1}{n} \cdot \sigma^2 + \frac{1}{n^2} \left( \sum_{k=1,l=1 \\k\neq l}^n \rho_{k,l} \right) \cdot \sigma^2$$

It is this sum of all the correlations $\rho_{k,l}$ that is being replaced by the average.

$$\sum_{k=1,l=1 \\k\neq l}^n \rho_{k,l} = \bar\rho \cdot n \cdot (n-1)$$

but I am not sure on how to compute the $\operatorname{Cov}\left(X_i, X_j\right)$.

See the definition of the Pearson correlation coefficient $\rho_{X,Y} = \frac{Cov(X,Y)}{\sqrt{Var(X)\cdot Var(Y)}}$

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