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I want to calculate the average value of $n_i$ time series each of length $n_t$, i.e. an average of $n_i \times n_t$ values, together with a measure of uncertainty. To be more concrete, I have $n_i\approx 6$ temperature measurements at different places for $n_t\approx 100$ different times. Each of the time series is auto-correlated but the different time series are also correlated. How can I estimate the variance of the average?

My approach so far was to treat each time series separately. Let $x^i_t$ be the measurement of time series $i$ at time $t$. Then I estimated the variance $\mathrm{Var}(\bar x^i)$ of the mean over time, $\bar x^i$, for a fixed $i$ to be $$ \mathrm{Var}(\bar x^i) = \frac{\mathrm{Var}(x^i_t)}{n_t} \frac{n_t-1}{n_t/\gamma_i-1} $$ with $\gamma_i$ derived from the auto-correlation $\rho^i_k$ of $x^i_t$: $$ \gamma_i = 1+ 2\sum_{k=1}^{k_\mathrm{max}}(1-\frac{k}{n_t})\rho^i_k \, . $$ While the linked article uses $k_\mathrm{max}=n_t-1$, I found it more useful and also convincing to sum up only the first values of the auto-correlation that are significantly different from $0$. (Please correct me if I am wrong here.)

How do I combine the variances of the single time series to the uncertainty of the total average $$ \bar{\bar x} = \frac{1}{n_i}\sum_i \bar x^i $$ taking into account the correlation between the time series as for the general case proposed here? Using $$ \mathrm{Var}(\bar{\bar x}) = \frac{1}{n^2} \sum_i \mathrm{Var}(\bar x^i) $$ ignores this correlation.

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  • $\begingroup$ What do you mean by uncertainty? Is it the variance? Do you want it to be conditional on something? $\endgroup$
    – Aksakal
    Jul 17, 2015 at 15:08
  • $\begingroup$ @Aksakal Yes, I mean variance (edited question). $\endgroup$
    – sebschub
    Sep 8, 2015 at 14:29

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The problem seems to be suited for some kind of PCA analysis. You'll represent your series $x_{it}$ where $i$ is the place and $t$ - time as follows: $$x_{it}=\sum_{j=1}^{n_i} f_{jt}C_{j}$$ where $f_{jt}$ are not correlated, i.e. $Corr[f_{jt},f_{kt}]=\delta_{jk}$

However, $f_{jt}$ are autocorrelated, i.e. $Cov[f_{jt},f_{j,t+1}]\ne 0$

Since, the factors are not correlated you can add up their variances, i.e. uncertainties in your language. So, this trick may allow you to keep treating the components separately without worrying about correlations.

PCA is often used to reduce the dimensionality of the problem. Your temperatures are very correlated, so you may not need all $n_i$ factors, but use first couple of them. However, that would be an additional benefit in your case, it seems. You will use PCA to deal with correlations.

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  • $\begingroup$ Thanks a lot! Based on your answer, I added another one with all the details. $\endgroup$
    – sebschub
    Jul 21, 2015 at 12:34
  • $\begingroup$ I am not sure any longer that this really helps. The problem is that the components of a PCA are not correlated but only if the values were centred before. Doing this, the average of the centred data and its transformation is zero. So you miss this centring in the equation above and with it, I cannot derive the expressions I need. $\endgroup$
    – sebschub
    Sep 8, 2015 at 14:10
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EDIT: I am not sure the equations below work because for the components of principle component analysis to be uncorrelated, the data has to to be centred first: $$ y_t^i = \sum_j \Gamma_{ij} (x_t^j - \bar{x}^j) $$ which leads to vanishing averages: $$ \bar y^i = 0 \, . $$ /EDIT

The following gives a detailed solution based on Aksakal's answer as a future reference (for me and maybe others).

Let $y_t^i$ be the components of a principle component analysis with $$ y_t^i = \sum_j \Gamma_{ij} x_t^j \, , \qquad x_t^i = \sum_j \Gamma_{ij}^{-1} y_t^j = \sum_j \Gamma_{ji} y_t^j \, . $$ Then the time average $\bar{x}^i$ of the $i$th series is $$ \bar{x}^i = \frac{1}{n_t} \sum_t x_t^i = \frac{1}{n_t} \sum_t \sum_j \Gamma_{ji} y_t^j = \sum_j \Gamma_{ji} \bar{y}^j $$ and with that, the total average $\bar{\bar{x}}$ is given by $$ \bar{\bar{x}} = \frac{1}{n_i} \sum_i \bar{x}^i = \frac{1}{n_i} \sum_i \sum_j \Gamma_{ji} \bar{y}^j \,. $$ Since the $\bar{y}^k$ are not correlated by design, $$ \mathrm{Var}(\bar{\bar{x}})=\sum_k \left(\frac{\partial \bar{\bar{x}}}{\partial \bar{y}^k}\right)^2 \mathrm{Var}(\bar{y}^k) = \frac{1}{n_i^2}\sum_k \left(\sum_i\Gamma_{ki}\right)^2 \mathrm{Var}(\bar{y}^k) \, . $$ The $y_t^i$ are still auto-correlated in terms of $t$, thus, $\mathrm{Var}(\bar{y}^i)$ is calculated similarly to $\mathrm{Var}(\bar{x}^i)$ in the original question.

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  • $\begingroup$ You basically copied my answer adding more details, then picked your own answer as a solution? That's an interesting move. $\endgroup$
    – Aksakal
    Aug 6, 2015 at 12:52
  • $\begingroup$ @Aksakal Yes, because I think that my answer adds important details and involves some math which might not be obvious. For example, I was not sure whether your idea works in my case, the given details would have helped me. A third person only needs to read my answer to get a recipe. If I had to choose exactly one answer to show someone, I would choose mine due to this. And this is AFAIK the criterion for the accepted answer. However, I do cite you (first! sentence), did upvote your answer and did not receive any points for the accepted answer. If you wish, I ask meta about it. $\endgroup$
    – sebschub
    Aug 6, 2015 at 15:29
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If you can do with an upper bound on the variance of the average, you may call upon Azuma's Inequality, which applies to averages of Martingale processes with sub-Gaussian margins.

Try following the details in Phillipe Rigollet's notes, and not the Wikipedia entry which discusses only bounded margins.

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  • $\begingroup$ Thanks! Could you be a bit more specific (e.g. which equation etc.). The lecture notes are currently a bit beyond me... $\endgroup$
    – sebschub
    Sep 9, 2015 at 7:50

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