# Why aggregation reduces the noise [closed]

I am looking for a proof of why aggregating data (summation) to a higher level (e.g., aggregating daily values of a variable to weekly or monthly) reduces the noise. Does anyone know how to prove this? Basically, why is more fine-grained data more wiggly and volatile than the aggregated, coarser data?

• "aggregation" does not necessarily imply taking an average. You could, for instance, inspect the minimum or maximum which can be more variable than a single observation from a cluster of IID RVs. Commented Jun 4 at 18:51
• Can you provide a more mathematical description of what you're asking? I'm not clear on what is being asked. Commented Jun 5 at 22:14
• Your sense of "wiggly" is scale-invariant, so the sum and the arithmetic mean always give identical results from this perspective. // It's also worth knowing that summing or averaging do not always reduce the "wiggliness:" when the noise has a sufficiently heavy-tailed distribution, summing makes the situation worse, not better.
– whuber
Commented Jun 8 at 12:25
• I'm sorry, you are simply incorrect. Perhaps you have forgotten that the signal-to-noise ratio is measured in terms of standard deviations? // The exceptional situations are characterized by an infinite variance and/or lack of independence of the observations that comprise the "signal."
– whuber
Commented Jun 8 at 13:03
• Re "confronted ... numerous times before:" Because this is your first post under your present account, that statement cannot possibly be true. Perhaps you have multiple accounts? If that is unintentional, please visit stats.stackexchange.com/help/merging-accounts at the first opportunity to merge them so that you aren't taken to be a troll trying to evade a system ban.
– whuber
Commented Jun 9 at 14:28

The reason is that "aggregating" is basically averaging, and averaging will reduce your noise by $$\frac 1{\sqrt n}$$ ($$n$$ being the number of samples averaged). This is a general property of averaging, used in signal processing, image processing and many other fields, not just for statistics. This is why in statistics, the standard error of the mean is $$\frac {\sigma}{\sqrt n}$$. You can find a good wiki page here; for your case, see the section called "Noise power for sampled signals". Or check here on CV , or here on wiki.

Note that, as there is no free lunch, you gain this noise reduction at the cost of reducing your time resolution: your weekly averages are less noisy, but you no longer "see" the daily variability.

Now, what happens when we sum the data (w/o dividing by $$n$$)? Exactly the same thing. The signal (the sum) grows like $$n$$, while the noise (sd) grows like $$\sqrt n$$. Hence your SNR grows like $$\sqrt n$$. And the noise is "reduced" (not per se, it also grows, but relative to the signal, it grows more slowly, by the same $$\sqrt n$$ factor). And the CV, which is the inverse of the SNR, shrinks by $$\frac {1} {\sqrt n}$$. So to say that summing decreases the noise is not "technically" correct, but given that the signal increases even faster than the noise, for all intents and purposes it is practically correct. You can find a good discussion of this here on CV.

• Aggregation is not averaging, it is summing. I understand how averaging reduces variance but my question was about summation.
– Amin
Commented Jun 5 at 21:45
• @AminShn An average is just a sum divided by N, so this argument still applies. Commented Jun 5 at 22:06
• @AminShn, if you were inquiring about sum, you should have said "sum". Aggregation icould be either, and most commonly means averaging. Having said this, I will edit the answer, and deal with specifically the sum. Commented Jun 5 at 22:19

Let $$y$$ denote the variable on which to calculate the total and $$y_j$$ for $$j = 1,\ldots, n$$, denote an individual observation on $$y$$. If no weights are specified, then $$w_j = 1$$ for all $$j$$. The sum of the weights is an estimate of the population size: $$\hat N = \sum_{j=1}^{n}w_j.$$ If the population values of $$y$$ are denoted by $$Y_j$$ for $$j = 1,\ldots,N$$, the associated population total is $$Y = \sum_{j=1}^{N}Y_j = N \bar y,$$ where $$\bar y$$ is the population mean. The total is estimated by $$\hat Y = \hat N \bar y$$ The variance estimator for the total is $$\hat V(\hat Y) = {\hat N}^2 \hat V(\bar y),$$ where $$\hat V(\bar y)$$ is the variance estimator of the mean. This means that the variance does grow with summation.

However, the coefficient of variation remains stable. Since $$V(\bar y) = \frac{\sigma^2}{N}$$ and $$CV = \frac{\sigma}{\mu}$$, then

$$CV_y = \frac{\sigma}{\mu}$$

$$CV_{\bar y} = \frac{ \frac{\sigma}{\sqrt{n}}}{\mu}$$

$$CV_{\hat Y} = \frac{n \cdot \frac{\sigma}{\sqrt{n}}}{n \cdot \mu} = \frac{ \frac{\sigma}{\sqrt{n}}}{\mu} = CV_{\bar y}$$

This confirms your intuition that aggregation makes data less wiggly, though only in relative and not absolute terms.

This derivation was for $$iid$$ variables. If there is some correlation $$\rho$$ between them, then

$$V\left(\bar y\right)=\rho\sigma^2+\frac{1-\rho}{n}\sigma^2$$

Everything above still goes through if we relax independence.

• please read the approved solution, I think you misunderstood the question.
– Amin
Commented Jun 8 at 8:01
• I'm not sure what I misunderstood since I find nothing to disagree with in the answer you selected, @Amin. In any case, I added some detail to the derivation that others may appreciate. Commented Jun 8 at 22:27
• Not to add to the confusion but Yule ( 1927 ) showed that summing data that is correlated can actually induce cycles. It could be argued that cycles are more noisy than the original data. But, this is only in a time-series context. In the current scenario, I think the OP is talking about an independent sample in a non-time series context, in which case, Yule's results don't apply. Commented Jun 9 at 0:43
• In case, anyone is interested. It's a pretty detailed and difficult read. royalsocietypublishing.org/doi/epdf/10.1098/rsta.1927.0007 Commented Jun 9 at 0:44
• Maybe you tried to be very concise, but I found your approach and terminology more confusing. You could have simply used the formula for the variance of sum of random variables which grows if the covariance terms are non-negative (usually true for time series) instead of your first 4 equations. I still don't know where that formula for variance estimator of yours comes from (would appreciate a reference). Then explain why noise grows slower which would explain why the summed time series look less noisy. I didn't find this very imporant part in your first answer.
– Amin
Commented Jun 9 at 8:07