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Say I have a time series, with actual measurements of a variable $a$ in different locations. If I want to know the average of $p$ from time $t_1$ to $t_n$, I could say that $\overline{a_{p}}=\frac{\sum\limits_{t=1}^n a_{{p}_t}}{n}$, where $p$ is the coordinate of the point of measurement. So far, so good. Let's generalize and call this averaging function $m_p=f(a_p,t)$, where $f(a,t)$ is some deterministic function and $m$ is its result.

Now, say I want to calculate $m$ for a location $o$, where $o$ is a points for which I don't have actual measurements and $a_o$ was estimated through a an interpolating function $g(a,o)$. My intuition tells me that in this case, we should first estimate every $a_{o_t}$ before applying $f(a,t)$. In other words, considering that $a_o = g(a,o)$, we get $m_o$ by doing:


But I have seen so many papers that do the other way around. They first calculate $f(a,t)$ for every known $a$ and then interpolate the results using the same g() even though there is no guarantee that $m$ behaves like $a$, or:


In some cases, they use the same logic when instead of $t$, there is some other variable $b$, i.e., it's not a time series, but $f(a,b)$ is a deterministic function.

Are the models in those papers conceptually wrong? If they are wrong, is there a techcnical term to call this kind of mistake?

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If your interpolation scheme is linear in the input $a$'s (and all interpolation schemes I can think of are), then there is no difference between the mean of interpolated values and the interpolation of the means.

Your notation is a little confusing, so let me put it this way. Suppose you know the values $a_{i,j} = a(p_i,t_j)$ for $m$ points $p_i$ at $n$ times $t_j$. The interpolated value at a point $o$ at any time $t_j$ can typically be expressed as a weighted sum of the values at known points, $$a(o,t_j) := w_1 a_{1,j} + w_2 a_{2,j} + \cdots + w_m a_{m,j}.$$ The mean of these interpolated values is $$\bar a(o) = \frac{1}{n}\sum_j a(o,t_j).$$ If you expand this out, you'll find that it's equal to $$\begin{align}\bar a(o) &= w_1 \frac{1}{n}\sum_j a_{1,j} + w_2 \frac{1}{n}\sum_j a_{2,j} + \cdots + w_m \frac{1}{n}\sum_j a_{m,j} \\ &= w_1 \bar a_1 + w_2 \bar a_2 + \cdots + w_m \bar a_m,\end{align}$$ which amounts to interpolating the means. Presumably this is better from a computational perspective because if you want to find $\bar a(o)$ at many different $o$, you can keep the $\bar a_i$'s precomputed.

Your strategy of taking the mean of the interpolated values would be necessary if the sample points were moving. In that case, there's no sensible way to interpolate the means, because the interpolation weights would have been changing.

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Rahul, I'm sorry for my confusing notation. I'm not very good at mathematics. You mean that if I am right in my suspicion if the data are to be interpolated using kriging? What if instead of just a simple average $m$ is something not linearly related to $a$? ($f(a,t)$ could be a hydrological model, for example). – Adriano Dec 27 '10 at 2:57
@Adriano: Sorry, I can't answer your first question because I'm not familiar with kriging at all (Wikipedia mentions that it is actually linear, but I can't comment on that). For your second question, I don't know about hydrological models, but in general, if you use a nonlinear averaging such as the median, then yes, the two approaches will give different answers. Deciding which of them would be the "right" way to do it, however, probably requires an involved statistical argument that I'm not qualified to make, so I'll stop here. Sorry if my answer turned out to be not that helpful for you. – Rahul Dec 27 '10 at 3:17
@Adriano: You could try asking your question at, which is more focused on statistics and data analysis. – Rahul Dec 27 '10 at 3:20
Rahul, your answer was very helpful to me. At least, now I know that there are situations in which averaging the interpolated values gives me the same result as interpolating the averages, as you kindly proved. I didn't know that there was Thank you for the tip and sorry for my illeteracy in mathematics. – Adriano Dec 27 '10 at 4:03

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