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I would like to know whether a weighted average can be defined as the product of the different values, to the power of their weights?

Basically, I have this formula and I have to describe it in one or two sentances:

$$E_{i,t}=\prod_{j=1}^{n}(S_{i,j,t}P_{i,t}/P_{jt})^{w_{i,j}}$$

E is the real effective exchange rate of country i at time t

S is the bilateral exchange rate between country i and country j

The two P's represent the price level in the relevant countries

Lastly, Wij is the weight of country j in the overall trade activities in country i.

I believe this can be described as: the Real effective exchange rate is the trade weighted average of bilateral real exchange rates between the home country and a basket of other countries. But, I am uncertain, since until now, I have seen weighted averages represented differently.

So does this formula correspond to a weighted average?

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  • $\begingroup$ There are many weighted averages. Which one do you have in mind? $\endgroup$
    – Ivan Neretin
    Commented Sep 21, 2015 at 19:15
  • $\begingroup$ Give a clear numerical example. $\endgroup$
    – NoChance
    Commented Sep 21, 2015 at 19:16
  • $\begingroup$ I did, thanks for the comment $\endgroup$
    – Docksan
    Commented Sep 22, 2015 at 6:46

1 Answer 1

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Yes, it is similar to the Weighted geometric mean.

If $\{x_1, x_2, ... x_n\}$ are the values, and $\{w_1, w_2, ... w_n\}$ are the weights, the weighted geometric mean is defined as

$$(\prod_{i=1}^{n}x_i^{w_i})^{1/\sum_{i=1}^{n}w_i}$$

Note that this version accounts for the sum of the $w_i$'s not being 1.

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  • $\begingroup$ To view the code which is required to display these formulas, just click the edit button below my post. $\endgroup$
    – Glorfindel
    Commented Sep 21, 2015 at 19:21
  • $\begingroup$ Hello, Glorfindel! Thank you for your comment! I have used your code in order to provide a formula in my initial question. I hope this makes it clear. It seems similar to what you are showing, but there are some differences, so I am not sure if it is the same $\endgroup$
    – Docksan
    Commented Sep 22, 2015 at 6:48
  • $\begingroup$ It looks similar, but my ... pardon ... Wikipedia's formula accounts for the fact that the $w_i$'s might not add up to 1. $\endgroup$
    – Glorfindel
    Commented Sep 22, 2015 at 6:54
  • $\begingroup$ So it is a weighted average, and if I discribe it in this way, it would not be a mistake? $\endgroup$
    – Docksan
    Commented Sep 22, 2015 at 6:57
  • $\begingroup$ It depends. If you want to compare your averages with each other, you need to be sure that they have the same total weight. $\endgroup$
    – Glorfindel
    Commented Sep 22, 2015 at 6:59

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