As already noticed by whuber in a comment, it is not clear if your procedure of setting weights is valid. Notice that in non-weighted RMSE larger areas already have greater weight on the estimate since they are larger, so they appear more often in your data. That is why, as suggested, people rather down-weight such subpopulations, so that the final estimate treats all the subpopulations more evenly.

However if you wanted to use weighted RMSE, then recall that RMSE is by design pretty close to standard deviation, so why not look at how weighted variance is calculated?

$$ \sigma^2 = \sum_{i=1}^n w_i (x_i - \bar x)^2 $$

where weights are non-negative and $\sum_{i=1}^n w_i = 1$. The same you can take weighted RMSE as

$$ \text{RMSE} = \sqrt{\sum_{i=1}^n w_i (\hat x_i - x_i)^2} $$

Notice that we take sum of weighted differences, not the mean. Unweighted mean is the same as weighted mean with weights that are all equal to $w_i = 1/n$, so if you took arithmetic mean, it would be like dividing RMSE by $n$ second time.

Check also:
Weighted Variance, one more time

As already noticed by whuber in a comment, it is not clear if your procedure of setting weights is valid. Notice that in non-weighted RMSE larger areas already have greater weight on the estimate since they are larger, so they appear more often in your data. That is why, as suggested, people rather down-weight such subpopulations, so that the final estimate treats all the subpopulations more evenly.

However if you wanted to use weighted RMSE, then recall that RMSE is by design pretty close to standard deviation, so why not look at how weighted variance is calculated?

$$ \sigma^2 = \sum_{i=1}^n w_i (x_i - \bar x)^2 $$

where weights are non-negative and $\sum_{i=1}^n w_i = 1$. The same you can take weighted RMSE as

$$ \text{RMSE} = \sqrt{\sum_{i=1}^n w_i (\hat x_i - x_i)^2} $$

Notice that we take sum of weighted differences, not the mean. Unweighted mean is the same as weighted mean with weights that are all equal to $w_i = 1/n$, so if you took arithmetic mean, it would be like dividing RMSE by $n$ second time.

Check also:
Weighted Variance, one more time

As already noticed by whuber in a comment, it is not clear if your procedure of setting weights is valid. Notice that in non-weighted RMSE larger areas already have greater weight on the estimate since they are larger, so they appear more often in your data. That is why, as suggested, people rather down-weight such subpopulations, so that the final estimate treats all the subpopulations more evenly.

However if you wanted to use weighted RMSE, then recall that RMSE is by design pretty close to standard deviation, so why not look at how weighted variance is calculated?

$$ \sigma^2 = \sum_{i=1}^n w_i (x_i - \bar x)^2 $$

where weights are non-negative and $\sum_{i=1}^n w_i = 1$. The same you can take weighted RMSE as

$$ \text{RMSE} = \sqrt{\sum_{i=1}^n w_i (\hat x_i - x_i)^2} $$

Notice that we take sum of weighted differences, not the mean. Unweighted mean is the same as weighted mean with weights that are all equal to $w_i = 1/n$, so if you took arithmetic mean of weighted differences, then in fact you would weight each difference by $w_i/n$ rather then by $w_i$.

Check also:
Weighted Variance, one more time

As already noticed by whuber in a comment, it is not clear if your procedure of setting weights is valid. Notice that in non-weighted RMSE larger areas already have greater weight on the estimate since they are larger, so they appear more often in your data. That is why, as suggested, people rather down-weight such subpopulations, so that the final estimate treats all the subpopulations more evenly.

However if you wanted to use weighted RMSE, then recall that RMSE is by design pretty close to standard deviation, so why not look at how weighted variance is calculated?

$$ \sigma^2 = \sum_{i=1}^n w_i (x_i - \bar x)^2 $$

where weights are non-negative and $\sum_{i=1}^n w_i = 1$. The same you can take weighted RMSE as

$$ \text{RMSE} = \sqrt{\sum_{i=1}^n w_i (\hat x_i - x_i)^2} $$

Notice that we take sum of weighted differences, not the mean. Unweighted mean is the same as weighted mean with weights that are all equal to $w_i = 1/n$, so if you took arithmetic mean, it would be like dividing RMSE by $n$ second time.

Check also:
Weighted Variance, one more time