Karol, although I do not have a solution for you, I do have a different way for you (and whomever requested this calculation) to think about the problem.

First, here is your original example, broken out just a bit more to provide additional clarity.

And, for the sake of discussion, let's add a fifth observation, with '5' for metric 1, and '2' for metric 2.

The method of achieving the result of 16 in the highlighted cell in your original example, and the 29 in the additional example, is actually a progressive weighted sum of the differences between the raw numbers in metric1 and metric2.

Let's take a look at the first example, with four observations, and calculate the results as a progressive weighted sum.

When viewed this way, we see that the earlier observations are weighted more heavily than later observations.  In the initial 4-observation example, the last observation of 8 (i.e. '9' for metric 1 minus '1' for metric 2) represents only 10% of the total weightings (1 out of 1+2+3+4), while the first observation represents 40% of the total weightings.

Now let's check the math with 5 observations.  Using your original calculation method, we saw that the final value '29' would be the result if we add an additional (5th) observation of '5' for metric1 and '2' for metric2.  Let's use the alternate progressive weighted sum calculation method and see if we get that same result.

Look at the weighting multiples (red numbers).  Those multiples come straight from the algebra required to solve the problem.  (I'm sure my 10th grade Algebra II teacher, Mr. Alspaugh, would be proud.)  If these 5 observations represented days, with observation #5 occurring today and observation #1 occurring 4 days ago, then the 4-day-old observation #1 would be weighted 5x as heavily as the observation taken today.

And now let's go one step further, and imagine that there are not just four or five observations of metric1 and metric2, but there are instead 30,000 observations, representing 30,000 chronologically arranged time periods, of any grain, anything from 30,000 microseconds up to 30,000 centuries.  The end product, that final number out at the end (the '16' in your four-observation example), will be a sum, not of four or five weighted numbers, but of 30,000 weighted numbers.  By extending the logic of the two progressive weighted sum examples shown above, we know that the first observation on that chronological continuum must receive a weighting multiple of 30,000, while the last observation on that continuum will receive a weighting multiple of 1.

I must admit that I'm intrigued by a model that, if extended to a rather silly but mathematically sound conclusion, could say this:  We have 30,000 observations.  Those observations represent one value from each day of the past 82+ years, starting on May 20, 1938 .  And, based on the unalterable reality of the mathematics, the date of May 20, 1938 contributes 30,000 times as much to the final product as does today, July 8, 2020.

Is that a silly example?  Of course it is.  But that's the mathematical reality of the algebra underlying the requirement.

As I mentioned, I have no OBIEE RPD solution for the problem.  I'll leave it to you to figure out how to generate the weighting multiples, and the rest will be quite simple, because at that point it will only require a single RollingPeriod function against the weighted component observations, i.e. rollingperiod((metric 1 minus metric 2) * weighting factor).

Before starting down that path: Are we sure that whoever made this request understands the implications of the math behind it?  Because frankly, I'm unable to imagine a scenario in which earlier observations should be weighted more heavily than later observations, and indeed that a daily observation taken on May 20, 1938 should contribute 30,000x as much to the final product as a daily observation taken on July 8, 2020.

But that's what the algebra says.