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I'm guessing this is not an error. If you divide 1/normal(0,1), the full distribution would range from -inf to inf, but the 95% output doesn't have to.

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SamBam 4 days ago | parent [-]

I don't quite understand, probably because my math isn't good enough.

If you're treating -1~1 as a normal distribution, then it's centered on 0. If you're working out the answer using a Monte Carlo simulation, then you're going to be testing out different values from that distribution, right? And aren't you going to be more likely to test values closer to 0? So surely the most likely outputs should be far from 0, right?

When I look at the histogram it creates, it varies by run, but the most common output seems generally closest to zero (and sometimes is exactly zero). Wouldn't that mean that it's most frequently picking values closest to -1 or 1 denoninator?

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blackbear_ 3 days ago | parent | next [-]

If X is normal and centered around 0, then the average of 1/X does not exist (math speak for "is infinity" in this case). In these cases Monte Carlo simulations are not reliable because they give high variance estimates (math speak for "the histogram varies run by run").

The actual distribution of 1/X is fairly interesting, see https://en.m.wikipedia.org/wiki/Inverse_distribution#Recipro...

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pyfon 4 days ago | parent | prev [-]

Only 1 percent of values would end up being 100+ on a uniform distribution.

For normal it is higher but maybe not much more so.

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etbebl 4 days ago | parent | next [-]

OK, but do we necessarily just care about the central 95% range of the output? This calculation has the weird property that values in the tails of the input correspond to values in the middle of the output, and vice versa. If you follow the intuition that the range you specify in the input corresponds to the values you expect to see, the corresponding outputs would really include -inf and inf.

Now I'm realizing that this doesn't actually work, and even in more typical calculations the input values that produce the central 95% of the output are not necessarily drawn from the 95% CIs of the inputs. Which is fine and makes sense, but this example makes it very obvious how arbitrary it is to just drop the lowermost and uppermost 2.5%s rather than choosing any other 95/5 partition of the probability mass.

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lswainemoore 4 days ago | parent | prev [-]

That may be true, but if you look at the distribution it puts out for this, it definitely smells funny. It looks like a very steep normal distribution, centered at 0 (ish). Seems like it should have two peaks? But maybe those are just getting compressed into one because of resolution of buckets?

	▲	blackbear_ 3 days ago \| parent [-]
		It does indeed have two peaks: https://en.m.wikipedia.org/wiki/Inverse_distribution#Recipro... As the mean of this distribution does not exist (it's "infinite") the Monte Carlo estimates aren't reliable