Comment by marcodiego
5 days ago
I put "1 / (-1~1)" and expected something around - to + infinty. It instead gave me -35~35.
I really don't known how good it is.
5 days ago
I put "1 / (-1~1)" and expected something around - to + infinty. It instead gave me -35~35.
I really don't known how good it is.
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.
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?
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...
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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