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On R, I have created a mixture distribution via a convex combination of a standard normal distribution, and a normal distribution with mean -3, variance 1 (i.e. subtracting 3 from a standard normal distribution).

In doing so, I have observed that the difference between absolute values of the upper quantiles (e.g. 95%) of the mixture distribution and a standard normal distribution are smaller than the same difference at lower quantiles (e.g. 5%).

I have an application for this result in economics, but other than looking at the plots of the distributions and using intuition, I have no way of explaning why this is the case. Is anyone able to offer an explanation as to why I observe this result mathematically?

Also, is this a result that holds generally when you form a mixture distribution via a convex combination of a standard normal distribution, and a normal distribution with lower mean?

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If you if you have a location-mixture (mixture of shifted versions of the same distribution) with a light tail, and the proportion of the rightmost component is not too small, then the upper tail of the mixture will mostly come from the upper tail of that component.

Consider the simple case of two normal densities with equal weight, that are well-separated (like your $\mu_1=-3,\mu_2=0,\sigma=1$). Then the 95th percentile of the mixture will be very close to the 90th percentile of the upper component (i.e. close to 1.28155).

cdf of 50-50 normal mixture where $\mu_2-\mu_1=3\sigma$

Similarly if it was say a 75%-25% mixture of a (-3,1) and (0,1) then the 95th percentile of the mixture would be close to a 1-.05/0.25 = 0.8 percentile of the upper component.

In the lower tail it will be mostly driven by the leftmost component, not the rightmost.

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  • $\begingroup$ This is good; I now understand the mechanics of what is going on. Do you think it is possible to generalise the result in some way? I am thinking that this will hold as long as the original distribution is symmetric and has thin tails. Is there some technical term that can be applied to such distributions? $\endgroup$
    – Alexander Whyte
    Commented Sep 14, 2017 at 21:00
  • $\begingroup$ Actually, even asymmetric would do if both tails weren't heavy. How thin the tails need to be really depends but I guess that asymptotically you'd be looking for (loosely) something like $\lim_{x\to\infty} P(X>x+t)/P(X>x) \to 0$ (but I've left that too vague for a formal definition). From the Wikipedia article on Heavy tailed distributions we'd be looking for something not long-tailed, because then as you went far out the proportion would go to 1 instead of 0. ... ctd $\endgroup$
    – Glen_b
    Commented Sep 14, 2017 at 22:16
  • $\begingroup$ ctd... Perhaps bounded by some small positive value would do, but for it to be practically useful (and again I'm only considering the right tail but similar things flipped around would apply on the left) you'd probably want that ratio going to zero pretty quickly. I wonder if something like $\lim_{x\to\infty} P(X>x+t)/P(X>x)\leq ke^{-ct}$ for some problem-dependent $k$ and $c$ would be enough for a practical definition? I guess I'd call that "asymptotically exponential tails"; I'm not sure that would be quite be light enough to be really practically useful, though it's perhaps in the area $\endgroup$
    – Glen_b
    Commented Sep 14, 2017 at 22:36

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