One way to think about my last post, about electoral coalitions and how narrow or broad its support base is to think about coalitions is to think of them in terms of their variances.
Let’s imagine two normal distributions. Both have means of 10 but one has variance of 1 and the other has variance of 10. For a given number x, you “catch” everything in the interval (x-1, x+1) under the curve.
For the high variance distribution, getting an x closer to the mean does not help you much. For x = 5, you get 7.4%, for x = 9, 23.6%, and for getting x exactly at the mean, 24.8%. Getting the “right answer” helps, but not dramatically so.
For the low variance distribution, getting the “righter” answer yields much larger dividends. For x = 5, you get essentially nothing. For x=9, you get 47.7%–nearly twice as under the high variance distribution, and for getting the mean exactly right, you get more than 68.2%–almost three times as much as with high variance distribution.
One might consider this the distributional consequences of policy chosen by a party, abstracted (that is, not strictly limited to economic benefits only). In other words, a high variance distribution provides some benefits to many, while a low variance distribution provides highly focused benefits to few. The support garnered by a party, in other words, would consist of the those to whom the benefits exceed the “needs”.
I’ve stolen the following diagram from wikipedia, but it illustrates the point: assuming that the distribution of “needs” is distributed uniformly over the interval (-π,π) and the probability density of party “policy” pdf is f(x), the party captures the votes where f(x)> 1/2π.
My math has gotten shaky enough that I can’t give a closed form answer to this problem off the bat, but the intuition is sufficiently captured by the diagram: the interval (-a,a) where f(x) > 1/2π for all x ∈ (-a,a) expands as the distribution of party policy becomes fatter, i.e. its variance increases. In a sense, this is not a surprise: it simply another way of saying that you can spread the loot more broadly if you give less to each recipient, on average. If the goal of politics is to build a broad electoral coalition to reliably win elections, it makes little sense to try to hoard the loot for the few.
Of course, the same high variance distribution makes it less profitable to identify the exact mean of the distribution of the party policy, or indeed, to identify where the exact mean is. Since a widely disparate groups of people are getting something, with little regard for their exact “ideologies,” it does not matter much either way. This is how a stable and “dominant” party, one that can win elections reliably, is born. But those in the middle of the distribution, so to speak (or who wish to move the middle near them, more accurately, in the spirit of the previous post) do not care for attracting voters if the electoral gains are achieved by losing their share of the loot. Reduced variance generates larger subsets of voters whose needs are unmet, or where f(x) < 1/2π. It is not difficult to conceptualize a situation where the party may move the distribution towards the “middle,” but with increasingly low variance, such that the share of the underserved voters, on BOTH ends of the distribution, increases.
In this sense, one might conceptualize the transition offered by Trump to the Republicans as creation of a high variance distribution–regardless of where his mean is. In so doing, he is demanding that the regular Republicans accept a far smaller share of the loot than they were with the conventional Republican distribution scheme. Clinton, in turn, is offering a smaller variance distribution that is steadily moving towards the regular Republicans–but perhaps still too far.
PS. Obviously, this is still a thoughtstream in progress. Suggestions are welcome.