0

Consider the following function defined on the probability distribution on [0,1] (whose cdf is denoted by $F$):

$$V(F)=\int_0^1 \frac{\int_0^xF(t)dt}{F(x)}dF(x).$$

Is it true that the function is lower-semi continuous?

It is certainly true that such a function is not upper-semi continuous, by consider that a probability distribution which puts mass points on only 0 and 1, and the nearby continuous distributions.

Higher levels
  • 56
  • It is functional. Meaning that it assign a number to each distribution function. I don't know how it is defined lower-semi continuity here. (Are there exact definitions?) – kolobokish Apr 22 '20 at 18:35
  • with respect to the weak topology on the probability space (or Levy metric) – Higher levels Apr 22 '20 at 18:52
  • I think for defining lower semi- continuity one needs some sort of order on argument space. The argument space here is space of functions. Some stochastic order you mean? – kolobokish Apr 23 '20 at 06:30
  • As long as we can define a metric (or just a topology) on the space, we can define the notation of continuity. In the case of space of probability distribution functions, its weak topology can be generalized by Levy metric, which is the natural metric we can use. – Higher levels Apr 28 '20 at 17:05

0 Answers0