I stumbled upon a problem in the context of static bayesian games. It pretty much boils down to the following question:
I have a cumulative distribution function $F(x)$ with support $[\underline{\nu}, \bar{\nu}]$ which is evaluated at a function $\nu^*(n)$ which is increasing in n. I am interested in the probability $1-F(\nu^*(n))^n$ and how it changes with n. Especially which characteristics of F determine the change in this probability, since as far as I can see, it is possible for it to increase or decrease.
I don't think it is of any importance, but from the original problem the value $\nu^*(1) = c \in (\underline{\nu},\bar{\nu})$ is known. Thanks for your answers!
Edit1: $\nu^*(n)$ is constructed such that: $\nu^*(n)*F(\nu^*(n))^{n-1} = \nu^*(1) = c$
Therefore $1-F(\nu^*(n))^n = 1- c\frac{F(\nu^*(n))}{\nu^*(n)}$. So is it about the average slope of the cdf up to the relevant points?
You can say
$F(x)$ is between $0$ and $1$
so $F(x)^n$ is a weakly decreasing function of $n$ and $1-F(x)^n$ is a weakly increasing function of $n$
and
$F(x)$ is a weakly increasing function of $x$
$\nu^*(n)$ is an increasing function of $n$
so $F(\nu^*(n))$ is a weakly increasing function of $n$ and $1-F(\nu^*(n))$ is a weakly decreasing function of $n$
but combining these means that you cannot say whether $F(\nu^*(n))^n$ or $1- F(\nu^*(n))^n$ are increasing or decreasing functions of $n$
A couple of examples with $\nu^*(n)=1-\frac1n$:
If $F(x)=x$ with $0 \le x \le 1$ then $1- F(\nu^*(n))^n= 1 - \left(1-\frac1n\right)^n$ which is an decreasing of $n$ on the positive integers
If $F(x)=1-e^{-x-1}$ with $-1 \le x$ then $1- F(\nu^*(n))^n= 1 - \left(1-e^{-2+\frac1n}\right)^n$ which is an increasing function of $n$ on the positive integers