Calculating limit of $E(X^{N+1}) / E(X^N)$ when N goes to infinity if $0 \leq X \leq 1$ and $P(X>1-\epsilon)>0$ for every $\epsilon>0$

Question

I have a question about a simple limit calculation problem. I want to calculate limit of $E(X^{N+1}) / E(X^N)$ when $N \to \infty$, $0 \leq X \leq 1$ and $\sup(X) = 1$

My naive thinking is as follows:

1. Expected Values: The expected value $ E[X^N] $ is defined as the mean value of $ X^N $ over its probability distribution. Since $ X $ is a random variable bounded between 0 and 1, $ X^N $ and $ X^{N+1} $ will also be bounded within this range.

2. Behavior for Large $ N $: As $ N $ becomes very large, $ X^N $ and $ X^{N+1} $ will approach 0 for all $ X $ in the interval $ (0, 1) $. However, at $ X = 1 $, both $ X^N $ and $ X^{N+1} $ will always be 1, regardless of $ N $.

3. Limit Calculation: The limit of the ratio of these expected values as $ N \to \infty $ will be determined largely by the behavior of the distribution of $ X $ near the upper bound of 1. Since the supremum of $ X $ is 1, there's a significant contribution to the expected values from values of $ X $ close to 1.

If we assume that the probability density function (PDF) of $ X $ does not degenerate or become a delta function at $ X = 1 $ (i.e., the distribution is not entirely concentrated at 1), then both $ E[X^{N+1}] $ and $ E[X^N] $ will tend to 0 as $ N \to \infty $, but they will do so at similar rates for large $ N $. This is because for any $ X $ in $ (0, 1) $, $ X^{N+1} $ will be smaller than $ X^N $, but the difference becomes less significant as $ X $ approaches 1.

Therefore, under the assumption of a non-degenerate distribution of $ X $ that does not overly concentrate near $ X = 1 $, the limit of the ratio of these expected values would be expected to approach 1.

But, can we get any rigorous proof? I also think that if pdf is weird, it might be not possible to obtain the limit, so you can assume some good condition of pdf.

Thank you.

Answer 1

There is a nice problem in Rudin, W., Real and Complex Analysis, 3rd edition, McGraw Hill, problem 23, pp. 74, that states that

If $(\Omega,\mathscr{F},\mu)$ is a finite probability space (a probability space for example) and $X\in L_\infty(\mu)$, then $$\lim_{p\rightarrow\infty}\frac{\|X\|^{p+1}_{p+1}}{\|X\|^p_p}=\|X\|_\infty$$ where $\|X\|_p=\Big(\int_\Omega |X|^p\,d\mu\Big)^{1/p}$, $1\leq p<\infty$, and $\|X\|_\infty=\inf\{t>0:\mu(|X|>t)=0\}$.

A solution to this exercise appears here at MSE.

In your case, $\|X\|_\infty=1$ since by assumption $0\leq X\leq 1$ and $P[X>1-\varepsilon]>0$ for all $\varepsilon$. Recall that $\|X\|_\infty=\inf\{t>0:P[X>t]=0\}$.

Answer 2

My wonderful labmate provided the answer (for sure he is a genius). For any $\epsilon > 0$, we have

$$\lim_{N \to \infty}\frac{E(X^{N+1})}{E(X^N)} = \lim_{N \to \infty} \frac{E(X^{N+1} \pmb{1}(X \geq 1- \epsilon))}{E(X^{N} \pmb{1}(X \geq 1- \epsilon))} \geq 1-\epsilon. $$