Mathematics Asked on November 26, 2021
Let $theta$ denote a smoothly distributed random variable with support $[0, 1]$. I am trying to evaluate
$$ lim_{n rightarrow infty} frac{mathbb{E}[theta^n]}{mathbb{E}[theta^{n-1}]}$$
I suspect, but cannot show, that the limit equals $1$. Does anyone know how to do this?
My attempts so far: Since $theta in [0, 1]$, it seems reasonably clear that both $mathbb{E}[theta^n] rightarrow 0$ and $mathbb{E}[theta^{n-1}] rightarrow 0$ as $n rightarrow infty$ (we are raising numbers that are less than $1$ to ever higher powers). Thus, we can apply L’Hopital’s rule to find that
$$ lim_{n rightarrow infty} frac{mathbb{E}[theta^n]}{mathbb{E}[theta^{n-1}]} equiv lim_{n rightarrow infty} frac{int_0^1 theta^nf(theta)dtheta}{int_0^1 theta^{n-1}f(theta)dtheta} = lim_{n rightarrow infty} frac{int_0^1 ln(theta)theta^nf(theta)dtheta}{int_0^1 ln(theta)theta^{n-1}f(theta)dtheta}$$
I am a bit unclear, however, how to proceed from this point (or whether better approaches are available).
We have
$$ mathbb{E}[theta^n]^{frac{n+1}{n}} leq mathbb{E}[theta^{n+1}] leq mathbb{E}[theta^n]. $$
Indeed, the first inequality is the consequence of the Jensen's inequality and the second inequality follows from $mathbb{P}(thetain[0,1])=1$. Dividing each side by $mathbb{E}[theta^n]$, we get
$$ mathbb{E}[theta^n]^{1/n} leq frac{mathbb{E}[theta^{n+1}]}{mathbb{E}[theta^n]} leq 1. $$
Now by noting that $mathbb{E}[theta^n]^{1/n} to | theta |_{infty} = 1$ as $ntoinfty$ by the assumption, the desired conclusion follows.
Answered by Sangchul Lee on November 26, 2021
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