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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

$$ \lim_{n \rightarrow \infty} \frac{\mathbb{E}[\theta^n]}{\mathbb{E}[\theta^{n-1}]} \equiv \lim_{n \rightarrow \infty} \frac{\int_0^1 \theta^nf(\theta)d\theta}{\int_0^1 \theta^{n-1}f(\theta)d\theta} = \lim_{n \rightarrow \infty} \frac{\int_0^1 \ln(\theta)\theta^nf(\theta)d\theta}{\int_0^1 \ln(\theta)\theta^{n-1}f(\theta)d\theta}$$

I am a bit unclear, however, how to proceed from this point (or whether better approaches are available).

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    $\begingroup$ The limit of $E[\theta^n]$ is not $0$ but $P(\theta=1)$. $\endgroup$
    – ECL
    Jul 8, 2020 at 12:18
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    $\begingroup$ yes but $P(\theta = 1) = 0$ right? (I am assuming that the random variable has a smooth distribution.) $\endgroup$
    – afreelunch
    Jul 8, 2020 at 12:19
  • $\begingroup$ @ECL I guess in the discrete case, your point reveals the limit is $P(\theta = 1)/P(\theta = 1) = 1$, which further suggests that the limit should be $1$ in my setting. $\endgroup$
    – afreelunch
    Jul 8, 2020 at 12:30
  • $\begingroup$ Oh yep sorry I had missed the "smoothly distributed" $\endgroup$
    – ECL
    Jul 8, 2020 at 13:02
  • $\begingroup$ Try these for uniform ,and for beta random variables? (Look up their moments from Wikipedia and ocw.mit.edu/courses/mathematics/… respectively). $\endgroup$ Jul 8, 2020 at 13:52

1 Answer 1

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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 $\mathbb{P}(\theta\in[0,1])=1$. 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 $n\to\infty$ by the assumption, the desired conclusion follows.

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  • $\begingroup$ Thanks for this answer! However, I have a quick question about the first inequality. By Jensen's inequality, $\mathbb{E}[f(\theta)] \leq f(\mathbb{E}[\theta])$ for any convex function $f$. Since $f(\theta) = \theta^n$ is convex on $[0, 1]$, I see that $\mathbb{E}[\theta^n] \leq \mathbb{E}[\theta]^n$. But doesn't this imply $\mathbb{E}[\theta^n]^{\frac{n+1}{n}} \leq \mathbb{E}[\theta]^{n+1}$, not $\mathbb{E}[\theta^n]^{\frac{n+1}{n}} \leq \mathbb{E}[\theta^{n+1}]$? Apologies if I am missing something obvious here! $\endgroup$
    – afreelunch
    Jul 23, 2020 at 16:35
  • $\begingroup$ @afreelunch, No worries. As for your question, Jensen's inequality tells that if $f$ is a convex function then $f(\mathbb{E}[X])\leq\mathbb{E}[f(X)]$. In this case, I applied this to $X=\theta^n$ and $f(x)=x^{(n+1)/n}$. $\endgroup$ Jul 23, 2020 at 16:42
  • $\begingroup$ Thanks for clarifying, that makes sense! $\endgroup$
    – afreelunch
    Jul 23, 2020 at 16:48
  • $\begingroup$ Apologies, I have one more question: why does $\mathbb{E}[\theta^n]^{1/n} \to \| \theta \|_{\infty} = 1$? (I guess this has something to do with $L_p$ norms?) $\endgroup$
    – afreelunch
    Jul 23, 2020 at 17:28
  • $\begingroup$ @afreelunch, The proof is quite standard, see this for instance. For any $0<r<1$, $\mathbb{P}(\theta>r)>0$, and so, $$r^n\mathbb{P}(\theta>r)\leq\mathbb{E}[\theta^n]\leq1.$$ Now by raising to the $1/n$-th power and taking $n\to\infty$, we have $$r\leq\liminf_{n\to\infty}\mathbb{E}[\theta^n]^{1/n}\leq\limsup_{n\to\infty}\mathbb{E}[\theta^n]^{1/n}\leq1.$$ So by letting $r\uparrow1$, the claim follows. In general, if $\mu$ is a finite measure, then $$\lim_{n\to\infty}\left(\int|f|^n\,\mathrm{d}\mu\right)^{1/n}=\|f\|_{\infty}=\operatorname{esssup}|f|.$$ $\endgroup$ Jul 23, 2020 at 17:38

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