# Finding the limit of $\mathbb{E}[\theta^n]/\mathbb{E}[\theta^{n-1}]$

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).

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

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.

• 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! Jul 23, 2020 at 16:35
• @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 $f(x)=x^{(n+1)/n}$. Jul 23, 2020 at 16:42
• Thanks for clarifying, that makes sense! Jul 23, 2020 at 16:48
• 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?) Jul 23, 2020 at 17:28
• @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 $$\lim_{n\to\infty}\left(\int|f|^n\,\mathrm{d}\mu\right)^{1/n}=\|f\|_{\infty}=\operatorname{esssup}|f|.$$ Jul 23, 2020 at 17:38