In Weak Convergence and Stochastic Processes, the authors introduce the following notation: $$\|\xi\|_{2,1} = \int_0^\infty \sqrt{P(\xi > x)}\,\mathrm dx$$ They then admit that this is technically not a norm but is equivalent to one. To substantiate this, exercise 1 in the chapter is the following: Show that for any $r > 2$ and random variable $\xi$, $$\frac12\|\xi\|_2 \leq \|\xi\|_{2,1} \leq \frac{r}{r-2}\|\xi\|_r$$ The first inequality can be shown as follows: $$\begin{align*}\|\xi\|_2^2 = \int P(\xi > x^2)\,\mathrm dx &= \int P(\xi > x)2x\,\mathrm dx \\ &\leq \int \sqrt{P(\xi > x)}\,\mathrm dx\cdot \sup_x 2x\sqrt{P(\xi>x)} \\ &\leq 2\|\xi\|_{2,1}\|\xi\|_2 \end{align*} $$ where the first inequality is Hölder and the second follows by Markov. However, I am having considerably more trouble showing the second part. My intuition has been to use Hölder (or perhaps reverse Hölder) again, but all of my efforts to do so lead to dead ends. For example, I have considered $$\|\xi\|_{2,1} = \frac{1}{(1-r)^{1/r}}\int P(\xi > x)^{1/r}(r-1)^{1/r}x\cdot \left(P(\xi>x)^{1/2-1/r}\right)\frac{1}{x} \,\mathrm dx$$ which can then be bounded by $\|\xi\|_r$ times a nasty integral by the use of Hölder's inequality. I have also considered trying something like a reverse Hölder inequality on an explicit expression for $\|\xi\|_r^r$, also with little progress. Any thoughts on where to proceed? Is Hölder even the right approach here?

  • $\begingroup$ Are you considering non-negative random variables $\xi$? Otherwise I would expect that $\xi$ needs to replaced by $|\xi|$ in many of your equations $\endgroup$ – saz Jan 20 at 8:02
  • $\begingroup$ Oh yeah sorry, I had been using the convention of "WLOG, we can consider $\xi$ positively supported" to conserve on notation, but I forgot to say this in my question. $\endgroup$ – stats_model Jan 20 at 20:06

Let's first consider the particular case that $\|\xi\|_r=1$, i.e. $E(|\xi|^r)=1$. By Markov's inequality, we then have

$$P(|\xi|>x) \leq \frac{1}{x^r} E(|\xi|^r) = \frac{1}{x^r}.$$

On the other hand, $P(|\xi| >x) \leq 1$ for all $x$. Combining both estimates we find that

\begin{align*} \|\xi\|_{2,1} &= \int_0^1 \sqrt{P(|\xi|>x)} \, dx + \int_1^{\infty} \sqrt{P(|\xi|>x)} \, dx \\ &\leq \int_0^1 \, dx + \int_1^{\infty} x^{-r/2} \, dx \\ &= 1 + \frac{1}{\tfrac{r}{2}-1} = \frac{r}{r-2} = \frac{r}{r-2} \|\xi\|_{r}. \end{align*}

(Here we have used that $r/2>1$; it ensures that the integral $\int_1^{\infty} x^{-r/2} \, dx$ is finite.)

Now for the general case we note that it follows from the definition of $\|\xi\|_{2,1}$ and a change of variables that

$$\|\lambda \xi\|_{2,1} = \lambda \|\xi\|_{2,1} \tag{1}$$

for any $\lambda>0$. If we define $\tilde{\xi} := \xi/\|\xi\|_r$, then $\|\tilde{\xi}\|_r=1$, and therefore we find from $(1)$ and the first part of this proof (applied for $\tilde{\xi}$) that

$$\|\xi\|_{2,1} = \bigg\| \|\xi\|_r \cdot \tilde{\xi} \bigg\|_{2,1} \stackrel{(1)}{=} \|\xi\|_r \|\tilde{\xi}\|_{2,1} \leq \|\xi\|_r \frac{r}{r-2}.$$


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