Do densities lie in $L^p$?

Question

Let $f$ be a probability density. We know $f \in L^1$ since by definition, it integrates to 1.

But is $f \in L^p$ for any $p > 1$?

It seems logical to assume so? We know that densities approach zero for $|x| >> 0$. So if you square it, or take it to some other power, it will become smaller and smaller, hence making it more and more integrable?

So all densiites are in $L^p$ for all $p > 1$!?

You should be concerned not what happens when $|x| >> 0$, but rather when $|x| <<1$. — Umberto P., Apr 15 '20 at 15:13
How about the density $f(x) = \frac 12 x^{-1/2} \mathbb{I}(x \in (0, 1])$? — pg_star, Apr 15 '20 at 15:16
@UmbertoP. Ok, so assume the density is bounded, which 99.999% of all densities used in practice are. Then my theorem holds. — saei, Apr 15 '20 at 19:14
What theorem? That a bounded function in $L^1$ belongs to $L^p$ for $p > 1$? — Umberto P., Apr 15 '20 at 21:22

score 1 · Answer 1 · 2020-04-15T15:35:56.943

1

Consider the pdf $$f(x)=\frac1{\sum_{j=2}^\infty \frac1{j\ln^2j}}\cdot\sum_{k=2}^\infty k\cdot 1_{\left[k,\,k+\frac1{k^2\ln^2 k}\right)}(x)$$

Then, $$\int_{-\infty}^\infty \lvert f(t)\rvert^p\,dt=\left(\sum_{j=2}^\infty\frac1{j\ln^2j}\right)^{-p}\sum_{k=2}^\infty\frac{1}{k^{2-p}\ln^2k}=\begin{cases}1&\text{if }p=1\\ \infty&\text{if }p>1\end{cases}$$

edited Apr 15 '20 at 15:35

answered Apr 15 '20 at 15:27

Do densities lie in $L^p$?

1 Answers1