0
$\begingroup$

Let $X\colon\Omega\to\mathbb R$ be a continuous random variable with density function $f_X$ and let $p,q\in[1,\infty)$ such that $p

We know that $$ \mathbb E(|X|^q)=\int_{-\infty}^\infty |x|^qf_X(x)\,\mathrm dx<\infty. $$ We can rewrite this as follows: $$ \mathbb E(|X|^q)=\int_{-\infty}^{-1} |x|^qf_X(x)\,\mathrm dx+\int_{-1}^1 |x|^qf_X(x)\,\mathrm dx+\int_1^\infty |x|^qf_X(x)\,\mathrm dx. $$ Because $f_X$ is a nonnegative function, and $|x|^q\geq|x|^p$ for $x\in\mathbb R\setminus(-1,1)$, we know that the integrand of the first and third integrals are greater dan those for $|X|^p$: $$ \int_{-\infty}^{-1} |x|^pf_X(x)\,\mathrm dx\geq\int_{-\infty}^{-1} |x|^qf_X(x)\,\mathrm dx<\infty $$ and $$ \int_1^\infty |x|^pf_X(x)\,\mathrm dx\geq\int_1^\infty |x|^qf_X(x)\,\mathrm dx<\infty. $$ Now because $|x|^qf_X(x)$ is an integrable function, it is by definition bounded, so we know for sure that $f_X$ is bounded. This yields $$ \int_{-1}^1 |x|^pf_X(x)\,\mathrm dx<\infty. $$ Is this the right way to solve the question? I would like to know if there is a neater solution to this.

  • 0
    Show the pointwise inequality $$|X|^p\leqslant|X|^q+1$$ and integrate it. – 2017-02-02
  • 0
    @Did I've never heard of the pointless inequality. I've looked up the definition, but I don't know how you came up with this. – 2017-02-02
  • 1
    "Pointwise", not pointless. You mean, you cannot show that for every nonnegative real number, $$x^p\leqslant x^q+1\ ?$$ Well... Assume first that $x\leqslant1$, then $x^p\leqslant$ $____$ hence we are done. Assume next that $x\geqslant1$, then $x^p\leqslant$ $____$ hence we are done. QED. – 2017-02-02
  • 0
    @Did Okay; $x^p\leq1\leq x^q+1$ if $x\leq 1$, and $x^p\leq x^q\leq x^q +1$ if $x\geq 1$. Oh, and I see how you came up with this, because we can split the integral into a real value and 1 (because it's the density function integrated over $\mathbb R$). – 2017-02-02

0 Answers 0