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Let $U$ be a standard Cauchy random variable, $f_U(x)=\dfrac{1}{\pi}\dfrac{1}{1+x^2}$, $x\in R$.

a) Show that $U$ and $1/U$ have the same distribution.

b) Show that $E|U|^\alpha\geq1$ for all $0<\alpha<1$. Hint: $1=U\frac{1}{U}$.

I did part (a) by showing their pdf are the same. i tried to use Cauchy-Schwarz Inequality, but I couldn't get anything. I'm asking for only some more hint (there's already a hint given).

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2 Answers 2

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$\newcommand{\E}{\operatorname{E}}$ $$ \E|XY| \le \sqrt{\E (X^2)\E(Y^2)} $$ and in particular $$ 1 = \E\left( |U^{\alpha/2}| \cdot \left|\frac 1 {U^{\alpha/2}} \right| \right) \le \overbrace{\sqrt{\E|U^\alpha|\E\left| \frac 1 {U^\alpha} \right| } = \sqrt{\E|U^\alpha|\E|U^\alpha|}}^{\text{Since $U$ and $1/U$ have the same distribution.}} = \E|U^\alpha|. $$

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Hint: $1 = |U|^{\alpha/2} \left|\frac{1}{U}\right|^{\alpha/2}$. Now apply the Cauchy-Schwarz inequality and part (a).

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  • $\begingroup$ That's what I did actually, but I didn't realize how I could use part (a) which is indeed so obvious now. After your hint, I understood that I'm on the correct way and used (a) and got the result! Thanks! $\endgroup$
    – user120005
    Sep 11, 2015 at 19:39
  • $\begingroup$ Glad I could help. Please accept this or Michael's solution so it is clear that the question got answered adequately. This helps keeping the list of problems organized. $\endgroup$
    – Dominik
    Sep 11, 2015 at 19:49

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