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Could anyone help prove the following claim?

For $Z$ a random variable, choose $X, Y$ appropriately in Holder's inequality to show that $$f(t) = \log(\mathbb{E}|Z^{t}|)$$ is a convex function on the interval of $t$ where $\mathbb{E}|X^{t}| < \infty$.

I'm confused on the part of choosing appropriate $X, Y$ in Holder's inequality. I tried using only X and Z and rearranging for the $\log(\mathbb{E}|Z^{t}|)$ term but it feels headed in the wrong direction.

jackyooo
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    You want to start by fixing $x,y$ and then we have for $\lambda \in (0;1)$ $$ f(\lambda x+(1-\lambda)y)=\log(\mathbb{E} [\vert Z\vert^{\lambda x} \vert Z\vert^{(1-\lambda)y}]).$$ Note that $$\frac{1}{1/\lambda} +\frac{1}{1/(1-\lambda)}=1.$$ Can you finish from here? – Severin Schraven Feb 08 '22 at 19:40

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Hint. You want to prove $$ f\bigl(\lambda t + (1-\lambda)s\bigr) \le \lambda f(t) + (1 - \lambda) f(s) $$ for $\lambda \in [0,1]$, $s, t \in I$ (the interval in question). That is $$ \def\E{\mathbf E} \log\E\def\abs#1{\left|#1\right|}\abs{Z}^{\lambda t + (1-\lambda)s} \le \lambda \log \E\abs{Z}^t + (1-\lambda)\log \E\abs{Z}^s $$ Using the properties of the logarithm, this translates to $$\log\E\abs{Z}^{\lambda t + (1-\lambda)s} \le \log (\E\abs{Z}^t)^{\lambda}(\E\abs{Z}^s)^{1-\lambda} $$ or - as $\log$ is increasing - $$\E\abs{Z}^{\lambda t + (1-\lambda)s} \le (\E\abs{Z}^t)^{\lambda}(\E\abs{Z}^s)^{1-\lambda} $$

Can you see Hölder now?

Another hint: Look at Hölder $$ \E\abs{XY} \le (\E\abs X^p)^{1/p}(\E\abs Y^q)^{1/q} $$ and compare this with the right hand side of the last equation: Let $1/p = \lambda$, then $1-\lambda = 1/q$.

martini
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