Is there any trivial way to prove/disprove the following two statements?
- Assume $a$, $b$, are two non-negative random variables.
$$ \mathbb{E}[a] \ge \mathbb{E}[b] \iff \mathbb{E}[\sqrt a] \ge \mathbb{E}[\sqrt b] $$
- Assume $x\in N(\mu,\sigma)$, then
$$ \mathbb{E}[xe^{x/2}] \ge \mathbb{E}[x]\mathbb{E}[e^{x/2}] $$
$1.$ The first one is false, for an example take $\mathbb{P}(a=25)=1/5$ and $\mathbb{P}(a=0)=4/5$ and $\mathbb{P}(b=4)=1$. Then $$E(a)=5,E(b)=4$$ and $$E(\sqrt{a})=1,E(\sqrt{b})=2$$
$2.$ Here denote $\mu$ the law of $X$. Since $x$ and $e^{x/2}$ are both increasing we have that $(y-x)(e^{y/2}-e^{x/2})\geq 0$. Therefore $\int\int (y-x)(e^{y/2}-e^{x/2})\text{d}\mu(y)\text{d}\mu(x) \geq 0$.
On the other hand we can calculate
$$ \int\int (y-x)(e^{y/2}-e^{x/2})\text{d}\mu(y)\text{d}\mu(x)= 2E(Xe^{X/2})-2E(X)E(e^{X/2})$$ So the second is true.