Minimisation over random variables

[TaPaKaH]

Suppose we have a function $F:\mathbb{R}_+\to\mathbb{R}_+$ such that $\frac{F(y)}{y}$ is decreasing.
Let $x$ and $y$ be some $\mathbb{R}_+$-valued random variables.
Would $\mathbb{E}x\leq\mathbb{E}y$ imply that $\mathbb{E}F(x)\leq\mathbb{E}F(y)$?

 

[jbunniii]

Suppose we take $F(y) = 1/y$. Then certainly $F(y)/y = 1/y^2$ is decreasing for positive $y$.

Now let $x$ be some random variable which is restricted to the interval $[1,2]$ and let $y$ be some other random variable restricted to the interval $[3,4]$. Thus $E[x] < E[y]$. But $F(x)$ is restricted to $[1/2, 1]$ and $F(y)$ is restricted to $[1/4, 1/3]$. So $E[F(y)] < E[F(x)]$.

 

[TaPaKaH]

Your example makes perfect sense.

But what if we assume that $F(x)$ is increasing in $x$ and $F(0)=0$?

 

[jbunniii]

Your example makes perfect sense.

But what if we assume that $F(x)$ is increasing in $x$ and $F(0)=0$?

What if we take
$$F(x) = \begin{cases}
\sqrt{x} & \text{ if }0 \leq x \leq 1 \\
1 & \text{ if } x > 1
\end{cases}$$
Then $F(x)/x$ is decreasing for all positive $x$.

Let $x$ be uniformly distributed over $[1,2]$. Let $y$ be 0 or 3, each with probability 1/2. Then $E[x] = E[y] = 1.5$.

But $F(x) = 1$ with probability 1, so $E[F(x)] = 1$. And $F(y)$ is 0 or 1, each with probability 1/2, so $E[F(y)] = 1/2$.

If you want $F$ to be strictly increasing, then give it a tiny positive slope for $x > 1$, and define $x$ and $y$ as above. The result will still be $E[F(y)] < E[F(x)]$.