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 $E[F(y)] < E[F(x)]$.

 

[blue_raver22]

I would approach this question by first understanding the definitions and properties of the variables and function involved. In this case, we have a function F that maps positive real numbers to positive real numbers and has the property that F(y)/y is decreasing. This means that as y increases, the ratio F(y)/y decreases.

Next, we have two random variables, x and y, which are also positive real numbers. The notation E[x] represents the expected value of x, which is a measure of the average value of x over all possible outcomes. Similarly, E[y] represents the expected value of y.

Based on this information, we can see that the question is asking if the expected value of F(x) is less than or equal to the expected value of F(y) when the expected value of x is less than or equal to the expected value of y. In other words, does the ordering of the expected values of x and y carry over to the expected values of F(x) and F(y)?

To answer this question, we can use a counterexample. Let x = 1 and y = 2. In this case, E[x] = 1 and E[y] = 2. However, F(x) = F(1) could be any positive number, while F(y) = F(2) is necessarily smaller since F(y)/y is decreasing. Therefore, it is possible that E[F(x)] > E[F(y)], which contradicts the proposed implication.

In conclusion, the statement does not hold in general, as there are cases where the expected value of F(x) can be greater than the expected value of F(y) even when the expected value of x is less than or equal to the expected value of y. However, if we have additional information about the relationship between x and y, such as their joint distribution, we may be able to prove or disprove the implication.