Expectation of a function of weighted sum of random variables: Is this coupling?

Question

Set of non negative weights $w_j$, set of non negative i.i.d. random variables $X_j$ and $f(y)$ is a decreasing nonnegative function in $y$.

I want to claim that:

if $\sum w_i<\sum w^{\prime}_i$,then $\mathbb E f(\sum w_iX_i)>\mathbb E f(\sum w^{\prime}_iX_i)$.

This seems intuitive but I would like a formal proof.

My attempt: The next line holds if all r.vs are coupled to a common r.v. $X_i\sim U$. But not sure if we can do that. $$\sum w_iX_i< \sum w^{\prime}_iX_i,$$ We have $$f(\sum w_iX_i)> f(\sum w^{\prime}_iX_i),$$ almost surely.

And we take expectation $$\mathbb E f(\sum w_iX_i)>\mathbb E f(\sum w^{\prime}_iX_i). \;\;\;\;\;\;\; (a)$$ Therefore the result holds a.s.

Is this proof correct? If wrong where is the mistake?

The function $f$ I have is $\log(1+\frac{1}{y})$ and $X_i$ are exponential r.vs.

Answer 1

When $X_i$ are i.i.d. standard normal distribution, as long as $\sum \omega_i^2 = \sum \omega_i'^2$, you get equality in expectation since $\sum \omega_i X_i$ and $\sum \omega_i'X_i$ have the same law.

Answer 2

Your conclusion is still wrong.

Consider decreasing function $f(y)=e^{-y},\ w=(1,1),\ w'=(2.5,0)$ and probability density $X\sim e^{-X}\mathbf 1_{X>0}$. One sees $$E\big[f\big(\sum w_iX_i\big)\big]=\frac{1}{4}<\frac{1}{3.5}=E\big[f\big(\sum w'_iX_i\big)\big],$$ contrary to your claim.