I'm curious about the following situation. Suppose we have two stochastic processes, $\{X_t\},$ and $\{Y_t\},$ which satisfy the expectation ordering $E[X_t] \leq E[Y_t],$ for all $t.$
What kind of operations $\phi$ may I apply to the processes such that the transformed processes $\{\phi(X_t)\}$ and $\{\phi(Y_t)\}$ satisfy $E[\phi(X_t)] \leq E[\phi(Y_t)].$
It is apparent that if $\{X_t\},$ and $\{Y_t\}$ were pointwise ordered, i.e. $X_t(\omega) \leq Y_t(\omega)$ for all $\omega$ in the sample space and all $t,$ then $\phi$ can be an arbitrary monotone increasing function.
Additionally, if $\phi$ is a sign-preserving linear map, linearity of expectation helps the inequality pass through.
One might expect there to be a class of $\phi$ which is more general than linear maps, but less general than arbitrary monotone increasing functions for which the above holds. As a particular example, does the above hold for $\phi(z) = \exp(z)$? I can't seem to determine this, either way.
Thanks in advance!
Without further assumptions on your stochastic processes, you cannot improve on the linear functions.
It suffices to look at one timestep; we want to show that if $$\mathbb{E}[X]\leq\mathbb{E}[Y]\Rightarrow\mathbb{E}[\phi(X)]\leq\mathbb{E}[\phi(Y)]$$ then $\phi$ is linear.
Suppose $Y=\mathbb{E}[X]$ a.s. Certainly $\mathbb{E}[X]\leq\mathbb{E}[Y]$, so that $$\mathbb{E}[\phi(X)]\leq\mathbb{E}[\phi(Y)]=\mathbb{E}[\phi(\mathbb{E}[X])]=\phi(\mathbb{E}[X])$$ Thus $\phi$ satisfies reverse Jensen's inequality, and so must be convex down (concave). Reversing the roles of $Y$ and $X$, the function $\phi$ must also be convex (up).
But the only function that is both convex up and convex down is linear.