Proving unique solution for $f$ when $f(P_a \cdot a+P_b \cdot b)=P_a \cdot f(a)+P_b \cdot f(b)$ This problem is about expected value, and it's a real world problem.
I know so far that $f$ is strictly increasing, if that makes the proof more concise (but if you can also prove it without this assumption, that would be awesome). Find all solutions for $f$ when $f(P_a \cdot a+P_b \cdot b)=P_a \cdot f(a)+P_b \cdot f(b)$ for all $a,b \in R$, and for $P_a,P_b \in R_+$. I'm 80% sure $f$ must be $f(x)=mx$ but I don't know how to prove it.
 A: By taking $P_b=0$, you have $f(P_a \cdot a) = P_a \cdot f(a)$ for any $a$ and $P_a$.
Taking $a=1$ shows that $f(P_a) = P_a \cdot f(1)$ for any $P_a$.
For purely cosmetic reasons you can rewrite the last result as $f(x) = x \cdot f(1)$ for all $x$.
Letting $m:=f(1)$ shows that $f(x) = mx$.
This shows that a necessary condition for $f$ to satisfy the original condition is "$f$ must be of the form $f(x)=mx$ where $m:=f(1)$." To show it is also a sufficient condition, you can directly check that the original condition holds with this kind of $f$.

The question was edited (after my original answer) to assert $P_a , P_b \ge 0$.
The same argument as before shows that $f(x) = f(1) \cdot x$ for $x \ge 0$.
Then by applying the original assumption, we have $f(-1) + f(1) = f(-1 + 1) = f(0) = 0$ and thus $f(-1) = f(1)$. Applying the same argument before shows $f(-x) = f(-1) \cdot x = f(1) \cdot x$ for $x \ge 0$. Thus $f(x)=f(1) \cdot x$ for all real $x$.
A: HINT: For all $P_a$, $P_b\in\mathbb{R}$ and all $\lambda\in[0,1]$ your condition gives that the fuction is convex and concave, so its graph is an interval with $x$-ends $P_a$ and $P_b$.
