I am stuck with a proof about positive homogeneity and have tried now for several days. Hopefully someone can help me out. Thanks in advance.
I am considering a risk measure $p=(x)$, where $x$ denotes a loss. $p$ satisfies the following properties:
P1: Normalization, $p(0)=0$
P2: Cash additive, $p(x+c)=p(x)+c$
P3: Monotonic, if $x$$\geq$$y$ then $p(x)$$\geq$$p(y)$
P4: Convexity, if $\lambda\in(0,1)$ then $p(\lambda$$x+(1-\lambda$$y$) $\leq$ $\lambda$$p(x)+(1-\lambda$$)p(y)$
P6: Subadditivity, $p(x+y)\leq$$p(x)+p(y)$
Also, we have that $\lambda\in(0,1)$ and $\lambda\in$Q
I proved that $p(\lambda$$x)\leq$ $\lambda$$p(x)$for $\lambda\in(0,1)$ and $\lambda\in$Q
Now, I want to prove that $p(\lambda$$x)\geq$ $\lambda$$p(x)$ for $\lambda\in(0,1)$ and $\lambda\in$Q as well.
This should be possible using the properties above and the fact that $\lambda\in$Q.
Thanks for anyone who can help me with this proof.
From P1 and P2 it follows that $$ p(y) = p(0+y) = p(0)+y = 0+y = y $$ has to be true for all $y\in\mathbb R$.
Now that you know the value of $p$ for all inputs, it is easy to see that $$p(\lambda x) = \lambda x = \lambda p(x)$$ has to hold for all $\lambda\in(0,1), x\in \mathbb Q$.