Is $(\mathbb{E}[[X-\mathbb{E}[X]]_+^p])^{1/p}$ convex? We say that a risk measure $\rho(\cdot)$ is convex if
$$
\rho(tX+(1-t)Y) \leq t\rho(X) + (1-t)\rho(Y)
$$
for all $X,Y \in L^p$ be random varaibles and $t\in [0,1]$.
With the definition above, I wonder if the following random function is convex or not: Let $X \in L^p$, consider
$$
\rho(X):=\left( \mathbb{E}\left[ \left( X-\mathbb{E}[X]\right)_+^p \right] \right)^{1/p}
$$
where $(a)_+:=\max\{a,0\}$ and $p \in [1,\infty)$ .
My attempt: If $X \leq \mathbb{E}[X]$, then $\left( X-\mathbb{E}[X]\right)_+ = 0$. Hence, $\rho(X)$ is convex trivially.
On the other hand, if $X > \mathbb{E}[X]$, then $\left( X-\mathbb{E}[X]\right)_+ = X-\mathbb{E}[X]$. Thus, in this case, we have
$$
\rho(X) = \mathbb{E}[(X - \mathbb{E}[X])^p]^{1/p}.
$$
If $X,Y$ are given and $t \in [0,1]$, then
$$\begin{align*}
\rho(tX+(1-t)Y) 
&=\mathbb{E}\left[\left( (tX+(1-t)Y) - \mathbb{E}[tX +(1-t)Y] \right)^p \right]^{1/p}\\
&\leq \|(tX+(1-t)Y) - \mathbb{E}[tX + (1-t)Y]\|_p \\
&= \|t(X - \mathbb{E}[X]) + (1-t) (Y - \mathbb{E}[Y])\|_p\\
& \leq t\|X-\mathbb{E}[X]\|_p + (1-t) \|Y-\mathbb{E}[Y]\|_p
\end{align*}
$$
Then I get stuck on how to achive $t\rho(X) + (1-t) \rho(Y)$ on the right hand side.
Any help is appreciated.
 A: Define $ a_- = | \min\{ a, 0 \} | $, so that for any real number $ a $, $ a = a_+ - a_- $. Define $ f(X) = (X - \mathbb{E}[X] )_+ $ and $ g(Y) = (\mathbb{E}[| Y |^p])^{1/p} $. $ f(X) $ is convex:
$\begin{align*}
& f(t X + (1-t) Y) \\
&= ( t (X - \mathbb{E}[X] ) + (1-t) (Y - \mathbb{E}[Y] ) )_+ \\
&= ( t (X - \mathbb{E}[X] )_+ - t (X - \mathbb{E}[X] )_- + (1-t) (Y - \mathbb{E}[Y] ) )_+ - (1-t) (Y - \mathbb{E}[Y] ) )_- )_+ \\
&\le t (X - \mathbb{E}[X] )_+ + (1-t) (Y - \mathbb{E}[Y] ) )_+ \\
&= t f(X) + (1-t) f(Y) ,
\end{align*}$
where $ 0 \le t \le 1 $.
Now, $ g(Y) $ is the $ L_p $ norm of $ Y $, and hence $ g(Y) $ is convex for $ p \ge 1 $. Further, if $ 0 \le Y_1 \le Y_2 $, then $ | Y_1 |^p \le | Y_2 |^p $, and hence $ g(Y_1) \le g(Y_2) $.
Note that $ \rho(X) = g(f(X)) $. Therefore,
$\begin{align*}
& \rho(t X + (1-t) Y) \\
&= g(f(t X + (1-t) Y)) \\
&\le g(t f(X) + (1-t) f(Y)) \;\; \text{because } 0 \le f(t X + (1-t) Y) \le t f(X) + (1-t) f(Y) \\
&\le t g(f(X)) + (1-t) g(f(Y)) = t \rho(X) + (1-t) \rho(Y) ,
\end{align*}$
which proves the convexity of $ \rho(\cdot) $.
