# Solve simple stochastic variational inequality

Let $$Z$$ be a random variable with finite mean $$\mathbb E[Z]$$ and let $$\phi:(-\infty,\infty) \rightarrow (-\infty,\infty]$$ be a convex l.s.c function which is differentiable at $$z=1$$ with gradient $$\phi'(1)=0$$.

# Question

How to find / characterize all the numbers $$c\in \mathbb R$$ verying the "stochastic" variational inequality $$\begin{split} &1 \in \mathbb E[\partial \phi^*(Z-c)],\\ &\phi'(1) = 0. \end{split}$$

N.B.: Here $$\phi^*(z) := \sup_t zt - \phi(t)$$ defines the usual convex conjugate and $$\partial \phi^*(z) := \{g \in \mathbb R | \phi^*(t) \ge \phi^*(z) + g(z-t)\;\forall t \in \mathbb R\}$$ is the subdifferential of $$\phi^*$$ at $$z$$.

# Some concrete examples

Z takes on finitely many values. Suppose $$Z$$ can only take one of $$n$$ distinct values $$z_1,\ldots,z_n$$ with probabilities $$p_1,\ldots,p_n$$.

$$\begin{split} &1 \in \sum_{i=1}^np_i\partial \phi^*(z_i-c),\\ &\phi'(1) = 0, \end{split}$$ which is just a classical variational inclusion problem. The particular case $$n=1$$ gives
$$\begin{split} &1 \in \partial \phi^*(z_1-c) \iff z_1-c \in \partial \phi(1),\\ &\phi'(1) = 0, \end{split}$$ giving the unique solution $$c=z_1$$.
Quadratic potential. Suppose $$\phi(z) = \alpha(z-1)^2/2$$, for some $$\alpha > 0$$. A simple computation gives, $$\phi^*(z) = \dfrac{z^2}{2\alpha} + z$$, from which it follows that the unique solution for our problem is $$c=\mathbb E[Z]$$.