# Derivative of sup(max) functions in distributionally robust optimization

In the distributionally robust optimization problem \begin{aligned} \min_{x\in X}\sup_{P\in\mathfrak{P}}\mathbb{E}_P[f(x,\xi)], \end{aligned} where $$f:\mathbb{R}^n\to\mathbb{R}$$ and $$P$$ is a distribution function over the measurable set $$\mathfrak{P}$$, and $$\mathbb{E}_P$$ denotes the expected value with respect to the distribution $$P$$.

For a differentiable function $$f$$, if we define $$g(x):=\sup_{P\in\mathfrak{P}} \mathbb{E}_P[f(x,\xi)]$$, is the function $$g$$ differentiable for the variable $$x$$ like $$\nabla g(x)=\sup_{P\in\mathfrak{P}}\mathbb{E}_P[\nabla_x f(x,\xi)]$$?

In general, I guess it is not true because the interchangeability of the differentiation and expectation operators cannot hold. But I am curious what is the condition that the following equation holds? \begin{aligned} \nabla g(x)=\sup_{P\in\mathfrak{P}}\mathbb{E}_P[\nabla_x f(x,\xi)]. \end{aligned}

• did you try the definition of expectation Jul 6 at 13:45
• While I am not attempting to answer your question in this comment, you may be interested to know that even sup and $\nabla$ cannot be interchanged. So the problem may be beyond just the issue of interchangeability between differentiation ($\nabla$) and expectation. Jul 7 at 1:48