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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}

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  • $\begingroup$ did you try the definition of expectation $\endgroup$
    – Qurious Cube
    Commented Jul 6, 2021 at 13:45
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    $\begingroup$ 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. $\endgroup$
    – batwing
    Commented Jul 7, 2021 at 1:48
  • $\begingroup$ I would answer this question by myself. If the following conditions hold, the equality $\nabla g(x)=\sup_{P\in\mathfrak{P}} \mathbb{E}_P[\nabla_{x} f(x,\xi)]$ holds: 1. the set $\mathfrak{P}$ is weakly compact; 2. the optimal element $P\in\mathfrak{P}$ is uniquely determined; 3. the function $f$ is regular. The condition might be stricter than I expected, though. $\endgroup$
    – Keith
    Commented Jan 19, 2022 at 7:26

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