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Let $\{ X_{1}, X_{2}, Y_{1}, Y_{2} \}$ be independent with $X_{i} =^{d} Y_{i}$ for $i \in \{ 1, 2\}$ and $\mathbb{E}(X_{1}), \mathbb{E}(X_{2}) < \infty$. Is it true that:

$$\mathbb{E}(\max\{(X_{1}+Y_{1}),(X_{2}+Y_{2}) \}) \geq \mathbb{E}( \max \{ (X_{1}+Y_{2}), (X_{2}+Y_{1}) \})$$

Any argument is helpful, even one that assumes that $X_{1}, X_{2}$ have densities and/or are nonnegative. (Although ultimately I don't want to make such assumptions.)

Edit:

With $Z_{1} = X_{1}+Y_{1}$, $Z_{2} = X_{2}+Y_{2}$, $Z_{3}=X_{1}+Y_{2}$, $Z_{4}=X_{2}+Y_{1}$ and assuming all r.v's are nonnegative for simplicity, and letting say $F_{X}$ denote the cdf of r.v. $X$, we want to examine:

$$\mathbb{E}( \max \{ Z_{1}, Z_{2} \}) - \mathbb{E}(\max \{Z_{3}, Z_{4} \}) =$$ $$\int_{0}^{\infty} 1-F_{Z_{1}}(t)F_{Z_{2}}(t)dt - \int_{0}^{\infty}1-F_{Z_{3}}(t)F_{Z_{4}}(t)dt =$$ $$\int_{0}^{\infty} F_{Z_{3}}(t)^{2}-F_{Z_{1}}(t)F_{Z_{2}}(t) dt$$

(Where $X_{i} =^{d} Y_{i} \implies Z_{3}=^{d}Z_{4} \implies F_{Z_{3}} = F_{Z_{4}}$ has been used)

Unclear how to proceed/if this is the right approach. Maybe use a convolution to try and get explicit expressions for $F_{Z_{i}}$ in terms of $F_{X_{i}}$? Alternatively maybe there's some way to integrate along the four sets of the form, for example $A_{1 \geq 2, 3 \geq 4} := \{ Z_{1} \geq Z_{2}, Z_{3} \geq Z_{4} \}$ and look at the differences restricted to those sets?

RobPratt
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Snidd
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  • You should show the ways you have tried to answer the question, or at least your thoughts about it, otherwise it seems as if you are asking other people to do your job. – Albert Jun 21 '24 at 12:27
  • @Albert

    Okay, sorry nothing I had seemed useful

    – Snidd Jun 21 '24 at 12:56
  • We all get blocked sometimes, but you can always show the ways in which you got blocked, since this shows---as your edit does---that you have given some thought to the question. :) – Albert Jun 21 '24 at 13:11
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    See: https://math.stackexchange.com/questions/2251364/prove-that-bbbex-y-le-bbbexy-for-i-i-d-x-and-y/2251389#2251389 – River Li Jun 22 '24 at 02:12
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    @RiverLi

    ty, this resolves the question

    – Snidd Jun 22 '24 at 02:17

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