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?
Okay, sorry nothing I had seemed useful
– Snidd Jun 21 '24 at 12:56ty, this resolves the question
– Snidd Jun 22 '24 at 02:17