Let $X_1,Y_1$ be two arbitrary random variables, with distribution functions $P(X_1=x)= P_X(x), P(Y_1=y)= P_Y(y)$, and let $X_2,Y_2$ be random variables with the same distribution respectively, i.e., $P(X_2=x)= P_X(x), P(Y_2=y)= P_Y(y)$. However, we don't assume anything about the join distribution of the variables. Therefore, they could have any arbitrary coupling, from begin indepdent, to being fully coupled. I would like to know if my analysis bellow about deviations of $X_1+Y_1$ from $X_2+Y_2$, in terms of their median, is correct: $$P(X_1+Y_1\ge t) = \int_s P(X_1=s,X_1+Y_1\ge t)\\ =\int_s P(X_1=s)P(X_1+Y_1\ge t\vert X_1=s)\\ \int_s P(X_1=s)P(Y_1\ge t-s) = \int_s P_X(X=s)P_Y(Y\ge t-s)]=\\ \int_s P(X_2=s)P(Y_2\ge t-s) = \int_s P(X_2=s)P(X_2+Y_2\ge t|X_2=s)=\\ \int_s P(X_2+Y_2\ge t) = P(X_2+Y_2\ge t) $$ Now, if we set $t=\mathrm{median}(X_1+X_2)$, the result above means that the median of the copies $X_2,Y_2$ is the same as the original $X_1,Y_1$. The result can be stated as follows ($\overset{\mathcal{D}}{\sim}$ means equal in distribution): $$ X_1\overset{\mathcal{D}}{\sim} X_2, Y_1\overset{\mathcal{D}}{\sim} Y_2 \Rightarrow P(X_1+Y_1\ge t)=P(X_2+Y_2\ge t) \quad \forall t\in \mathbb{R} $$
I've written every step in detail because for me the conclusion looks really strange. Particularly, because I haven't assumed anything about the joint probability distribution, asied from a loose condition on the marginals. If it's not correct, could you point out the flaw, and if it is, can you perhaps give some intuition why the marginal distribution is enough for preserving the median?
