Does $X_1 + Y_1$ have the same distribution as $X_2+Y_2$? Suppose $X_1$ and $X_2$ have the same distribution.
Suppose $Y_1$ and $Y_2$ have the same distribution.
Does it follow that $X_1 + Y_1$ have the same distribution as $X_2+Y_2$?
Find a counterexample or prove it. Please confirm my counterexample. Let $(0,1)$ be the domain of $X_1, X_2, Y$ with Lebesgue measure.
My intuition tells me yes but I think this is false because we can consider $X_1 = 1$ on $(0,.5)$ and $0$ elsewhere, and do the flip thing for $X_2$.  Then if $Y = -1$ on $(0,.5)$ and $0$ elsewhere, then $P(X_1 + Y = 0)=1$ and $P(X_2+Y = 0) = 0$.  
 A: The answer is no and the problem is dependence.  Suppose Alice and Bob each flip a coin.  Let $X_1 = X_2 = Y_1$ be equal to $1$ if Alices gets Heads, and $0$ if Alice gets Tails.  Let $Y_2$ equal $1$ if Bob gets Heads and $0$ if Bob gets Tails.  $X_1$, $X_2$, $Y_1$, and $Y_2$ all have the same distribution.  $P(X_1 + Y_1 = 1) = 0$, but $P(X_2+Y_2 = 1) = 1/2$.
Sorry for not confirming your counterexample, instead I answered your initial question using the simplest counterexample I could think of.
A: This answer contains a counterexample on its own and not a check of yours.
Let $Z$ and $-Z$ have the same distribution with $P\left\{ Z=0\right\} \neq1$.
For instance standard-normal.
Taking $X_{1}=X_{2}=Y_{1}=Z$ and $Y_{2}=-Z$ the mentioned conditions
are satisfied. 
But $X_{1}+Y_{1}=2Z$ and $X_{2}+Y_{2}=0$ have different
distributions.
A: Suppose $X_1,X_2$ both be from $U(0,1)$
Now let $Y_1=2X_1$ and $Y_2$ from $U(0,2)$ but independent of $X_2$.
Now distribution of $X_1+Y_1$ is different from $X_2+Y_2$
$Y_1 \sim U(0,2),Y_2 \sim U(0,2)$
$P(X_1+Y_1<2)=P(3X_1<2)=\frac{2}{3}$
$P(X_2+Y_2<2)=\int\limits_{0}^{1}\int\limits_{0}^{2-x}\frac{1}{1}.\frac{1}{2}dydx=\frac{3}{4}$
A: Your intuition is a good one, simply the way you formulate it can be made more precise. You first consider a standard Bernoulli random variable $X_1$, that is, any random variable such that $P[X_1=0]=\frac12$ and $P[X_1=1]=\frac12$. Then you define $X_2=1-X_1$ and $Y_1=Y_2=-X_1$. 
One then sees that, indeed, $X_1$ and $X_2$ have the same distribution, $Y_1$ and $Y_2$ have the same distribution (since $Y_1=Y_2$), and $X_1+Y_1=0$ while $X_2+Y_2=1-2X_1$. Hence, the distributions of $X_1+Y_1$ and $X_2+Y_2$ do not coincide, for example because $P[X_1+Y_1=1]=0$ and $P[X_2+Y_2=1]=P[X_1=0]=\frac12$.
