# Deriving joint distribution when one-to-one mapping does not exist

Let $x_1, x_2, x_3$ be iid random variables, and $$y_1=x_1+x_2+x_3,\qquad y_2=x_1+x_2$$

How can we derive the joint distribution of $y_1$ and $y_2$. In this case a one-to-one mapping does not exist so usual method involving jacobian cannot be used.

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I assume you mean $y_1 = x_1 + x_2 + x_3$ and $y_2 = x_1 + x_2$. Hint: first find the distribution of $y_2$, then you have a one-to-one mapping $(y_2, x_3) \to (y_1, y_2)$.
How are $y_2$ and $x_3$ dependent? $y_2$ only depends on $x_1$ and $x_2$, not $x_3$. –  Robert Israel Nov 8 '12 at 6:57
You are right, I read too fast. Sorry about the noise. (The last $y_3$ should be $y_1$, right?) –  Did Nov 8 '12 at 7:03
Compute the distribution of $(y_1,y_2,y_3)$ where $y_3=x_1$, by the usual change of variable method using the Jacobian, then marginalize this 3D didstribution to the 2D distribution of the coordinates $(y_1,y_2)$.