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Suppose that we have two IID random variables, $X_1, X_2$, carried by a triple $(\Omega,\mathcal{F},P)$.

While solving an exercise I ended to a point that I had to see that,

$$ \iint\limits_D x_1 P_X(\mathrm dx_1)P_X(\mathrm dx_2) = \iint\limits_D x_2 P_X(\mathrm dx_2)P_X(\mathrm dx_1), $$ where $D= \{ (x_1,x_2) \in \mathbb{R}^2 : x_1 + x_2 = c \} $ and $ P_X$ is the law of $X_1, X_2$, i.e. the measure induced by $P$

The only way that I could identify it was by a change of variables.

Anyone can give me some intuition about this equality? Anyone can tell me something more simple than change of variables?

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Identically distributed means $P_{X_1} = P_{X_2}$!!! The only difference is the dummy variable name: $x_1$ or $x_2$!! –  André Caldas Nov 2 '12 at 21:03
1  
Hint: $\int f(a,b) da db = \int f(x,y) dx dy$. –  André Caldas Nov 2 '12 at 21:11
    
Thank you, I see your point. –  noob-mathematician Nov 3 '12 at 13:58
    
You are welcome. It was a bit easy... but don't feel intimidated. Keep working! –  André Caldas Nov 5 '12 at 1:32

1 Answer 1

Both sides are $\displaystyle\iint_D u P_X(\mathrm du)P_X(\mathrm dv)$, or $\displaystyle\iint_D s P_X(\mathrm ds)P_X(\mathrm dt)$, or...

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