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Consider $X_1,X_2$ i.i.d. standard normal random variables(mean 0, variance 1). Are the random variables $Y=X_1+X_2$ and $Z=X_1-X_2$ dependent? I am not sure how to prove this one way or the other.

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    @Dilip: my apologies once again. Edited.2011-09-27

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$X_1$ and $X_2$ are independent standard normals, so $(X_1, X_2)$ has rotationally symmetric density, namely $ {1 \over 2\pi} \exp(-(x_1^2 + x_2^2)/2). $ If you change coordinates with $u = (x_1 + x_2)/\sqrt{2}, v = (x_1 - x_2)/\sqrt{2}$ (so the change from $(x_1, x_2)$ to $(u,v)$ is area-preserving) then this becomes $ {1 \over 2\pi} \exp(-(u^2+v^2)/2). $ That is, the random variables $U = (X_1 + X_2)/\sqrt{2}$ and $V = (X_1 - X_2)/\sqrt{2}$ are also independent standard normals. Your random variables are $Y = U \sqrt{2}$ and $Z = V \sqrt{2}$, so they're independent normals with mean 0 and SD $\sqrt{2}$.

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    Thanks. This is what happens when I write quickly.2011-09-28
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If you are familiar with the concept of characteristic function, it is easiest to compute characteristic function for $(Y, Z)$. For independent variables, the characteristic function would factor into a product:

$ \begin{eqnarray} \mathbb{E}\left( \exp( i t_1 Y + i t_2 Z ) \right) &=& \mathbb{E}\left( \exp( i (t_1+t_2) X_1 + i (t_1-t_2) X_2 ) \right) \\ & = & \exp\left( -\frac{1}{2} \left(t_1+t_2\right)^2 \right) \cdot \exp\left( -\frac{1}{2} \left(t_1-t_2\right)^2 \right) \\ &=& \exp\left( -t_1^2 \right) \cdot \exp \left(-t_2^2 \right) \end{eqnarray} $

Hence the $Y$ and $Z$ are independent normal with mean 0 and standard deviation of $\sqrt{2}$.