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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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    Use the fact that sums of jointly Gaussian random variables are Gaussian (exercise) and that two jointly Gaussian random variables are independent if and only if they have zero covariance (exercise). Note that the latter statement is false in general.2011-09-27
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    @QiaochuYuan A minor quibble: "two Gaussians are independent of and only if they have zero covariance" is not quite right since, as has been discussed elsewhere on math.stackexchange.com, joint Gaussianity is required for uncorrelated Gaussian random variables to be independent. Here of course, $Y$ and $Z$ _are_ jointly Gaussian and so the issue does not arise.2011-09-27
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    @Dilip: you're right, of course. Corrected.2011-09-27
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    We want $E(YZ)-E(Y)E(Z)$. But $YZ=X_1^2-X_2^2$.2011-09-27
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    So covariance is zero. So they are independent?2011-09-27
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    For *general* distributions, covariance $0$ does not imply independence. However, please see comments of Quiaochu Yuan and Dilip Sarwate.2011-09-27
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    @QiaochuYuan .... and if I might request yet another minor edit to your comment, "sums of Gaussians are Gaussian" is not quite correct; once again, joint Gaussianity (which fortunately holds in the question asked) is required.2011-09-27
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    @Dilip: my apologies once again. Edited.2011-09-27

2 Answers 2

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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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    Corrected some typos. Please check the result suits you.2011-09-28
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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 $\sqrt{2}$.