# Prove that if $X$ and $Y$ are Normal and independent random variables, $X+Y$ and $X-Y$ are independent

If $X \sim \mathrm{Normal}(\mu,\sigma^2)$ and $Y \sim \mathrm{Normal}(\mu,\sigma^2)$ are independent random variables, how do I prove that $X+Y$ and $X-Y$ are also independent?

What happens with the independence between $X+Y$ and $X-Y$ when $X \sim \mathrm{Normal}(\mu_x,\sigma_x^2)$ and $Y \sim \mathrm{Normal}(\mu_y,\sigma_y^2)$

Thank you

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Given that $X$ and $Y$ are independent normal, $X+Y$ and $X-Y$ are independent if and only if $\sigma_X^2=\sigma_Y^2$. –  Did Oct 3 '12 at 6:21
You only need to prove their covariance is 0 since uncorrelation implies independence for normal distribution. –  Patrick Li Oct 3 '12 at 6:21
@Patrick: that is not true. If $X$ has a standard normal distribution and $Y=X$ when $|X| \gt k \approx 1.538$ and $Y=-X$ when $|X|\le k$ then correlation is $0$ and $Y$ also has a standard normal distribution but they are not independent. –  Henry Oct 3 '12 at 6:42
@Henry Thank you for pointing it out. They need to be jointly normal distributed. –  Patrick Li Oct 3 '12 at 11:23

HINT:

1. Independent Gussian random variables makes Gaussian random vector.
2. Affine transform $Y=A X + b$ of Gaussian random vector $X$ is Gaussian.
3. Distribution of Gaussian random vector is determined by its mean vector, and covariance matrix.
4. If components $X_i$ and $X_j$ of the Gaussian random vector are independent, then $\mathbb{Cov}(X_i, X_j) = 0$.

Combining facts given above, it follows that evaluation of $\mathbb{Cov}(X+Y,X-Y)$ will help establish the result needed.

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No, you've got your "if" and "then" in the wrong order to get the conclusion you want. You want to prove independence. You need to show that IF the covariance is $0$ THEN they're independent, NOT that IF they're independent THEN the covariance is $0$. –  Michael Hardy Oct 3 '12 at 4:58
@MichaelHardy Yes, I agree. I was saying that since the mean vector and covariance matrix determine the distribution of a Gaussian vector, zero cross-covariance $\mathbb{Cov}(X_i,X_j)=0$ implies independence of $X_i$ and $X_j$. –  Sasha Oct 3 '12 at 5:29
Hence one should add 0. Independent Gaussian random variables make Gaussian vectors. –  Did Oct 3 '12 at 5:34
@did Thanks. Added. –  Sasha Oct 3 '12 at 5:38