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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

Michael Hardy
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Nick
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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 03 '12 at 06:21
  • You only need to prove their covariance is 0 since uncorrelation implies independence for normal distribution. – JACKY88 Oct 03 '12 at 06:21
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    @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 03 '12 at 06:42
  • @Henry Thank you for pointing it out. They need to be jointly normal distributed. – JACKY88 Oct 03 '12 at 11:23
  • https://math.stackexchange.com/q/3831387/321264 – StubbornAtom Sep 18 '20 at 18:13

1 Answers1

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HINT:

  1. Independent Gussian random variables make a 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.

EpsilonDelta
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Sasha
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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 03 '12 at 04:58
  • @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 03 '12 at 05:29
  • Hence one should add 0. Independent Gaussian random variables make Gaussian vectors. – Did Oct 03 '12 at 05:34
  • @did Thanks. Added. – Sasha Oct 03 '12 at 05:38