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Suppose that $X$ and $Y$ are independent standard random variables. Define $U=\frac{1}{\sqrt{2}}(X+Y)$ and $V=\frac{1}{\sqrt{2}}(X-Y)$.

  1. What are the distributions of $U$ and $Y$. What are their parameters?
  2. Are $U,V$ independent?

My Attempt

  1. I know that the sum of two independent normal random variables is normal, and the mean is equal to the sum of the two means, and the variance is the sum of the two variances. We are given that $X\sim\mathcal{N}(0,1)$ and $Y\sim\mathcal{N}(0,1)$. The parameters give us the mean and standard deviation. So we know $\mathbb{E}[X]=\mathbb{E}[Y]=0$ and $\sqrt{Var(X)}=\sqrt{Var(Y)}=1$. Then $$U\sim\mathcal{N}\left(\frac{1}{\sqrt{2}}(0+0),\frac{1}{\sqrt{2}}(\sqrt{1+1})\right)=\mathcal{N}(0,1)$$ $$V\sim\mathcal{N}\left(\frac{1}{\sqrt{2}}(0-0),\frac{1}{\sqrt{2}}(\sqrt{1-1})\right)=\mathcal{N}(0,0)$$ We can see that since the parameters for $U$ is $0,1$, it is a standard normal distribution. I have some concerns about $V$ because I've never seen a normal distribution with $0$ for both parameters. Did I take the right approach in finding $U$ and $V$?
  2. For this part, I think I just need to show that $Cov(U,V)=0$ However, to do this, I would need to find $\mathbb{E}[UV]$, which I am not sure how to approach.

Edit: Using Erick Wong's comment, and the formula for Covariance, I arrived at $Cov(U,V)=\frac{1}{2}Var(X)-\frac{1}{2}Var(Y)=1-1=0$. Thus, $U$ and $V$ are independent.

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    $\begingroup$ It will still be $\sqrt{1+1}$ in the expression for $V$. The mean gets subtracted because when you take $-V$ in place of $V$ the mean changes sign (of course, since the normals are standard we are in mean zero territory so nothing changes). But the variance doesn't change at all, it's still $1$. $\endgroup$ Jun 5 at 4:20
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    $\begingroup$ Use the property of covariance $Cov(U, V)=Cov\left(\frac 1 {\sqrt 2}(X+Y),\frac 1{\sqrt 2}(X-Y)\right)$ $\endgroup$
    – Jellyfish
    Jun 5 at 4:29
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    $\begingroup$ $\mathbb{E}(UV)$ is literally $\frac12\mathbb{E}(X^2 - Y^2)$. You can just use linearity of expectation to show these are uncorrelated (this isn’t the same as independent, so you have to rely on other results to deduce independence from being uncorrelated). $\endgroup$
    – Erick Wong
    Jun 5 at 5:10
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I don't know if this theorem is in your book; if it isn't, you can just ignore this.

If $A$ is an orthogonal matrix, $X_1, X_2, \dots, X_n$ are iid standard normal random variables, then the components of $Y=AX$ are iid standard normal random variables.

$$A=\begin{pmatrix}\frac 1{\sqrt 2}&\frac 1 {\sqrt 2}\\ \frac 1{\sqrt 2}&-\frac 1 {\sqrt 2}\end{pmatrix}$$

This matrix $A$ is orthogonal because $AA'=A'A=I$. It can also be verified by showing that the sum of squares in each row is 1 and the inner product between any two rows is 0.

From this theorem, we have that $Y_1=\frac 1{\sqrt 2}(X_1+X_2)$ and $Y_2=\frac 1{\sqrt 2}(X_1-X_2)$ are standard normal random variables and are independent, answering the two questions.

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