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I am struggling with the following problem: $X_1, X_2 \sim N(0, 1)$ are independent random variables. Let $Y_1 = \frac{1}{\sqrt{2}}(X_1 + X_2)$ and $Y_2 = \frac{1}{\sqrt{2}}(X_1 - X_2)$. Show that $Y_1, Y_2$ are independent, and have $N(0, 1)$ distibution. So $$Y_1 ∼N\left(0, \left(\frac{1}{\sqrt{2}}\right)^2\times 1 + \left(\frac{1}{\sqrt{2}}\right)^2\times 1\right) = N(0, 1)$$ Same goes for $Y_2$. I calculated their covariance to be 0, and now I want to use the general property that when $X_1,\ldots,X_n$ have joint normal distribution, then $X_1, \dots, X_n$ are uncorrelated $\iff$ $X_1, \dots, X_n$ are independent. So my goal is to show that $(Y_1, Y_2)$ has a normal join distribution. I do not know how to do that though.

Edit with solution:

Since $X_1, X_2$ are independent, their joint distribution densitiy is $f(x_1, x_2) = f(x_1)*f(x_2) = \frac{1}{2\pi}\exp(-\frac{1}{2}(x_1^2+x_2^2)$, so $(X_1, X_2) ∼ N_2(0, I)$. Now $$\begin{bmatrix}Y_1\\Y_2\end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix}X_1\\X_2\end{bmatrix}$$Using theorem provided by @utobi, $(Y_1, Y_2) ∼ N_2\left(0, \begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{bmatrix}^2\right) = N_2(0, I)$. From this, and the fact that $Cov(Y_1, Y_2)=0$ follows that $Y_1, Y_2$ are independent.

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Yours is a particular case of the following theorem.

Theorem. Let $X\,\sim\, \text{N}_p(\mu, \Sigma)$, $\underset{q\times p}{A}$ a and $\underset{q\times 1}{c}$ a fixed matrix and a fixed vector, respectively, and let $Y = AX+c$. Then

$$Y\,\sim\, \text{N}_q(A\mu+c, A\Sigma A^\top).$$

Proof. We will use the characteristic function, which for a random $p$-vector $X$, is defined as $$\phi(t) = E(e^{i t^\top X}),\quad t\in\mathbb{R}^p.$$

First note that if $V\sim \text{N}(\mu, \sigma^2)$, then $\varphi_V(t) = \exp(it\mu - t^2\sigma^2/2)$ and if $W\sim\,\text{N}_p(\mu, \Sigma)$, $\varphi_W (\underset{p\times 1}{s}) = \exp(i s^\top\mu-s^\top \Sigma s/2)$.

Then \begin{eqnarray*} \varphi_{Y}(\underset{q\times 1}{u}) & =& \mathbb{E}\{e^{iu^\top Y}\} = \mathbb{E}\{e^{iu^\top(AX+c)}\} = e^{iu^\top c}\mathbb{E}\{e^{i (A^\top u)^\top X}\}\\ && \,\,\color{gray}{\text{($y=Ax+c)$}}\\ &=& e^{iu^\top c} \mathbb{E}\{e^{i s^\top X}\} = e^{iu^\top c}\varphi_X(s) = e^{iu^\top c}e^{i s^\top\mu-\frac{1}{2}s^\top \Sigma s}\\ &=& e^{iu^\top c} e^{i(u^\top A \mu) - \frac{1}{2} u^\top A \Sigma A^\top u}\\ &\overset{(s=A^\top u)}{=}&\exp\{iu^\top(A \mu + c) - \frac{1}{2} u^\top (A \Sigma A^\top ) u\}. \end{eqnarray*}

Thus, since $\varphi_Y(t)$ the c.f. of a random vector $Y$ has the form of a c.f. of a multivariate normal distribution, by the properties of the c.f. (check the wiki link if you do not know these properties), we have proved that

$Y\sim \text{N}_q(A\mu+c, A\Sigma A^\top)$.

Now look for $A$ and $c$ in your particular case and you are done.

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    $\begingroup$ +1. I know this has been written numerous times in various posts here but again this is comprehensive. $\endgroup$ May 31, 2023 at 8:41
  • $\begingroup$ Thank you. Edited my question with solution. $\endgroup$
    – Kombajn
    May 31, 2023 at 9:59
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  1. $C(Y_1,Y_2) = E[Y_1Y_2] - E[Y_1]E[Y_2] = E[Y_1Y_2]$
  2. $E[2Y_1Y_2] = E[X_1^2-X_2^2] = 0$ So $C(Y_1, Y_2) = 0$
  3. $f_{X_1,X_2}(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2)$ so idp.
Borrowing from wikipedia: $f_{X_1, X_2}(x,y) =$ $$ \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp \left( -\frac{1}{2\left[1 - \rho^2\right]}\left[ \left(\frac{x-\mu_X}{\sigma_X}\right)^2 - 2\rho\left(\frac{x - \mu_X}{\sigma_X}\right)\left(\frac{y - \mu_Y}{\sigma_Y}\right) + \left(\frac{y - \mu_Y}{\sigma_Y}\right)^2 \right] \right) $$

For the MVN $$ f(\mathbf{x})= \frac{1}{\sqrt { (2\pi)^k|\boldsymbol \Sigma| } } \exp\left(-{1 \over 2} (\mathbf{x}-\boldsymbol\mu)^{\rm T} \boldsymbol\Sigma^{-1} ({\mathbf x}-\boldsymbol\mu)\right) $$ $\Sigma$ becomes diagonal when correlations are zero so you get $f(x) = \prod_j f_j(x_j)$

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