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.


2 Answers 2


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.

  • 1
    $\begingroup$ +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
  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 $f(x) = \prod_j f_j(x_j)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.