# Calculate joint distribution from marginal distributions

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.

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. I know this has been written numerous times in various posts here but again this is comprehensive. Commented May 31, 2023 at 8:41
• Thank you. Edited my question with solution. Commented 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 $$f(x) = \prod_j f_j(x_j)$$