# Bivariate normal distribution from independent random variables

Let $$X_1$$ and $$X_2$$ random variables such that $$X_1+X_2$$ and $$X_1-X_2$$ have independent standard normal distributions. Show that $$x=(X_1, X_2)$$ has a bivariate normal distribution.

My work:

Since $$X_1+X_2$$ and $$X_1-X_2$$ have independent standard normal distributions, so

$$X_1+ X_2 \sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2 )$$

$$X_1- X_2 \sim N(\mu_1-\mu_2,\sigma_1^2+\sigma_2^2 )$$

Also $$Cov(X_1,X_2)=0$$.

I don't know how to relate these to prove my result. I was wondering if you could help me! Thank you so much for your time.

• You’ve made a mistake with the variance of the $X_1 - X_2$ distribution. Variances add. I’ve also edited what looked like a typo that confused the HECK out of me for a minute. – Dave Oct 14 '19 at 2:40
• @Dave corrected! But how can I use these info? – Barsal Oct 14 '19 at 2:42
• Write the joint distribution of $(X_1+X_2,X_1-X_2)$ then do a change of variable. You should have an explicit bivariate normal. – Glen_b -Reinstate Monica Oct 14 '19 at 3:03
• @Glen_b $Cov(X_1,X_2)=0$. Can I write this directly from the given info? – Barsal Oct 14 '19 at 3:17
• A covariance is not a joint distribution. Specifically, write the joint density function. – Glen_b -Reinstate Monica Oct 14 '19 at 3:19

Let $$Z = X_1 + X_2$$ and let $$W = X_1 - X_2$$. From the question, both $$Z$$ and $$W$$ are standard normals. It is straight forward to show that

$$X_1 = \dfrac{Z+W}{2}$$ $$X_2 = \dfrac{Z-W}{2}$$

The mean of $$X_1$$ and $$X_2$$ is 0 (why?)

Because $$Z$$ and $$W$$ are standard normal and are assumed independent, then

$$\operatorname{Var}(X_1) = 0.25(\operatorname{Var}(Z) + \operatorname{Var}(W) + \operatorname{Cov}(Z,W)) = 0.25(1 + 1 + 0)= 0.5$$

A similar argument can be made for $$X_2$$.

We know $$X_1$$ and $$X_2$$ must be normal through the properties you've listed (that is, because $$Z$$ and $$W$$ are normal and $$X_1$$ and $$X_2$$ can be obtained from $$Z$$ and $$W$$ through addition/subtraction).

So we know three things:

• $$X_1$$ and $$X_2$$ are normal
• The mean of $$X_1$$ and $$X_2$$ is 0
• The variance $$X_1$$ and $$X_2$$ is 0.5

Can you finish from here?

Alternatively, consider $$Z$$ and $$W$$ as defined above and that they are standard normal. By definition of multivariate standard normals, the joint distribution of $$Z$$ and $$W$$ is multivariate standard normal. One can show that $$x = Ay$$ where $$x$$ is defined as above, $$y = (Z,W)$$ and

$$A = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix}$$

By definition, $$x$$ is multivariable normal with mean $$(0,0)$$, and with covariance matrix $$\Sigma = AA^T$$. If I remember correctly, $$A$$ is called a Cholesky Factor of $$\Sigma$$.

• @Demitri Can I write then, probability density function: $f(x,y)=\frac{1}{2\Pi(.5)(.5)} e^{-\frac{x_1^2}{.5}-\frac{x_2^2}{.5}} ?$ – Barsal Oct 14 '19 at 3:11
• @Barasal yea, good enough. If you found this answer helpful, please consider accepting it. – Demetri Pananos Oct 14 '19 at 3:33
• Thank you so much! – Barsal Oct 14 '19 at 3:36