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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.

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  • $\begingroup$ 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. $\endgroup$
    – Dave
    Oct 14 '19 at 2:40
  • $\begingroup$ @Dave corrected! But how can I use these info? $\endgroup$ Oct 14 '19 at 2:42
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Oct 14 '19 at 3:03
  • $\begingroup$ @Glen_b $Cov(X_1,X_2)=0$. Can I write this directly from the given info? $\endgroup$ Oct 14 '19 at 3:17
  • $\begingroup$ A covariance is not a joint distribution. Specifically, write the joint density function. $\endgroup$
    – Glen_b
    Oct 14 '19 at 3:19
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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 $\Sigma$.

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

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