4
$\begingroup$

  1. From here, it says that, linear combination of two Gaussian distribution, are always Gaussians.

  2. However, Let 𝑋 be standard normal and 𝜀=±1 with probability 1/2 each, independently of 𝑋. Let 𝑌=𝜀𝑋. Then 𝑌 is also standard normal, but 𝑍=𝑋+𝑌 is exactly equal to zero with probability 1/2 and is equal to 2𝑋 with probability 1/2.

But (2) contradicts with (1). Am I missing anythings?

Improve this question
$\endgroup$
  • $\begingroup$ interesting... You logics seem to be correct, but I still think both (1) and (2) are true.. $\endgroup$ – kuku Nov 30 '20 at 1:57
2
$\begingroup$

The simplest case is when you take linear combinations of independent Gaussian random variables, these will be Gaussian. However, $X$ and $Y$ are not independent. A more general case in which linear combinations of correlated Gaussian random variables are Gaussian is if $X$ and $Y$ are bivariate Gaussian (aka jointly Gaussian). You can read more about this difference between (1) $X$ and $Y$ are both Gaussian and (2) $X$ and $Y$ are bivariate (jointly) Gaussian here: https://en.wikipedia.org/wiki/Multivariate_normal_distribution

Essentially, there are many ways for variables to correlated. Being jointly Gaussian random variables is a special way for Gaussian variables to be correlated that implies this property we've discussed. In fact, jointly Gaussian variables can actually be defined by this property that all linear combinations are normally distributed.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Wait... (1) also holds for non independent cases. $\endgroup$ – Nygen Patricia Nov 30 '20 at 2:02
  • 1
    $\begingroup$ This holds if either independent random variables or if $X$ and $Y$ are bivariate Gaussian. Check out: en.wikipedia.org/wiki/… $\endgroup$ – Math Helper Nov 30 '20 at 2:04
  • 1
    $\begingroup$ Yes, I think you change the word "independent" into "bivariate" or "jointly" or "multivariate" because as it stands, the reasoning in your answer is not correct. (Linear combinations of non-independent Gaussians is also sometimes a Gaussian.) $\endgroup$ – E-A Nov 30 '20 at 2:06
  • 1
    $\begingroup$ Just a technicality, "You need" should read "In case" $\endgroup$ – VictorZurkowski Nov 30 '20 at 2:07
  • $\begingroup$ Thanks @E-A and @ VictorZurkowski, I've incorporated your feedback! $\endgroup$ – Math Helper Nov 30 '20 at 2:08
0
$\begingroup$

Linear combinations of jointly Gaussians (also known as multivariate Gaussians) are always Gaussian; however, X and Y are not jointly Gaussian. (One of the easiest ways to define a joint Gaussian $X$ is to say that $X = AZ + b$ where $A$ is some matrix, $b$ is some vector and $Z$ is a vector of i.i.d. Gaussians.)

Improve this answer
$\endgroup$
  • $\begingroup$ Sorry, what do you mean by jointly Gaussian. ...? $\endgroup$ – Nygen Patricia Nov 30 '20 at 2:05
  • $\begingroup$ I wrote one way in the parantheses. The canonical way to define a Gaussian is by its characteristic function, but you can use the alternate definition I wrote in the paranthesis. When the random variables admit a joint density, you can use the fact that the density of it has a certain form) $\endgroup$ – E-A Nov 30 '20 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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