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I have the intuition that it is, but I have no idea on how to show it.

More formally, let $X$, $Y$ be two independent random variables, with

$X \sim \mathcal{N}(\mu_1,\,\sigma_1^{2})$

$Y \sim \mathcal{N}(\mu_2,\,\sigma_2^{2})$

and in general $\mu_1 \neq \mu_2$, $\sigma_1 \neq \sigma_2$, and all of them are different from zero.

Let $Z$ be defined as $Z = f(X,Y)$, with $f:\,R^2 \rightarrow R$.

Is the linear combination (i.e. $f(X,Y) = aX + bY$, with $a,b \in R$) the only $f$ for which $Z \sim \mathcal{N}(\mu_3,\,\sigma_3^{2})$ ?

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    $\begingroup$ You mean linear combinations, not just the sum, right? $\endgroup$ Aug 28, 2018 at 11:47
  • $\begingroup$ yes, I should have specified, my bad... I guess I meant more the type of operation. $\endgroup$
    – J. R. C.
    Aug 28, 2018 at 11:49
  • $\begingroup$ @J.R.C. Then please edit your question. $\endgroup$ Aug 28, 2018 at 11:52
  • $\begingroup$ is it ok like this? (sorry, I'm quite inexperienced here) $\endgroup$
    – J. R. C.
    Aug 28, 2018 at 11:55
  • $\begingroup$ @J.R.C. Sure. But the answer to your original question is "no", since a linear combination of X and Y is normally distributed as well. $\endgroup$ Aug 28, 2018 at 11:57

1 Answer 1

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The answer is Yes.

Assume that $X$ and $Y$ are two independent and normal random variables that are not necessarily identically distributed. Then they are jointly normal (because they are independent), which is important here.

Then the only function of $X$ and $Y$ leading to a normal random variable has to be linear; i.e has the form $$ f(X,Y) : = aX + bY$$ for some constants $a$ and $b$.

This result is known as Cramér–Wold theorem (see for example: this lecture note). It is a consequence of the uniqueness of characteristic functions and the particular form of the characteristic function of normal distributions.

Another way to see this is to note that if $f (X,Y)$ is normal with positive variance, you must be able to write it in terms of a standard normal random variable $N$ by whiting (scaling and shift). But you can always do this also for the given random variables $X $ and $Y$.

For some counter examples when $X$ and $Y$ are not jointly normal, see the answer of this question: Are any linear combination of normal random variables, normally distributed?

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  • $\begingroup$ I don't understand how to apply the Cramér-Wold theorem in relation to this result. If I understand correctly, the theorem states that the distribution of a random n-vector is completely determined by all its one-dimensional projections. So in this case, because I have a vector of two random variables I have to consider all combinations $t_1X + t_2Y$, which we know that are normally distributed. But how does this help me to show that only the linear combination preserves the normal distribution? $\endgroup$
    – J. R. C.
    Aug 28, 2018 at 15:26
  • $\begingroup$ I can see your point regarding using only Cramer-Wold theorem. But If normal distributions are closed under linear maps, doesn't this mean that we do not need any other functions to describe normal variables, because any other function should be linear by closure? $\endgroup$
    – user144410
    Aug 28, 2018 at 16:00
  • $\begingroup$ Maybe another approach could be helpful. The question is if there any nonlinear function that will lead to a normal variable. Assume that you have only one normal random variable $X$ and assume an arbitrary function $f$, such that $Z= f(X)$. Assume that $Z$ is normal. Then you can show a contradiction by applying the change of variable formula: en.wikipedia.org/wiki/… because the determinant of the derivative will be independent of $Z$ only if $f$ is linear. Otherwise, its clear that you get a contradiction. $\endgroup$
    – user144410
    Aug 28, 2018 at 16:24
  • $\begingroup$ That approach seems more helpful to me indeed. I was going to reply to your previous comment saying that I'm not sure how to show that the other side of the arrow follows: you are saying that linear transformation $\Rightarrow$ closed, but why does that imply closed $\Rightarrow$ linear transformation? For example, a set can be part of different groups under different operations, right? $\endgroup$
    – J. R. C.
    Aug 28, 2018 at 16:31
  • $\begingroup$ You are right; what I wrote means that linear functions are sufficient. the second arrow only refer to existence.. there exists a linear transformation. $\endgroup$
    – user144410
    Aug 28, 2018 at 17:48

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