Equal in distribution, linear combination of random variables

Question

I have a problem that I am working on, and I am trying to figure out if there is more information I can say about it.

I have that $X, Y, Z$ are iid and that $\frac{X + Y+ Z}{\sqrt{3}} \overset{d}{=} X$, and $E(X^2) = 1$. I need to find the distribution of X.

It seems clear that

$$\frac{X + Y+ Z}{\sqrt{3}} \overset{d}{=} X$$ $$\Rightarrow X + Y+ Z \overset{d}{=} \sqrt{3}X$$

Hence, for expected value we have $$E(X) = \mu = E(Y) = E(Z)$$ $$E(X+Y+Z) = \sqrt{3} E(X)$$ $$3\mu= \sqrt{3} \mu \Rightarrow \mu = 0$$

And for the Variance of X we can see that $$Var(X) = E(X^2) - (E(X))^2$$ $$Var(X) = 1 - \mu^2 = 1$$ $$Var(X) = Var(Y) = Var(Z) = 1$$ $$Var(\frac{X+Y+Z}{\sqrt{3}}) = \frac{1+1+1}{3} = 1$$

Therefore $E(X) = 0,$ and $Var(X) = 1.$ It seems like $X$ could be a standard normal distribution, but clearly we do not know that for sure. Is there anything else that I can say about it?

Answer 1

Use CLT. Let $X_1,X_2,\ldots$ is a sequence of iid random variables with the same distribution as $X$, $S_n=X_1+\ldots+X_n$. Then $$ \frac{S_3}{\sqrt{3}} \overset{d}{=} X, $$ $$ \frac{S_9}{\sqrt{9}} = \frac{\frac{X_1+X_2+X_3}{\sqrt{3}}+\frac{X_4+X_5+X_6}{\sqrt{3}}+\frac{X_7+X_8+X_9}{\sqrt{3}}}{\sqrt{3}} \overset{d}{=} X, $$ and so on. So, for any $n$, $$ \frac{S_{3^n}}{\sqrt{3^n}} \overset{d}{=} X. $$ CLT implies that the distribution of l.h.s. converges to standard normal. Since it coincides with the distribution of $X$, then $X\sim N(0,1)$.

Answer 2

I have not full solution to this problem. I will appreciate if you solve this to the end.

Let $\phi(t)$ - characteristic function of $X$. We have:

$$ X + Y + Z = \sqrt{3} X$$

Hence:

$$\phi^3(t) = \phi(\sqrt{3}t) \ \ (*)$$

So we need to solve the functional equation for characteristic function, but I have no idea how to solve it. Anyway, we can see that $\phi(t) = e^{-\frac{1}{2}t^2}$ - is solution $(*)$. So one of the solutions is standard normal distribution.