I'm trying to solve the following exercise:
Let $\mu$ be a probability distribution on $\mathbb{R}$ having second moment $\sigma^2<\infty$ such that if $X$ and $Y$ are independent with law $\mu$ then the law of $(X+Y)/\sqrt{2}$ is also $\mu$. Show that $\mu =\mathcal{N}(0,1)$ Hint: apply the central limits theorem to packs of $2^n$ variables
My attempt:
So let $Z_n=(X_1,Y_1)+\cdots+(X_n,Y_n)$, then $\mathbb{E}(Z_n)=n\mathbb{E}(Z_n)$ then for $n\to \infty$ $$T_n=\frac{Z_n-n\mathbb{E}(Z_n)}{\sqrt{\sigma^2n}}\xrightarrow{\mathcal{D}}\mathcal{N}(0,1)$$ converges in distribution to the normal distribution.
Now I don't see the connection how to proof $\mu=\mathcal{N}(0,1)$. I also do not understand what "packs" of $2^n$ variables are. Is it $Z_n=(X_n,Y_n)+\dots$?
I think it must be proved that $\mu=\mathcal N(0,\sigma^2)$ but for convenience I will also preassume that $\sigma=1$
If $\phi$ denotes the characteristic function then:$$\phi(t)=\phi\left(\frac{t}{\sqrt2}\right)^2$$
Note that this can be repeated to arrive at $\phi(t)=\phi(\frac{t}2)^4$ and can be repeated again.
Actually with this it can be shown that $X$ and $2^{-\frac12n}(X_1+\cdots+X_{2^n})$ have equal distribution if the $X_i$ are independent and all mentioned random variables have distribution $\mu$.
It is not really necessary to use the characteristic function to come to this conclusion. You could just observe that $([X_1+Y_1]/\sqrt2+[X_2+Y_2]/\sqrt2)/\sqrt2$ again has $\mu$ as distribution, and so on - if the $X_i$ and $Y_i$ are independent and have $\mu$ as distribution.
Also we have $0$ as expectation, since $\nu=(\nu+\nu)/\sqrt2$ implies $\nu=0$.
Applying the CLS on the $X_i$ you will find that the constant $\mu$ must convergence in distribution to $\mathcal N(0,1)$.
This can only be the case if $\mu=\mathcal N(0,1)$.