$X$ and $Y$ are independent, identically distributed random variables with mean $0$, variance $1$ and characteristic function $\phi$, If $X+Y$ and $X-Y$ are independent, prove that $$\phi(2t)=\phi(t)^3\phi(-t).$$By making the substitution $\gamma(t)=\phi(t)/\phi(-t)$ or otherwise, show that, for any positive integer $n$,$$\phi(t)={\left\{1-\frac {1}{2}{\left(\frac{t}{2^n}\right)}^2+o\left({\left[\frac t{2^n}\right]}^2\right)\right\}}^{4^n}$$ Hence, find the common distribution of $X$ and $Y$.
I know how to do the first part. However for the second part, I have no idea at all. Can anyone teach me? Thanks.
Best Answer
From independence we have $\varphi_{X+Y+(X-Y)}=\varphi_{2X}$ and thus $$ \varphi(2t) = \varphi(t)^2(\varphi(t)\varphi(-t)) = \varphi(t)^3\varphi(-t). $$ Set $\gamma(t) = \varphi(t)/\varphi(-t)$, then $$ \gamma(2t) = \frac{\varphi(2t)}{\varphi(-2t)} = \frac{\varphi(t)^3\varphi(-t)}{\varphi(-t)^3\varphi(t)} =\gamma(t)^2. $$ It follows that $\gamma(t)=\gamma(t/2)^2$ and by induction, $\gamma(t) = \gamma(t/2^n)^{2^n}$ for nonnegative integers $n$. Moreover, $\varphi'(0)=i\mathbb E[X]=0$ and $\varphi''(0)=i^2\mathbb E[X^2] = -1$, so as $h\to0$ we have by Taylor expansions $$ \gamma(h) = \frac{\varphi(h)}{\varphi(-h)} = \frac{1-\frac12h^2+o(h^2)}{1-\frac12h^2+o(h^2)} = 1 + o(h)^2. $$ It follows that for large $n$, $$\gamma(t) = (1+o(t^2/2^{2n})^{2^n}\stackrel{n\to\infty}\longrightarrow1,$$ and so $\gamma\equiv1$, yielding $\varphi(t)=\varphi(-t)$. Therefore $\varphi(2t)=\varphi(t)^4$, hence $\varphi(t)=\varphi(t/2)^4$ and by induction, $$\varphi(t)=\varphi(t/2^n)^{4^n},n\geqslant 1.$$ Using Taylor expansions we see that for large $n$, $$ \varphi(t) = \left(1-\frac12\left(\frac t{2^n}\right)^2+o\left(\left(\frac t{2^n}\right)^2\right)\right)^{4^n}\stackrel{n\to\infty}\longrightarrow \exp\left(-\frac12 t^2\right) $$ and hence $X$ and $Y$ have standard normal distribution.