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$X$ 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.

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    $\begingroup$ Not sure yet how to solve it but you can start from the observation that for every positive integer $n$, $\gamma(2^n t) = \gamma(t)^{2n}$ $\endgroup$
    – Ant
    Aug 15, 2018 at 9:08
  • $\begingroup$ @Ant Just realize how you get that observation and remember it appeared somewhere in the textbook. I will see how... $\endgroup$ Aug 15, 2018 at 9:19
  • $\begingroup$ For the case of a symmetric distribution this is fairly easy. Expanding the characteristic function around $0$ and taking the limit shows that the common distribution is normal. How one can use $\gamma $ to handle non-symmetric case beats me. $\endgroup$ Aug 15, 2018 at 9:31
  • $\begingroup$ @Ant I figured out the answer. Thanks. $\endgroup$ Aug 15, 2018 at 10:21

2 Answers 2

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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.

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    $\begingroup$ For the part to show that $\gamma(t)=1+o(t^2)$ you can actually differentiate it once and plug in zero to show that it does not have t term. $\endgroup$ Aug 15, 2018 at 10:27
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The substitution $\gamma(t)$ has the following properties:$$\gamma(t) = \phi(t)/\phi(-t)=\frac{\phi^3(t/2)\phi(-t/2)}{\phi(t/2)\phi^3(-t/2)}=\gamma^2(t/2)$$ therefore,$$\gamma(t)=\lim_{n\to\infty}{\gamma^{2^n}(t/2^n)}$$ It is easy to show that $$\gamma(0)=1$$ $$\gamma'(0)=0$$ That means the maclaurin expansion as $t\to0$ will be $$\gamma(t)=1+o(t^2) $$ thus $$\gamma(t)=\lim_{n\to\infty}{{\left[1+o(t^2/4^n)\right]}^{2^n}}=1$$ Hence, $$\phi(t)=\phi(-t)$$ And further that $$\phi(2t)=\phi^4(t)$$ Therefore, $$\phi(t)=\phi^{4^n}(t/2^n)$$ We know that $$\phi(t)=1+E(X)ti-\frac{E(X^2)}2t^2+o(t^2)=1-\frac{t^2}2+o(t^2)$$ then we get that long expression. By taking limit on it we get $$\phi(t)=\lim_{n\to\infty}{\left\{1-\frac12\frac{t^2}{4^n}\right\}^{4^n}}=e^{-\frac12t^2}$$

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  • $\begingroup$ Very nice work! :-) $\endgroup$
    – Ant
    Aug 15, 2018 at 10:23

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