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Let $X$, $Y$ be i.i.d, that $X+Y$ and $X-Y$ are independent, and that $\varphi_{X}(2u)=(\varphi_{X}(u))^{3}\varphi_{X}(-u)$. Also, let $E\{X\}=0$ and $E\{X^{2}\}=1$. Show that $X$ is Normal $N(0,1)$.

We are given a hint that reads as follows: "Show that for some $\alpha>0$, we have $\phi(u)\neq 0$ for all $u$ with $|u|\leq a$. Let $\psi (u)= \frac{\varphi(u)}{\varphi(-u)}$ for $|u|\leq a$, and show $\psi(u)=\{\psi(u/2^{n})\}^{2^{n}}$; then, show this tends to 1 as $n \to \infty$. Deduce that $\varphi(t)=\{\varphi(t/2^{n})\}^{4^{n}}$, and let $n \to \infty$.

Now, I've done all this, and you wind up with $\varphi(t)\to 1$ as $n \to 0$ by definition of characteristic functions.

WHAT I WANT TO KNOW IS THIS: how does doing all of this tell us that $X$ is Normal $N(0,1)$? At this point, this is the only part of the proof that I have left.

1 Answers1

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I have to disagree that $\varphi(t) = 1$ for any $t$, since this would be the characteristic function for $X = 0$ a.e. However, suppose that you can conclude that for every $n$, $\varphi(t) = \big( \varphi(t/2^n) \big)^{4^n}$. Perhaps it might be helpful to realize that $\varphi'(0) = 0$ and $\varphi''(0) = -1$. Then $$ \varphi(t/2^n) = \varphi(0)+ \varphi'(0)\frac{t}{2^n} +\frac{1}{2}\varphi''(0) \frac{t^2}{4^n} + O(8^{-n}) = 1 - \frac{t^2}{2\cdot 4^n} + O(8^{-n}) $$ and therefore $$ \varphi(t) = \lim_{n \to \infty} \left( 1 - \frac{t^2}{2\cdot 4^n} + O(8^{-n}) \right)^{4^n} $$ This limit should result in the characteristic function for $N(0,1)$.

Tom
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  • Do you mean for every $n$ that $\psi(t)=\left[\varphi\left(\frac{u}{2^{n}}\right)\right]^{4^{n}}$ rather than $\varphi(t)=$? –  Nov 12 '13 at 02:30
  • No. I'm writing $\varphi$ for the characteristic function $\varphi_X$. You want $\varphi_X(t) = e^{-\frac{1}{2}t^2}$, which will prove that $X =_d N(0,1)$. So, as you wrote above in the question, if you can deduce that $\varphi_X(t) = [\varphi_X(t/2^n)]^{4^n}$ you can follow my suggested solution to finish what remains. – Tom Nov 12 '13 at 02:35
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    Also, $\varphi_{X}^{\prime\prime}(0)=-1$, not $1$. –  Nov 12 '13 at 02:36
  • You're right! Sorry, the characteristic function is $e^{-\frac{1}{2}t^2}$ not $e^{\frac{1}{2}t^2}$. I'll edit and fix the mistake! – Tom Nov 12 '13 at 02:39
  • I'm afraid I can't tell exactly what you did wrong from that one expression. If for every $n$ it happened that $\psi(u) = \big(\varphi(u/2^n) \big)^{4^n}$ then you could apply my answer to $\psi(u)$, but you want $\psi(u) = 1$. – Tom Nov 12 '13 at 03:02
  • It's okay; I figured it out and fixed it. I'm really sorry to keep bothering you, but there's just one last thing I can't get to work, and that's showing that $\lim_{n \to \infty}(1-\frac{u^{2}}{2\cdot 4^{n}}+O(8^{-n}))=exp(-\frac{1}{2}t^{2})$? I tried using logs, and then L'Hopital, but I wound up with a big mess in the denominator. Any hints? –  Nov 12 '13 at 04:08
  • No worries. Consider $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{c}{x^{3/2}} \right)^{x}$$. Before doing this, use l'Hopital's rule applied to $$x \log\left(1 + \frac{k}{x} + \frac{c}{x^{3/2}} \right)$$ then, deduce from this that $$\lim_{x \to \infty} \left( 1 + \frac{k}{x} + \frac{c}{x^{3/2}} \right)^{x} = e^k$$. Now, since $4^n \to \infty$ as $n \to \infty$, replace $x$ with $4^n$ to conclude your result. – Tom Nov 12 '13 at 04:17
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    I got it to work! Thank you so much; you're the coolest! –  Nov 12 '13 at 14:37