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$X$ and $Y$ are independent and identically distribued (i.i.d.), $X+Y$ and $X-Y$ are independent, $\mathbb{E}(X)=0$ and $\mathbb{E}(X^2)=1$. Show that $X\sim N(0,1)$.

We should use characteristic functions to prove this. Any ideas?

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  • $\begingroup$ Sure. And yours? $\endgroup$ – Did Nov 7 '13 at 20:41
  • $\begingroup$ A good reference to someone with a lot of good ideas!? Bogachev! ..in his book Gaussian Measures; particularly section 1.9 $\endgroup$ – Tom Nov 7 '13 at 20:48
  • $\begingroup$ What if $X$ and $Y$ are discrete? $\endgroup$ – Michael Hoppe Nov 7 '13 at 21:56
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Denote by $\Phi(t) = \mathbb{E}e^{\imath \, t \cdot X}$ the characteristic function of $X$. We have

$$X = \frac{1}{2} \big((X+Y)+(X-Y) \big).$$

Thus,

$$\begin{align*} \Phi(t) &= \mathbb{E}e^{\imath \, \frac{t}{2} (X+Y)} \cdot \mathbb{E}e^{\imath \, \frac{t}{2} (X-Y)}= \left( \mathbb{E}e^{\imath \, \frac{t}{2} X} \right)^2 \cdot \mathbb{E}e^{\imath \, \frac{t}{2} Y} \cdot \mathbb{E}e^{-\imath \, \frac{t}{2} Y} \end{align*}$$

where we used the independence of $X-Y$ and $X+Y$ as well as the independence of $X$ and $Y$. By assumption, $X \sim Y$; therefore

$$\Phi(t) = \Phi \left( \frac{t}{2} \right)^3 \cdot \Phi \left( - \frac{t}{2} \right). \tag{1} $$

This shows that it suffices to determine $\Phi$ on $B(0,\varepsilon)$ for some $\varepsilon>0$. Since $\Phi(0)=1$, we can choose $\varepsilon>0$ such that $\Phi(B(0,\varepsilon)) \cap \{x+\imath \, y; x \leq 0, y \in \mathbb{R}\} = \emptyset$. For $t \in B(0,\varepsilon)$ we define

$$\psi(t) := \log \Phi(t)$$

Then $(1)$ reads

$$\psi(t) = 3 \psi \left( \frac{t}{2} \right) + \psi \left( - \frac{t}{2} \right). \tag{2} $$

Applying this to $-t$ yields

$$\psi(-t) = 3 \psi \left( - \frac{t}{2} \right) + \psi \left( \frac{t}{2} \right).$$

Subtracting the last two equalities we obtain

$$\delta(t) := \psi(t)-\psi(-t) = 2 \psi \left( \frac{t}{2} \right) - 2 \psi \left( - \frac{t}{2} \right) = 2 \delta \left( \frac{t}{2} \right).$$

Consequently,

$$\frac{\delta(t)}{t} = \frac{\delta \left( \frac{t}{2^n} \right)}{\frac{t}{2^n}}. \tag{3}$$

Note that $\Phi$ is twice differentiable and $\Phi'(0)=0$, $\Phi''(0)=-1$ since $\mathbb{E}X=0$, $\mathbb{E}(X^2)=1$. Therefore, $\delta$ and $\psi$ are also twice differentiable and we can calculate the deriatives at $t=0$ explicitely. From $(3)$, we find

$$\frac{\delta(t)}{t} \to \delta'(0) = 0\qquad \text{as} \,\, n \to \infty,$$

i.e. $\delta(t)=0$. By the definition of $\delta$ and $(2)$,

$$\psi(t) = 4 \psi \left( \frac{t}{2} \right).$$

By Hôpital's theorem,

$$\frac{\psi(t)}{t^2} = \frac{\psi \left( \frac{t}{2^n} \right)}{\left( \frac{t}{2^n} \right)^2} \to \frac{1}{2} \psi''(0) = - \frac{1}{2} \qquad \text{as} \,\, n \to \infty.$$

Hence, $$\psi(t) = - \frac{t^2}{2}.$$

Reference:

  • Rényi, A.: Probability Theory. (Chapter VI.5 Theorem 1)
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  • $\begingroup$ "From Φ(0)=1 we conclude by the intermediate value theorem that Φ(t)>0 for any t∈R"... Except that Φ is complex-valued, not real-valued a priori. $\endgroup$ – Did Nov 9 '13 at 1:24
  • $\begingroup$ @Did You are right. Should be fine now... $\endgroup$ – saz Nov 9 '13 at 7:41
  • $\begingroup$ I fail to see the construction of $(t_n)$. That $\Phi(t)^3\Phi(-t)$ is in $\mathbb R_-$ does not imply a priori that $\Phi(t)$ or $\Phi(-t)$ is in $\mathbb R_-$. // Note that the identity you are trying to use is $\Phi(2t)=\Phi(t)^2|\Phi(t)|^2$. $\endgroup$ – Did Nov 9 '13 at 7:50
  • $\begingroup$ @Did Yes, of course, sorry. $\endgroup$ – saz Nov 9 '13 at 9:33
  • $\begingroup$ Rrrright... +1. $\endgroup$ – Did Nov 9 '13 at 9:45

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