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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?

Maximilian Janisch
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user106240
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1 Answers1

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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)
saz
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  • "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. – Did Nov 09 '13 at 01:24
  • @Did You are right. Should be fine now... – saz Nov 09 '13 at 07:41
  • 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 $\Phi(2t)=\Phi(t)^2|\Phi(t)|^2$. – Did Nov 09 '13 at 07:50
  • @Did Yes, of course, sorry. – saz Nov 09 '13 at 09:33
  • Rrrright... +1. – Did Nov 09 '13 at 09:45
  • What does $B(0, \epsilon)$ mean? –  Nov 10 '13 at 20:29
  • @user100463 It denotes the open ball of radius $\varepsilon$ centered at $0$, i.e. $B(0,\varepsilon) = {x \in \mathbb{R}; |x|<\varepsilon}$. – saz Nov 10 '13 at 20:37
  • is it possible to show that $\phi(t) > 0$, $\forall t\in R$ already at stage (1)? It seems reasonable looking at the conclusion and it would avoid the construction of the interval – Vittorio Apicella Aug 24 '14 at 13:55
  • @VittorioApicella If we knew that $\phi(t)$ is real-valued (...which is equivalent to $X \sim -X$...), then it $\phi(t)>0$ would follow easily from $(1)$. But I don't see how to prove this. – saz Aug 24 '14 at 14:44
  • @saz Why not simply wipe out that $\log$ part? Let $\phi(t)$ denote the characteristic function of $X$. Clearly $\phi(t)$ is non zero. Define $\gamma(t)=\frac{\phi(t)}{\phi(-t)}$. $$\gamma(t)=\gamma(\tfrac{t}{2})^2=\gamma(\tfrac{t}{2^n})^{2^n}\to \exp[t\lim_{x\to 0} \frac1x(\gamma(x)-1)]=1$$ Differentiability of $\gamma$ follows from that of $\phi$. Hence $\gamma(t)\equiv 1$. Then just observe $$\phi(t)=\phi(\tfrac{t}2)^4=\phi(\tfrac{t}{2^n})^{2^{2n}}\to \exp[t^2\lim_{x\to 0} \frac1{x^2}(\phi(x)-1)]=e^{-\frac{t^2}2}$$

    L'Hospital rule was used in both the ta calculate the limits.

     – Sayan Nov 02 '14 at 04:38
  • @Sayan Sounds correct. If you like to, you can present this approach in an answer; if this is not the case, I can add a short remark to my answer. – saz Nov 02 '14 at 07:27