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Question

Let $X$ and $Y$ be i.i.d with means $0$ and variances $1$. Let $\phi(t)$ be their common characteristic function and suppose that $X+Y$ and $X-Y$ are independent. Show that $\phi(2t)=\phi(t)^3\phi(-t)$ and deduce that $X$ and $Y$ are standard normal random variables.

The above question is from Grimmett and Stirzaker.

My attempt

I was able to show the first part but unable to fully justify the second part. For the first part here is a proof. Note that $$ \begin{align} \phi(2t)=Ee^{it2X}&=E\exp\{it(X+Y+X-Y)\}\\ &=E\exp\{it(X+Y\}E\exp\{it(X-Y\}\\ &=E\exp(itX)E\exp(itY)E\exp(itX)E\exp(-itY)\\ &=\phi(t)^3\phi(-t) \end{align} $$ by the independence assumptions.

My Problem

For the second part, it is easy to show that $e^{-t^2/2}$ (the cf of a standard normal) satisfies the equation $\phi(2t)=\phi(t)^3\phi(-t)$ but I am unable to show that this is the only choice of $\phi$ that satisfies this equation. I tried taking derivatives and setting up a differential equation but it got messy.

Any help is appreciated.

Sri-Amirthan Theivendran
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  • Note that $e^t$ is also a function that satisfies that functional equation. So you'll have to use other properties of a characteristic function to narrow down the set of possibilities. – zoidberg Dec 10 '18 at 00:49
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    A hint: First try to show that $\phi(t) = \phi(-t)$ by considering the ratio $\rho(t) := \phi(t)/\phi(-t).$ The natural strategy to do this also implies a strategy for showing the normality. Further hint: The relation $\rho(2t) = \rho^2(t)$ is obvious. This implies that $\rho(t) = (\rho(t/2^n))^{2^n}$. Now what can you say about $\rho(h)$ for small values of $h$? – stochasticboy321 Dec 10 '18 at 02:54

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