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Suppose that $\phi$ is a characteristic function.

How can I prove that $\phi^2$ and $|\phi|^2$ are characteristic functions?

I use the definition of characteristic:

$\phi_x^2(t)=(E(e^{itx}))^2=var(e^{itx})+E(e^{itx})$.

nikola
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  • Do you know the theorem that says something is a characteristic function if and only if......? – Shashi Dec 26 '17 at 12:49
  • your ask not clear? – nikola Dec 26 '17 at 13:21
  • I ask how would you show this? Do you know any theorems that may help? – Shashi Dec 26 '17 at 13:22
  • I guess @Shashi is referring to Bochner’s theorem, but I’ve never seen that actually being useful (compare the accepted answer to mine in the similar question at https://math.stackexchange.com/questions/2482460/show-that-int-01-phiyt-dy-is-a-characteristic-function-where-phi/2482788#2482788) – Dap Dec 26 '17 at 13:39
  • @Dap actually there are properties of the characteristic function that characterize the charecteristic function. The theorem goes like $\phi$ is a characteristic function if and only if ...... – Shashi Dec 26 '17 at 13:48
  • @Dap never mind. It is called Bochener's theorem, I see it. I was asking the question to OP to let the OP have a look at the theorems for characteristic function since the question lacked effort... – Shashi Dec 26 '17 at 13:52

1 Answers1

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$(E(e^{itX}))^2$ is $E(e^{it(X+X’})$ where $X’$ is an independent copy of $X.$ So the first one is the characteristic function of $X+X’.$

The second one is similarly the characteristic function of $X-X’.$

Dap
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  • thnx alot , but how you assume that $x^-$ is independent copy of x this allow?? – nikola Dec 26 '17 at 14:21
  • @nikola: yes, in measure theory terms this is justified by considering the product measure space. See also https://math.stackexchange.com/questions/250145/existence-of-iid-random-variables – Dap Dec 26 '17 at 14:53
  • just there exist question on my mind whom people answer the problem in this site ? what nautral of this site @Dap – nikola Dec 26 '17 at 19:32