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I am now reading the following theorem

$\phi$ is the characteristic function and $\phi (t)=E(e^{itX})$, If $\phi ^{(2k)}(0)$ exists and is finite, then $E(|X|^{2k})< \infty $

Proof: We will prove this for $k=1$ and use the induction on $k$. Suppose $\phi ''(0)=A$ for some finite $A$. Then $$A=\lim_{h \to 0} \frac{1}{h^{2}}(\phi (h)-1+ \phi(-h))$$

Could someone help me explain how we get this formula? I know $\phi (0)=1$, but I don't understand how to use the definition of derivative to get this.

Thanks in advance!

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  • The statement as such is wrong. You want a -2 in that limit instead of -1. The correct statement can be proven using L'Hospital. It's an exercise in Rudin, so I'm sure it's been done on this site before. – Thorgott Jan 13 '20 at 03:34
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    Read up the limit formula (via central differences) for the second derivative of a function. See here e.g. : https://math.stackexchange.com/questions/210264/second-derivative-formula-derivation and furthermore correct the error , it is $-2$ not $-1$ in the limit above. – Sarvesh Ravichandran Iyer Jan 13 '20 at 03:37

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