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Determine the distribution of $X$ if

$$\varphi(t) = \frac{2}{3e^{it} - 1},$$

for $t\in \mathbb{R}$ is its characteristic function.

I tried to use the inverse formula, which says that the density is given by

$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-itx} \cdot \frac{2}{3e^{it} - 1} \mathop{dx} $$

$$= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{-itx}}{3e^{it} - 1} \mathop{dx} $$

But I have no idea how to compute this integral. Can someone please help me?

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$\phi (t)=\frac 2 3 e^{-it} (1-\frac 1 3 e^{-it})^{-1}$. Expand this as $\sum\limits_{k=0}^{\infty} \frac 2 3 e^{-it} (\frac 1 3 e^{-it})^{n}=\frac 2 3(e^{-it}+\frac 1 3 e^{-2it}+\frac 1 {3^{2}} e^{-3it}+...$). From this you can conclude the random variable takes the values $-1,0,1,...$ with probabilities $\frac 2 3 , \frac 2 {3^{2}},...$.

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