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The characteristic function of a random variable $X$ is given as $$\phi_X(t)=\frac{3+\cos(t)+\cos(2t)}{5}; $$ what is the distribution of $X$? I was thinking of the discrete random variable $X=0,1,2$ with mass $\frac{3}{5},\frac{1}{5},\frac{1}{5}$ respectively, but it's not true. Is the distribution of $X$ related to this discrete random variable? Thanks

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  • $\begingroup$ In complex exponential form you have $\phi(t) = \frac{3}{5} + \frac{e^{it} + e^{-it} + e^{2it} + e^{-2it}}{10}$. Does that help? $\endgroup$ – Ian Oct 20 '14 at 14:22
  • $\begingroup$ @Ian Thank you so much. It's still a discrete random variable, x=0,1,2,-1,-2 $\endgroup$ – LPS Oct 20 '14 at 14:52
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If you manage to write the characteristic function as a linear combination of exponentials, that is, $$\phi_X(t)=\sum_{j= 1}^N p_j e^{i\theta_j t} $$ then you have $\mathbb P (X=\theta_j)=p_j$ for each $j\in\{1,\dots,N\}$.

Using the formula in Ian's comment $$\phi_X(t) = \frac{3}{5} + \frac{e^{it} + e^{-it} + e^{2it} + e^{-2it}}{10},$$ we derive that $\mathbb P(X=0)=3/5$, $$\mathbb P(X=1)=\mathbb P(X=-1)=\mathbb P(X=2)=\mathbb P(X=-2)=\frac 1{10}. $$

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