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Let $X$ be a random variable and $X \in\mathbb Z.$ Show that $$\varphi_X(2\pi k) = \operatorname E\left(e^{i2\pi kX}\right)=1 $$ for $k \in\mathbb Z$

I tried to expand the expected value by its definition but it didn't help. Can anyone give me any idea or hint about this proof? Thanks in advance.

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    $\begingroup$ Since $X\in \mathbb Z$, $e^{i2\pi kX}$ is constant and equal to $1$. The expectation is therefore $1$. $\endgroup$ – Gabriel Romon May 3 '18 at 11:10
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Hint: $E[e^{i2\pi k X}] = E[\cos(2\pi k X) + i \sin (2\pi k X)]$.

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Recall

  1. $$e^{i2\pi kX} = \cos(2\pi kX) + i \sin(2\pi kX)$$

  2. $\cos$ and $\sin$ are periodic, i.e. $\forall k \in \mathbb Z$, $$\cos(0) = \cos(0 + 2\pi k)$$ and $$\sin(0) = \sin(0 + 2\pi k)$$

  3. $\mathbb Z$ is closed under multiplication, i.e. $$k, X \in \mathbb Z \to kX \in \mathbb Z$$

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