The random variables $X_1,...,X_n,...,$ are i.i.d. and have the p.m.f. $p_X(-1)=\frac{1}{4}$, $p_X(0)=\frac{1}{2}$, and $p_X(1)=\frac{1}{4}$.

We define the random variable $N$ by $N=\mathrm{min}(n|X_n=0)$.

I have successfully proven that $N\in \mathrm{Fs}(1/2)$.

Now I want to show that the characteristic function of $S_N=\sum_{k=1}^{N}X_k$ is $\phi_{S_N}(t)=\frac{1}{2-\mathrm{cos}(t)}$.

My first thought was to use $\phi_{S_N}(t)=g_N(\phi_{X}(t))$, but this composition formula requires that $N$ is independent of $X_k\forall k\geq 1$.

But since $N$ is dependent on the $X$'s I don't know how to proceed. Any recommendations?


1 Answer 1


We have \begin{align*} \mathbb{E}\exp(it S_N) & = \sum_{n=1}^{\infty} \mathbb{E}(1_{\{N=n\}} \exp(itS_n)) \\ &= \sum_{n=1}^{\infty} \mathbb{E} \left(e^{it X_n} 1_{\{X_n=0\}} \prod_{k=1}^{n-1} 1_{\{X_k \neq 0\}} \exp(itX_k)\right). \end{align*}

By the independence of the random variables, this gives

$$ \mathbb{E}\exp(it S_N) =\sum_{n=1}^{\infty} \underbrace{\mathbb{P}(X_n=0)}_{1/2} \prod_{k=1}^{n-1} \underbrace{\mathbb{E}(\exp(it X_k) 1_{\{X_k \neq 0\}})}_{=\frac{1}{2} \cos(t)} = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{\cos(t)}{2} \right)^{n-1}.$$

Finally, note that the series on the right-hand side is a geometric series; it can be computed explicitly.

  • $\begingroup$ Thank you! Such a clever way to work around the dependency issue! $\endgroup$
    – lognormal
    Sep 10, 2019 at 15:42

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