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I've been trying to prove the following and I'm really stuck.

Consider $N\sim \mathrm{Geometric}\left(p\right)$, and let $X=(-1)^N$. Show that the probability mass function of $X$ is given by

$$ p_X\left(k\right) = \begin{cases} \frac{1}{2-p}, \qquad k=1,\\ \frac{1-p}{2-p},\qquad k=-1. \end{cases} $$

I've managed to compute the expected value of $X$, $\mathbb{E}\left[X\right] = \frac{p}{2-p}$, and the variance as well, $\mathrm{Var}\left(X\right) = \frac{4\left(1-p\right)}{\left(2-p\right)^2}$. But I don't know if these values are helpful or not.

I am looking for hints. Thank you very much in advance.

air
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MrPoisson
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  • Did you calculate the expected value and variance from the definition of $X$ or from the pmf? – air Sep 08 '15 at 23:36

1 Answers1

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$X=-1$ when $N$ is odd and $1$ when $N$ is even. $N$ is odd if it is $1,3,5,\dots=2k-1$ for a $k \in \mathbb{N}$. So using the geometric PMF:

$$P[N \text{ is odd }]=\sum_{k=1}^\infty (1-p)^{2k-2} p.$$

Now this is a geometric sum (after you factor out some terms). (I hope my convention for the geometric distribution coincides with yours...)

Ian
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