Bayesian inference with geometric distribution, verify $p(y > r|\theta) = (1-\theta)^r \text{ , where } r = 1,2,...$

Question

The geometric distribution is given by: $$p(y|\theta) = (1-\theta)^{y-1}\theta \text{ , where } y=1,2,..., 0 < \theta < 1$$ Verify that: $$p(y > r|\theta) = (1-\theta)^r \text{ , where } r = 1,2,...$$

How does one actually go about doing this? I tried a few integrals with different limits across $ (1-\theta)^{y-1}\theta$ but that did not make too much progress. I can't find any similar examples in course literature. Any hints to get started appreciated.

Answer 1

Guide:

Simplify\begin{align} P(Y>r|\theta) &= \sum_{y=r+1}^\infty (1-\theta)^{y-1} \theta \end{align}

perhaps using geometric series.