Partial sum of geometric distribution

Question

I am stuck with the following exercise, from Probability by Grimmett and Welsh.

If $X$ is a discrete random variable having the geometric distribution with parameter $p$, show that the probability that $X$ is greater that $k$ is $(1-p)^k$.

My idea: the required probability should be given by: $$ P=\sum_{i=k}^\infty p(1-p)^i $$ but how to sum this series? The only thing that came to my mind is: $$ P = p\sum_{i=k}^\infty (1-p)^i = p \left(\frac{1}{p}-\sum_{i=0}^{k-1}(1-p)^i\right). $$ But still I do not know how to sum the right hand side series.

Answer 1

$1-p<1$ then as you wrote

$$ p \sum_{i=k}^{+\infty}\left(1-p\right)^{i}=p\frac{\left(1-p\right)^k}{1-\left(1-p\right)}=p\frac{\left(1-p\right)^{k}}{p}=\left(1-p\right)^{k} $$

Reminder : If $\left|x\right|<1$

$$ \sum_{n=p}^{+\infty}x^n=\frac{x^p}{1-x} $$ It comes from $$ \sum_{n=p}^{N}x^n=\frac{1-x^{N-p+1}}{1-x}x^p \ \text{ (Partial Sum of geometric sequences )} $$ And $$ x^{N-p+1} \underset{ N \rightarrow+\infty}{\rightarrow}0 \text{ because } \left|x\right|<1 $$ Hence the result.

Answer 2

A trick for summing geometric series when the index does not begin at 0: Factor out the first term.

$\Sigma_{i=k}^{\infty} r^i=r^k+r^{k+1}+r^{k+2}+...$

Factoring out the $r^k$ turns this into:

$\Sigma_{i=k}^{\infty}r^i=r^k\left(1+r+r^2+...\right)$

The sum in the parentheses is your standard geometric series, which sums to $\frac{1}{1-r}$.

The value of the sum is thus: $\Sigma_{i=k}^{\infty}r^i=\frac{r^k}{1-r}$

All of this of course assumes that $-1<r<1$, which is definitely true in your case, since $r=1-p$