Geometric Distribution - Probability How we can proof that if $X$~$Geo(p)$
Than: $Pr(X>a) = (1-p)^a$
How can we proof it formally? There is an example for that to understand it intuitively?
 A: The geometric distribution gives the probability that the first occurrence of success requires $a$ independent trials, each having a success probability of $p$. Then the probability that the $a$th trial is the first success is $$P(X=a)=(1-p)^{a-1}p, $$
by independence.
Thus, \begin{align}
P\left(X>a\right)&=\sum_{k=a+1}^\infty P(X=k)\\ &=\sum_{k=a+1}^\infty (1-p)^{k-1}p \\ &=\sum_{k=1}^\infty (1-p)^{k-1}p \ - \sum_{k=1}^a(1-p)^{k-1}p  \\ &= \frac{p}{1-(1-p)}-\frac{p\left(1-(1-p)^a\right)}{1-(1-p)} =(1-p)^a.
\end{align}
A: To proof it formally you should know the formula for the partial sum of a geometric series. The pmf of X is
$$P(X= a)= p(1-p)^{a-1}= pq^{a-1} \quad \forall \ \ a=1,2, \dots$$
And the cdf is
$$P(X \leq a) = p\sum_{i=1}^a q^{i-1}\quad \forall \ \ a=1,2, \dots$$
For a discrete random variable $X$ we have $P(X>a)=1-P(X\leq a)$.
Thus $P(X>a)=1-p\sum\limits_{i=1}^a q^{i-1}=1-\frac{p}q\sum\limits_{i=1}^a q^{i}$
The sum is a partial sum of a geometric series: $\sum\limits_{i=1}^a q^{i}=q\cdot \frac{q^a-1}{q-1}$. Use this and the fact that $p=1-q$ to obtain $P(X>a)=(1-p)^a$
A: What does $\{X>a\}$ mean? It means that the first $a$ trials were failures. Also you know that the trials were independent. Therefore
$$
Pr(X>a) = (1-p)^a.
$$
If you want to derive it from the probability mass function: $Pr(X=k)=p(1-p)^{k-1}$ for $k=1,2,\ldots$, then
$$
Pr(X>a) =p \sum_{k=a+1}^{\infty}(1-p)^{k-1} = (1-p)^a.
$$
