# Geometric distribution, showing that $P(X>x)=(1-p)^{x}$ from the pmf definition

This may be a silly question but I can't seem to solve this for the life of me!

The question states:

Let the random variable $X$ have a Geometric distribution with parameter $p$ and probability mass function:

$p(x)=p(1-p)^{x-1}$ ; $x=1,2,3,...$

$X$ can represent the number of the trial on which the first success occurs in an infinite sequence of independent Bernoulli trials each with parameter $p$.

So from the definition of the probability mass function, I must show that:

$P(X>x)=(1-p)^{x}$ ; $x=1,2,3,...$

• Use the result $$1+y+y^2+\cdots = \frac{1}{1-y}~\text{for}~|y|<1$$ to sum $p(x+1)+p(x+2)+\cdots$. You will need to pull out a common factor to write the sum in a form where the formula above is visibly applicable. – Dilip Sarwate Oct 24 '13 at 1:41
$$P(X > x) = P(\hbox{all fails for the first x trials}) = (1 - p)^x.$$
\begin{align}\mathsf P(X>x) &=\mathbf 1_{x\in\Bbb Z^+}\sum_{k=1}^\infty \mathsf p(x+k)\\[1ex] &=\mathbf 1_{x\in\Bbb Z^+}\sum_{k=1}^\infty p~(1-p)^{x+k-1}\\[0ex] &~~\vdots\end{align}