# cumulative distribution function , cdf problem

I cannot understand how step 2 transformed to step 3, anybody help me please ???

• Geometric series. That's a result in mathematics, rather than a statistical one. Jan 10 '14 at 10:28

It should be written for clarity with the sings reversed:

$$F(i) = P(X=1) + ...+P(X=i) = p+ p(1-p)+p(1-p)^2+...+p(1-p)^{i-1} =$$ $$=p\Big[1+ (1-p)+(1-p)^2+...+(1-p)^{i-1}\Big]$$

I guess you recognize the well known formula, that for $0<a<1$

$$1+a+a^2 +...+a^{k-1} = \frac {1-a^k}{1-a}$$

which gives, since here $a= 1-p$,

$$F(i) = p\frac {1-(1-p)^i}{1-(1-p)} = p\frac {1-(1-p)^i}{p} = 1-(1-p)^i$$

The trick is to show that

$\left(1+ (1-p) + \cdots + (1-p^{i-1})\right)= \frac{(1-p)^i-1}{(1-p)-1}$.

In order to see that, consider the term

$\left(1+(1-p) + \cdots + (1-p)^{i-1} \right)\left((1-p)-1\right)$.

Expanding this product yields that every summand in the left bracket times $(1-p)$ cancels with its successor times $1$. All what remains is the first summand times $-1$ plus the last summand times $(1-p)$, i.e. $(1-p)^i-1$.

Edit: I wrote this when there was no answer given. It's nice to see that my answer neatly covers the well-known part of Alecos Papadopoulos' answer. I've deliberately explained this in an explicit way instead of citing Calculus.

• Glad to hear that. If you like a particular answer, upvoting it is the best way to that. If you think that an answer is 'the answer to your questions', you can accept it as an answer. Jan 10 '14 at 10:54