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

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  • 1
    $\begingroup$ Geometric series. That's a result in mathematics, rather than a statistical one. $\endgroup$
    – Glen_b
    Commented Jan 10, 2014 at 10:28

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Roland
    Commented Jan 10, 2014 at 10:54

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