I have a parameter $q$ which is the probability of selecting a vertex (among $n$ vertices...) to be in a certain set.

I am constructing the sets in an iterative way, having the vertex $v_i$ be in the set with probability $\leq \left(1 - q\right)^i$ (I'm not going into details about the order of the vertices for which this is performed, as I don't think its vital).

By that I can tell that the expected size of my set is $\Sigma_i \left(1-q\right)^i$.

Now the part which baffles me is where it says: $$\Sigma_i \left(1-q\right)^i = O\left( \frac{1}{q} \right)$$

So I know $q$ represents the probability for a vertex to be selected. So $q<1$.... But as I recall the formula for the sum of a geometrical series is:

$$\frac{a_1 \cdot \left( \left(1-q\right)^i -1\right)}{1-q-1} = \frac{\left(1-q\right) \cdot \left( \left(1-q\right)^i -1\right)}{-q}$$

Seems more like $O\left( q^i \right)$ than $O\left( \frac{1}{q} \right)$.

This is a silly question, but I would like to know where I got confused...


1 Answer 1


There is an abuse of notation in your question: you are using $i$ as both the index for the summation and as the total number of terms in the sum (in the closed formula).

Anyway, the ratio of the geometric series is $1-q$, but you mixed it up with $q-1$ in the closed formula. For simplicity consider the infinite sum $\sum_{i=0}^{\infty} (1-q)^i$, which is clearly an upper bound to the quantity you are after (and only affect the result by a multiplicative constant): $$ \sum_{i=0}^{\infty} (1-q)^i = \frac{1}{1 - (1-q)} = \frac{1}{q}. $$


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