## 3 Materials and Methods
### Theoretical Study  
#### 3.2.1 Expected Number of Impressions  
In this section, we focus on investigating the probability of getting new impressions. This probability is crucial because it controls the advertiser's chances of sending a new impression to the user. If this probability is always high, we will be able to send many impressions to the user, while if the probability is small, the user will receive few impressions, and the advertiser may not be able to send him/her more. Note that this probability is out of the advertiser's control, and it could become even equal to zero. The parameter the advertiser can configure to make this probability equal to 0 (stop sending impressions) is the \(FC\).  
Let us study the most basic possible scenario where we assume that the probability of getting a new impression is constant (\(p_{I}=c\)). Under this assumption, we will understand the number of impressions that each user is expected to receive.  
If we call \(X\) the variable of the number of received impressions. Then, \(P(X)\) is the probability of getting \(X\) impression, and we have \[P(X\leq 1) =(1-p_{I})\] \[P(X\leq 2) =(1-p_{I})+p_{I}(1-p_{I})\] \[P(X\leq 3) =(1-p_{I})+p_{I}(1-p_{I})+p_{I}^{2}(1-p_{I})\] \[P(X\leq k) =(1-p_{I})\sum_{i=0}^{k-1}p_{I}^{i}=(1-p_{I})\frac{1-p_{I}^{k-1}}{ 1-p_{I}}=1-p_{I}^{k-1}\;\;\forall k\geq 1\]  
using the complementary rule, we find that  
\[P(X\geq 1)=1\]  
\[P(X\geq k)=1-P(X\leq k-1)=1-(1-p_{I}^{k-2})=p_{I}^{k-2}\;\;\forall k\geq 2\]  
by definition, the expected value is  
\[E(X)=\sum_{i=1}^{\infty}x_{i}p_{i}\]  
where \(x_{1},x_{2},\cdots\) are the set of outcomes, occurring with probabilities \(p_{1},p_{2},\cdots\), respectively.  
Then, in our case, we have that  
\[E(X)=1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)+4\cdot P(X=4)+\cdots\]  
that is the same as  
\[E(X)=P(X\geq 1)+P(X\geq 2)+P(X\geq 3)+P(X\geq 4)+\cdots\]  
\[E(X)=\sum_{i=1}^{\infty}P(X\geq i)\]  
if we make the substitution, we have that  
\[E(X)=1+\sum_{i=2}^{\infty}p_{I}^{i-2}=1+\sum_{i=0}^{\infty}p_{I}^{i}=1+\frac{1 }{1-p_{I}}\]  
This is for the case when we do not define any \(FC\). In case we want to study this scenario with a \(FC\) value, we have the following formula:  
\[E(X)=1+\sum_{i=0}^{n}p_{I}^{i}=1+\frac{1-p_{I}^{n}}{1-p_{I}}\]  
To illustrate the differences between the two cases we can use a few examples. For instance, in the first example, we assume that the probability of the next impression is very high (\(p_{I}=0.9\)) and we will compare the cases of \(FC=0\), \(FC=10\), and \(FC=45\). For each of these values, we will have an expected number of impressions equal to  
\[\text{For }FC=0\text{: }E(X)=1+\frac{1}{1-0.9}=1+\frac{1}{0.1}=11\text{ Impressions}\]  
\[\text{For }FC=10\text{: }E(X)=1+\frac{1-0.9^{10}}{1-0.9}=7.51\text{ Impressions}\]  
\[\text{For }FC=45\text{: }E(X)=1+\frac{1-0.9^{45}}{1-0.9}=10.91\text{ Impressions}\]  
As we observe, considering a very high new impression probability of 0.9, the number of impressions we expect a user receives is 11 in case the \(FC\) is not used; therefore, most users will not receive a massive number of impressions. Our example also shows that if we use high \(FC\) values such as 45, then the expected number of impressions users receive is almost the same as if we were not setting up the \(FC\). However, the number of impressions considerably decrease with a \(FC=10\).  
Let us use another scenario where the probability of new impressions is small (\(p_{I}=0.1\)) and comparing the cases of not having \(FC\) and \(FC=10\)  
\[\text{Not }FC:E(X)=1+\frac{1}{1-0.1}=1+\frac{1}{0.9}=2.11\text{ Impressions}\] \[FC=10:E(X)=1+\frac{1-0.1^{10}}{1-0.1}=2.11\text{ Impressions}\]  
We can see that in this case, the number of expected impressions is so small that a \(FC\) equal to 10 will not restrain the number of impressions that users will receive on average.