[1]

Steemit and Friendship Paradox

Hello steemeans! Today's topic is related to Steemit!

Why do the people we follow have more followers than us?

Most people in Steemit get frustrated or depressed when they encounter the fact that

The people we follow have more followers than we have

People who are not in this sad situation are typically whales or moderators, having enormous followers compared to average steemeans. This sad phenomenon is called Friendship paradox, because it seems like our friends are always famous compared to us.

Then why do most people suffer from this bad situation? Well, the mathematics behind this paradox is really! simple. Now let's analyze this statement in mathematical sense.

1. Modeling Steemit using Directed Graphs

First, let's assume there are total $n$ number of steemeans, and label them using $1, 2, ..., n-1, n$ . Consider each person as a vertex of a large graph $G$ .

Then, if a person $i$ follows person $j$ , draw an arrow from $i$ to $j$ .

Conversely, draw an arrow from $j$ to $i$ if $j$ follows $i$ .

Then, theoretically, all the follow, follwer relationships in Steemit are recorded in single graph $G$ , something like this.

For example, 1 follows 17, but 17 follows no one; whereas 8 and 13 follows each other.

2. Adjacent Matrix

Now, we assign a number to each pair of people $(i, j)$ , such that $a_{ij} = 1$ if $i$ follows $j$ , and 0 otherwise. Note that $a_{ii} = 0$ for all $i=1,2,...,n-1,n$ because following themselves is impossible (obvious). Then we can construct a $n\times n$ square matrix

$\mathbf{A} = (a_{ij})$

containing all the variables $a_{ij}$ . Now the question is

How many follwers does person $i$ have?

Well, the people following $i$ are the ones having $a_{ij} = 1$ , so the total number of followers can be calculated as

$\sum_{j=1}^{n} a_{ij}$

For convenience, define this variable as

$d_{i}^{\text{in}} := \sum_{j=1}^{n} a_{ij}$

(in comes from the fact that arrows come towards $i$ ) .

3. Friendship Paradox Revisited - [2]

Friendship paradox does not apply to all the steemit users, so we need to modify the statement a little bit. How about taking the average? Since the number of users $n$ , is sufficiently large, the average is a good representative value for the number of followers. So what we are trying to verify is the following.

$\text{Average \&hash; of followers of Steemit Users}$

is less or equal to

$\text{Average \&hash; of followers of our followers have}$

3-1.

What is average number of followers of Steemit Users? This is simple, since we already defined the variable $d_{i}^{\text{in}}$ for each $i$ . Thus it is

$\frac{d_{1}^{\text{in}} + d_{2}^{\text{in}} + ... + d_{n}^{\text{in}}}{n} = \frac{1}{n} \sum_{i} d_{i}^{\text{in}}$

3-2.

What is average number of followers of our followers have? This is little bit tricky. First, fix $i$ . Any person $j$ following $i$ should satisfy

$a_{ji} = 1$

and such person $j$ has $d_{j}^{\text{in}}$ number of followers. So,

$\sum_{j=1}^{n} a_{ji}d_{j}^{\text{in}}$

is the total number of followers of followers of $i$ have. Since $i$ runs over $1,2,...,n-1, n$ , the grand total would be

$\sum_{i=1}^{n}\sum_{j=1}^{n} a_{ji}d_{j}^{\text{in}}$

Now what is average? Many people think it should be divided by $n$ , but it's not. Followers of follwers are determined by the relationship $(i, j)$ where $a_{ij}$ or $a_{ji}$ equals to 1. Single steemean alone can not determine this value, since we should specify who are followers or not! Thus, the grand total should be divided by total number of incoming arrows, which is

$d_{1}^{\text{in}} + ... + d_{n}^{\text{in}} = \sum_{i=1}^{n} d_{i}^{\text{in}}$

Therefore, average number of followers of our followers have is

$\frac{\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ji} d_{j}^{\text{in}}}{\sum_{i=1}^{n} d_{i}^{\text{in}}}$

4. Awesome Trick

Now, we are faced with a problem,

$\frac{1}{n} \sum_{i=1}^{n} d_{i}^{\text{in}} \leq \frac{\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ji} d_{j}^{\text{in}}}{\sum_{i=1}^{n} d_{i}^{\text{in}}}$

which is equivalent to

$\left( \sum_{i=1}^{n} d_{i}^{\text{in}} \right)^2 \leq n \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ji}$

Uhh... the double sum is annoying us. Here the trick called double counting - [3], comes into account. Since the sum is finite, we can change the order of $\sum$ notation.

$\sum_{j}\sum_{i} a_{ji} d_{j}^{\text{in}}$

and using the definition $\sum_{i} a_{ij} = d_{j}^{\text{in}}$ , we have

$\sum_{i=1}^{n}\sum_{j=1}^{n} a_{ji }d_{j}^{\text{in}} = \sum_{j=1}^{n} (d_{j}^{\text{in}})\left( \sum_{i=1}^{n} a_{ji} \right ) = \sum_{j=1}^{n } (d_{j}^{\text{in}})^2$

So the inequality reduces to

$\left( \sum_{i=1}^{n} d_{i}^{\text{in}} \right )^2 \leq n \sum_{j=1}^{n } (d_{j}^{\text{in}})^2 = \left( \sum_{i=1}^{n} 1^2 \right ) \left(\sum_{j=1}^{n } (d_{j}^{\text{in}})^2 \right)$

which is true by Cauchy-Schwartz Inequality . So on average,

the people we follow have at least as many followers as us !

5. Conclusion

What we've showed not only applies to Steemit but also to any social networks having following and follwer system. Now the last question is

Can we make a social network that every user does not suffer friendship paradox?

This is equal to saying that equality

$\left( \sum_{i=1}^{n} d_{i}^{\text{in}} \right )^2 = \left( \sum_{i=1}^{n} 1^2 \right ) \left(\sum_{j=1}^{n } (d_{j}^{\text{in}})^2 \right)$

holds, not $<$ . But you know, in order to have equality - [4], one should satisfy

$d_{1}^{\text{in}} = d_{2}^{\text{in}} = ... = d_{n}^\text{in}$

; all Steemit users having equal followers, which is essentially impossible... :(. However, this does not imply that you are really in bad situation, on average most people suffer it not only you! So cheer up!

6. Citations

[1] Youtube Cover Image

[2] https://arxiv.org/pdf/1807.01496.pdf (Proof)

[3] https://en.wikipedia.org/wiki/Double_counting_(proof_technique)

[4] https://math.stackexchange.com/questions/279056/when-does-schwarz-inequality-become-an-equality

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steemstem (73) 6 years ago

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[Steemit & Math #13] Friendship Paradox - Is it bad or not?