Question

Social network densities. Social networking sites for business professionals are used to promote one’s business. Each social network involves interactions (connections) between members of the network. Researchers define network density as the ratio of actual network connections to the number of possible one-to-one connections.

For example, a network with 10 members has a$$\begin{gathered} \left(10 \\ 2\right)\mathbf{=}\mathbf{45} \end{gathered}$$ total possible connections. If that network has only 5 connections, the network density is $\frac{\mathbf{5}}{\mathbf{45}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{111}$. Sociologists at the University of Michigan assumed that the density x of a social network would follow a uniform distribution between 0 and 1 (Social Networks, 2010).

a. On average, what is the density of a randomly selected social network?

b. What is the probability that the randomly selected network has a density higher than .7?

c. Consider a social network with only 2 members. Explain why the uniform model would not be a good approximation for the distribution of network density.

Short answer

a. On average, the density of a randomly selected social network is 0.5.

b. The probability that the randomly selected network has a density higher than .7 is .3.

c. There is only one possible connection for a social network with only 2 members. The network density for no connection is $\frac{0}{1}=0$and for one connection is$\frac{1}{1}=1$ . In this case, it takes only two possible densities, 0 and 1; since it does not take the values between 0 and 1, the uniform model would not be a good approximation for the distribution of network density.

Step-by-step solution

Given information

The density x of a social network follows a uniform distribution between 0 and 1.

Define the probability density function

The probability density function of a random variable x is:

$f\left(x\right)=1;0⩽x⩽1$

Computing the mean value

a.

The mean of a random variable x is obtained as:

$\begin{array}{rcl}\mu & =& \frac{1}{2}\\ & =& 0.5\end{array}$.

Since the mean of a uniform distribution is: $\mathit{\mu }\mathbf{=}\frac{\mathbf{c}\mathbf{+}\mathbf{d}}{\mathbf{2}}$ .

Therefore, on average, the density of a randomly selected social network is 0.5.

Computing the required probability

b.

The probability that the randomly selected network has a density higher than .7 is obtained as:

$\begin{array}{rcl}P\left(x>.7\right)& =& P\left(.7<x<1\right)\\ & =& \frac{1-.7}{1-0}\\ & =& \frac{.3}{1}\\ & =& .3\end{array}$

For a uniform distribution: $\mathit{P}\left(a<x<b\right)\mathbf{=}\frac{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{d}\mathbf{-}\mathbf{c}}\mathbf{,}\mathit{c}\mathbf{⩽}\mathit{a}\mathbf{<}\mathit{b}\mathbf{⩽}\mathit{d}$.

Hence, the probability that the randomly selected network has a density higher than .7 is .3.

Stating the reason why a uniform model would not be a good approximation

c.

There is only one possible connection for a social network with only 2 members.

The network density for no connection is $\frac{0}{1}=0$and for one connection is$\frac{1}{1}=1$ .

In this case, it takes only two possible densities, 0 and 1; since it does not take the values between 0 and 1, the uniform model would not be a good approximation for the distribution of network density.