Farmers’ Social Networks and the Fluctuation in Their Participation in Crop Insurance: The Perspective of Information Diffusion

ABSTRACT

By investigating the diffusion of information among social networks, and by presenting an analysis of apple growers in terms of the mechanism that underlies insurance decision-making, this paper first examined the relationship between information diffusion in social networks and the volatility that is observed in the crop insurance purchasing rate in China. In accordance with the mean-field theory of social networks, a mathematical model of infectious disease dynamics was employed to construct an insurance purchasing model, and the convergence value of the crop insurance purchasing rate was derived. Second, this paper analyzed research data related to apple insurance, and verified some of the properties related to purchasing rate fluctuations. The results show that the purchasing rate of crop insurance has a unique convergence point. The greater the average degree among the social network, the quicker the crop insurance purchasing rate reached the convergence point, and the lower its volatility.

KEY WORDS:

• crop insurance purchasing rate
• infection model
• information diffusion
• social network

JEL:

• G22
• Q12
• Q18

Acknowledgments

We thank International Science Editing (www.internationalscienceediting.com) for editing this manuscript. The authors accept all errors.

A-1: Proof of Theorem 1

Because $\sigma \le 1$, hence $1-\frac{1}{\sigma }\le 0$ and $i(t)-(1-\frac{1}{\sigma })\ge 0$. Since $\delta >0,i(t)>0$, we obtain $\frac{di(t)}{dt}\le 0$, the application ratio is a decreasing function, until $i(\mathrm{\infty })=0$

A-2: Proof of Theorem 2

Consider the following differential function assuming the initial condition, we assume $\delta ,\mu$ is constant for convenient.

$\frac{di(t)}{dt}=-\delta i(t)[i(t)-(1-\frac{1}{\sigma })],i(t_0)=i_0$

Since $0<\delta <1,0<i(t)<1$, we obtain:

A{\hbox{-}1 $\left\{\begin{array}{c}-\delta i(t)[i(t)-(1-\frac{1}{\sigma })],i(t)>(1-\frac{1}{\sigma })\\ -\delta i(t)[i(t)-(1-\frac{1}{\sigma })],i(t)<(1-\frac{1}{\sigma })\end{array}\right.$A{\hbox{-}1

So, when the initial value $i(t)>(1-\frac{1}{\sigma }),i(t)$ is continuous and decreasing; whereas when $i(t)<(1-\frac{1}{\sigma }),i(t)$ is continuous and increasing. Hence, whatever the value of $i_0$, the unique asymptotical convergence $i^{\ast }$ exists, and $i^{\ast }=1-\frac{1}{\sigma }$.

A-3: Proof of Corollary 1

Consider a scenario where time is discrete. The equation then changes as follows:

(A{\hbox{-}2) $\left\{\begin{array}{c}N[i(t+1)-i(t)]=\delta Ns(t)i(t)-\mu Ni(t)\\ s(t)+i(t)=1\\ i(0)=i_0>0\end{array}\right.$(A{\hbox{-}2)

We then obtain $i(t+1)=i(t)-\delta i(t)[i(t)-(1-\frac{1}{\sigma })]$ .

Let $i_0\in (0,1],\mathrm{\Delta }t>0$, set $i_x=i(x\mathrm{\Delta }\mathrm{t})$,

Note

$\begin{array}{rl}& \rho (i_x,i_y)=\left|i_x-i_y\right|\\ & =\left|i_0-\delta i_{0}x\mathrm{\Delta }\mathrm{t}\left(i_0-\left(1-\frac{1}{\sigma }\right)-\left[{\mathrm{i}}_0-\delta {\mathrm{i}}_0\mathrm{y}\mathrm{\Delta }\mathrm{t}\left({\mathrm{i}}_0-\left(1-\frac{1}{\sigma }\right)\right)\right]\right.\right|=\left|(\mathrm{x}-\mathrm{y})\delta {\mathrm{i}}_0\mathrm{\Delta }\mathrm{t}\left({\mathrm{i}}_0-\left(1-\frac{1}{\sigma }\right)\right)\right|\end{array}$

Since $0<\delta i_0<1,0<\mathrm{\Delta }t<1$, when $\sigma >1,0<\left|i_0-(1-\frac{1}{\sigma })\right|<1$, so $0<\left|\delta i_0\mathrm{\Delta }\mathrm{t}\left(i_0-\left(1-\frac{1}{\sigma }\right)\right)\right|<1$.

Let $\alpha =\left|\delta i_0\mathrm{\Delta }\mathrm{t}\left(i_0-\left(1-\frac{1}{\sigma }\right)\right)\right|\in (0,1)$. Thus we have $\rho (i_x,i_y)<\alpha \rho (x,y)$ $p\left(i\left(x\right),i\left(y\right)\right)\le \alpha \rho \left(x,y\right)$.

Then $i_x:R\to R$ is contraction map.

For any $x_0$:

$\begin{array}{rl}& \rho ({i_{x_0}}^{n+p},{i_{x_0}}^n)\le {\alpha }^n\rho ({i_{x_0}}^p,x_0)\\ & \le {\alpha }^n({\alpha }^{n-1}+{\alpha }^{n-2}+\cdots +1)\rho (i_{x_0},x_0)\\ & \le \frac{{\alpha }^n}{1-\alpha }\rho (i_{x_0},x_0)\end{array}$

since $0<\alpha <1$,$\{i_{x_n}\}$ is a Cauchy sequence.

And, as R is complete, from the nature of the contraction map, we obtain the unique-fixed point $1-\frac{1}{\sigma }$.

A-4: Proof of Corollary 3

Consider the scenario of equilibrium ${i_k}^{\ast }(t)$: under the extreme value, we have $\frac{d{i}_k(t)}{dt}={\delta }_k(t)i_k(t)(1-i_k(t))-\mu i_k(t)=0$

${\delta }_k(t)=\upsilon +\frac{i_k(t)}{1-i_k(t)}\theta e^{c-\eta t}q$

We obtain the following:

$[\upsilon +\frac{i_k(t)}{1-i_k(t)}\theta e^{c-\eta t}q]i_k(t)(1-i_k(t))-\mu i_k(t)=0$

After rearranging, we obtain the following:

(A{\hbox{-}3) ${i_k}^{\ast }(t)=\frac{\upsilon -\mu }{\upsilon -\theta e^{c-\eta t}q}=\frac{\upsilon -\mu }{\upsilon -\theta e^{c-\eta t}\frac{1}{⟨k⟩}\sum _{k}kp(k){i_k}^{\ast }(t)}$(A{\hbox{-}3)

since $q^{\ast }=\frac{1}{⟨k⟩}\sum _{k}kp(k){i_k}^{\ast }(t)$.

Completing the square of both sides of the above equation, and by calculating the sum of all of k, we obtain the following:

$q^{\ast }=\frac{1}{⟨k⟩}\sum _{k}kp(k){i_k}^{\ast }(t)=\frac{1}{⟨k⟩}\sum _k\left[kp(k)\frac{\upsilon -\mu }{\upsilon -\theta e^{c-\eta t}{q}^{\ast }}\right]=\frac{\upsilon -\mu }{\upsilon -\theta e^{c-\eta t}{q}^{\ast }}\frac{1}{⟨k⟩}\sum _{k}kp(k)$

Because $\frac{1}{⟨k⟩}\sum _{k}kp(k)=1$, we have$q^{\ast }=\frac{\upsilon -\mu }{\upsilon -\theta e^{c-\eta t}{q}^{\ast }}$.

After rearranging, we obtain the following equation:

$F(t,q^{\ast })=(q^{\ast }{)}^2\theta e^{c-\eta t}+(1-q^{\ast })\upsilon -\mu =0$, and $\theta ,c,\eta ,\upsilon ,\mu$ are all constant.

From the function, we can obtain the following:

• (i) $F(t,q^{\ast })$ is continuous to t and$q^{\ast }$in the domain of definition;

• (ii) When $q^{\ast }=1,t_0=\frac{c-ln\left(\frac{\mu }{\theta }\right)}{\eta }$ and $F(t,q^{\ast })=0$;

• (iii) $F(t,q^{\ast })=-(q^{\ast }{)}^2\theta \eta e^{c-\eta t}$, the partial derivative of $F(t,q^{\ast })$, to $t$in the domain of definition; and

• (iv) $F_t(t_0,{q_0}^{\ast })=F\left(1,\frac{c-ln\left(\frac{\mu }{\theta }\right)}{\eta }\right)=-\theta \eta e^{c-\eta \frac{c-ln\left(\frac{\mu }{\theta }\right)}{\eta }}=-\eta \mu \ne 0$.

From the theorem of implicit function, the unique implicit function $t=f(q^{\ast })$ exists, and we have the following:

$f^{\prime }(q^{\ast })=-\frac{F_{q^{\ast }}(t_0,{q_0}^{\ast })}{F_t(t_0,{q_0}^{\ast })}$

And because

A{\hbox{-}5 $\left\{\begin{array}{c}{F}_t(t,q^{\ast })=-(q^{\ast }{)}^2\theta \eta e^{c-\eta t}\\ F_{q^{\ast }}(t,q^{\ast })=2q^{\ast }\theta \eta e^{c-\eta t}-\upsilon \end{array}\right.$A{\hbox{-}5

hence .

$f^{\prime }(q^{\ast })=-\frac{F_{q^{\ast }}(t_0,{q_0}^{\ast })}{F_t(t_0,{q_0}^{\ast })}=\frac{2q^{\ast }\theta \eta e^{c-\eta t}-\upsilon }{{(q^{\ast })}^2\theta \eta e^{c-\eta t}}$

From $f^{\prime }(q^{\ast })=0$, we can obtain the following:

$2q^{\ast }\theta \eta e^{c-\eta t}-\upsilon =0$

We can input this into the following:

$q^{\ast }=\frac{1}{⟨k⟩}\sum _{k}kp(k){i_k}^{\ast }(t)$

We can obtain the following:

(A{\hbox{-}6) ${\mathrm{t}}^{\ast }=\frac{c-ln\frac{\upsilon }{2q^{\ast }\theta }}{\eta }=\frac{c-ln\frac{⟨k⟩\upsilon }{2\theta \sum _{k}kp(k){i_k}^{\ast }(t)}}{\eta }$(A{\hbox{-}6)

Hence, when$⟨k^{\prime }⟩>⟨k⟩$, we have ${t^{\prime }}^{\ast }<t^{\ast }$.

Notes

1. The elaboration likelihood model was proposed by psychologists Richard E. Petty and John T. Cacioppo. It is the most influential theoretical model about consumer information-processing behavior. Consumers can take the information provided by advertisements seriously when forming their attitude toward advertising brands. They will search, analyze, and use the information related to advertised products carefully, which ultimately leads to the formation or transformation of their attitudes.

Additional information

Funding

This work was supported by the National Natural Science Foundation of China [71403215, 71503203, 71603190, 71673206, 71873103]; The Fundamental Research Funds for the Central Universities [No. 2015RWYB07].