# Dynamics of a Stochastic Intraguild Predation Model
## 3 Stochastic Persistence and Stochastic Extinction
 \tag{31}\]  
We consider the following two cases:  
Case 1: if \(\limsup_{t\to+\infty}(\ln P(t)/\ln t)<0\;a.s.\), then \(\lim_{t\to+\infty}P(t)=0\;\;a.s.\)Case 2: if \(\limsup\limits_{t\rightarrow+\infty}(\ln P(t)/\ln t)\geq 0\)_a.s._, then by Equation (7), for sufficiently large \(t\), we get  
\[ \begin{split}\frac{1}{t}\ln\frac{R(t)}{R(0)}\leq&\frac{ L_{1}-M_{1}}{a_{mn}a_{pp}+e_{np}a_{np}^{2}}+\varepsilon-\frac{L}{a_{mn}a_{pp}+e_{ np}a_{np}^{2}}[R(t)]\\ &+\frac{\alpha_{r}B_{r}(t)-\mu_{1}\alpha_{n}B_{n}(t)-\mu_{2}a_{p} B_{p}(t)}{t}.\end{split}  \tag{32}\]  
It follows from Lemma 1 and the arbitrariness of \(\varepsilon\) that  
\[[R]^{*}\leq(L_{1}-M_{1})/L. \tag{33}\]  
On the other hand, a direct calculation also shows that  
\[a_{rr}\delta_{9}/\delta_{2}+e_{rp}a_{rp}\delta_{8}/\delta_{2}-e_{rn}a_{rn}=0, \ -a_{rp}\delta_{9}/\delta_{2}+a_{pp}\delta_{8}/\delta_{2}-a_{np}=0,\]  
where \(\delta_{9}=a_{rn}a_{np}-a_{rp}a_{nn}\). For sufficiently large \(t\), multiplying both sides of three equalities of Equation (21) by \(-\delta_{9}/\delta_{2}\), \(-1\) and \(\delta_{8}/\delta_{2}\), respectively, and then adding these three equalities, we obtain  
\[ \begin{split}\frac{1}{t}\ln\frac{N(t)}{N(0)}=&- \frac{\delta_{9}}{\delta_{2}t}\ln\frac{R(t)}{R(0)}+\frac{\delta_{8}}{\delta_{2 }t}\ln\frac{P(t)}{P(0)}+\frac{L_{2}-M_{2}}{\delta_{2}}-\frac{L}{\delta_{2}}[N( t)]\\ &+\frac{\alpha_{n}B_{n}(t)+\delta_{10}/\delta_{2}\alpha_{r}B_{r}( t)-\delta_{8}/\delta_{2}\alpha_{p}B_{p}(t)}{t}.\end{split}  \tag{34}\]  
Here, we have \(\limsup\limits_{t\rightarrow+\infty}(\ln R(t)/\ln t)\geq 0\)_a.s._ In fact, if \(\limsup\limits_{t\rightarrow+\infty}(\ln R(t)/\ln t)<0\)_a.s._, then \(\lim\limits_{t\rightarrow+\infty}R(t)=0\)_a.s._, which implies that \(\lim\limits_{t\rightarrow+\infty}N(t)=0\)_a.s._ and \(\lim\limits_{t\rightarrow+\infty}P(t)=0\)_a.s._ This is a contradiction. By Equation (7), for sufficiently large \(t\), we obtain  
\[ \begin{split}\frac{1}{t}\ln\frac{N(t)}{N(0)}\leq&\frac {L_{2}-M_{2}}{\delta_{2}}+\varepsilon-\frac{L}{\delta_{2}}[N(t)]\\ &+\frac{\alpha_{n}B_{n}(t)+\delta_{10}/\delta_{2}\alpha_{r}B_{r}( t)-\delta_{8}/\delta_{2}\alpha_{p}B_{p}(t)}{t}.\end{split}  \tag{35}\]  
From Lemma 1, we get  
\[[N]^{*}\leq(L_{2}-M_{2})/L. \tag{36}\]  
since \(\varepsilon\) is arbitrary. For the third equality of Equation (21) and sufficiently large \(t\), combining Equation (33) with Equation (36) gives  
\[ \begin{split}\frac{1}{t}\ln\frac{P(t)}{P(0)}&\leq- d_{p}-\frac{a_{p}^{2}}{2}+\varepsilon+e_{rp}a_{rp}[R]^{*}+e_{np}a_{np}[N]^{*}-a_{pp}[P( t)]+\frac{\alpha_{p}B_{p}(t)}{t}\\ &\leq\frac{a_{pp}(L_{3}-M_{3})}{L}+\varepsilon-a_{pp}[P(t)]+ \frac{\alpha_{p}B_{p}(t)}{t}.\end{split}  \tag{37}\]  
Then, \(\lim\limits_{t\rightarrow+\infty}P(t)=0\)_a.s._ if \(\varepsilon\) is sufficiently small. Combining case 1 with case 2 gives \(\lim_{t\to+\infty}P(t)=0\)\(a.s.\) The first equality of Equation (21) multiplied by \(e_{rn}a_{rn}\) plus the second equality of Equation (21) multiplied by \(a_{rr}\) gives  
\[ \begin{split}\frac{a_{rr}}{t}\ln\frac{N(t)}{N(0)}&=- \frac{e_{rn}a_{rn}}{t}\ln\frac{R(t)}{R(0)}+\delta_{5}-\delta_{3}-\delta_{1}[N (t)]-(e_{rn}a_{rn}a_{rp}+a_{rr}a_{np})[P(t)]\\ &\quad+\frac{e_{rn}a_{rn}\alpha_{r}B_{r}(t)+a_{rr}\alpha_{n}B_{n} (t)}{t}\\ &\geq\delta_{5}-\delta_{3}-2\varepsilon-\delta_{1}[N(t)]+\frac{e _{rn}a_{rn}\alpha_{r}B_{r}(t)+a_{rr}\alpha_{n}B_{n}(t)}{t}\end{split}  \tag{38}\]  
for sufficiently large \(t\) and sufficiently small \(\varepsilon\) since Equation (7) and \(\lim_{t\to+\infty}P(t)=0\)\(a.s.\) hold. It follows from Lemma 1 and the arbitrariness of \(\varepsilon\) that  
\[[N]_{*}\geq(\delta_{5}-\delta_{3})/\delta_{1}\)\(a.s. \tag{39}\]  
By applying the above inequality and \(\lim_{t\to+\infty}P(t)=0\)\(a.s.\) into the first equality of Equation (21), we get  
\[