Linear fractional differential equations in bank resource allocation and financial risk management model

This article examines the credit process of nine rural commercial banks in the western region. As shown in Figure 1, the entire credit process can be divided into three parts. That is, before, during and after the loan [3]. The pre-loan part is mainly divided into customer application, customer application acceptance, loan investigation and evaluation. The loan part includes loan review, loan review and approval, loan contract signing and loan issuance. The post-loan part mainly includes management after loan issuance, loan recovery and loan file management.

Fig. 1

Compared with large commercial banks, rural commercial banks in the western region started late, and credit risk evaluation and management are also relatively backward. Coupled with the shortage of historical data, it will become extremely difficult for credit risk evaluation models to quantify data [7]. The nine rural commercial banks in the western region often adopt subjective judgements and adopt a simplified model to increase assumptions when measuring credit risk. Currently, the credit risk evaluation models of rural commercial banks in western China mainly include expert analysis methods, loan rating methods, financial analysis methods and credit scoring methods.

After combining multiple analysis methods, it is found that only efficient and scientific credit scoring methods can effectively assess credit risk in the market environment of financial innovation reform. Therefore, the evaluation and management of credit risk of rural commercial banks in western China need to adopt modern credit risk evaluation methods [8]. The modern risk evaluation model calculates the loaner's default probability, standard deviation, default loss rate, and distribution and then uses.

In this part, we suggest some methods for solving linear fractional differential equations. tcD0+af(t)+af(t)=0, t>0tcD0+a|t=0=C \matrix{ {_tcD_{0 + }^af(t) + af(t) = 0,\quad t > 0} \hfill \cr {_t^cD_{0 + }^a\left| {_{t = 0} = C} \right.} \hfill \cr } The article uses Laplace inverse transformation to get f(t)=Ct−aEa,a(−at) f(t) = C{t^{ - a}}{E_{a,a}}\left( { - a\sqrt t } \right) , F(p)=cp12+aC=tcD0+−12f(t)|t=0, f(t)=Ct−12E12,12(−at) \matrix{ {F(p) = {c \over {{p^{{1 \over 2}}} + a}}} \cr {C = _t^cD_{0 + }^{ - {1 \over 2}}f(t)\left| {_{t = 0}} \right.} \cr } ,\quad f(t) = C{t^{ - {1 \over 2}}}{E_{{1 \over 2},{1 \over 2}}}\left( { - a\sqrt t } \right) tcD0+Qf(t)+tcD0+qf(t)=h(t), 0<q<Q<1 _t^cD_{0 + }^Qf(t) + _t^cD_{0 + }^qf(t) = h(t),\;0 < q < Q < 1 uses Laplace transform to get (pQ+pq)F(p)=C+H(p)C=(tcD0+q−1f(t)+tcD0+Q−q)|t=0 \matrix{ {\left( {{p^Q} + {p^q}} \right)F(p) = C + H(p)} \cr {C = \left( {_t^cD_{0 + }^{q - 1}f(t) + _t^cD_{0 + }^{Q - q}} \right)\left| {_{t = 0}} \right.} \cr } .

Among them, F(p)=C+H(p)pQ+pq=C+H(p)pq(pQ−q+1)=(C+H(p))p−qpQ−q+1 F(p) = {{C + H(p)} \over {{p^Q} + {p^q}}} = {{C + H(p)} \over {{p^q}({p^{Q - q}} + 1)}} = \left( {C + H(p)} \right){{{p^{ - q}}} \over {{p^{Q - q}} + 1}} . Let a = Q − q, β = Q, we use Laplace inverse transformation to get f(t)=CG(t)+∫0tG|(t−τ)h(τ)dτ C=(tcD0+q−1f(t)+tcD0+Q−1f(t))|t=0G(t)=tQ−1EQ−q,Q(−tQ−q) tcD0+Qy(t)−λy(t)=h(t), t>0 tcD0+a−ky(t)|t=0=bk, k=1,2,3,⋯,n \matrix{ {f(t) = CG(t) + \int_0^t G|(t - \tau )h(\tau )d\tau } \hfill \cr {\;\;\;\;C = \left( {_t^cD_{0 + }^{q - 1}f(t) + _t^cD_{0 + }^{Q - 1}f(t)} \right)\left| {_{t = 0}} \right.} \hfill \cr {G(t) = {t^{Q - 1}}{E_{Q - q,Q}}\left( { - {t^{Q - q}}} \right)} \hfill \cr {\;\;\;\;\;\;_t^cD_{0 + }^Qy(t) - \lambda y(t) = h(t),\quad t > 0} \hfill \cr {\;\;\;\;\;\;_t^cD_{0 + }^{a - k}y(t)\left| {_{t = 0}} \right. = {b_k},\quad k = 1,2,3, \cdots ,n} \hfill \cr } Using Laplace transformation, we can get: paY(p)−λY(p)=H(p)+∑k=1nbkpk−1Y(p)=H(p)pa−λ+∑k=1nbkpk−1pa−λ \matrix{ {{p^a}Y(p) - \lambda Y(p) = H(p) + \sum\limits_{k = 1}^n {b_k}{p^{k - 1}}} \cr {Y(p) = {{H(p)} \over {{p^a} - \lambda }} + \sum\limits_{k = 1}^n {b_k}{{{p^{k - 1}}} \over {{p^a} - \lambda }}} \cr } . y(t)=∑k=1nbkta−kEa,a−k+1(λta)+∫0t(t−τ)a−1Ea,a(λ(t−τ)a)h(τ)dτ. y(t) = \sum\limits_{k = 1}^n {b_k}{t^{a - k}}{E_{a,a - k + 1}}\left( {\lambda {t^a}} \right) + \int_0^t {(t - \tau )^{a - 1}}{E_{a,a}}\left( {\lambda {{(t - \tau )}^a}} \right)h(\tau )d\tau .

First, it is random whether each loan in the loan portfolio defaults. Second, the probability of default for each loan is very small [9]. The third is that the probability of default between each loan is independent of each other. The occurrence of default events obeys the Poisson distribution Pi=λei−λ/i! {P_i} = \lambda _e^{i - \lambda }/i! (i indicates that i debtors defaulted and λ is the average value of the expected default rate).

The calculation process of the linear fractional differential equation model can be divided into three steps: The first step is the classification of risk exposure frequency bands. Assuming that a loan portfolio has a total of N loans, the risk exposure scale of the largest loan in this N loan divided by the risk exposure frequency band value L is the risk exposure frequency band classification. We denote the risk exposure frequency band thus divided as V₁, V₂, V₃, ······V_n (n = 1, 2, 3······). There is a total of n frequency bands, and the amount of risk exposure frequency band corresponding to each frequency band is L₁, L₂, L₃, ······L_n and L_n = L × n. Then the risk exposure of each loan is calculated and divided into various frequency bands. Among them, risk exposure = the scale of the lender's risk exposure, default rate and default loss rate.

The second step is the calculation of the default probability of N loan portfolio. After the first frequency band division, we divided N loan into n frequency bands [10]. If the number of loans in each frequency band is N_i, then N₁, N₂, N₃, ······N_n = N. If the number of default loans in Nloan is k, then p is the average default probability. Because the model only considers the two states of default and non-default, and the occurrence of each event is independent of each other, we believe that k obeys the Bernoulli distribution. K ~ B(N, p), then: (1) P(k)=CNkpk(1−p)N−k P(k) = C_N^k{p^k}{(1 - p)^{N - k}} From mathematical knowledge, it can be known that when N is very large and p is very small, the Bernoulli distribution approximately obeys the Poisson distribution. From this, we can see that we set μ as the expected number of defaults in 1 year, i.e., μ = N × p. When the debtor's default probability is very small and the number of defaults in the loan portfolio is n, we can get the probability of n default events occurring within a year as: (2) P(n)=μne−μn! P(n) = {{{\mu ^n}{e^{ - \mu }}} \over {n!}} Suppose the probability occurrence function is F(z) and is defined by an auxiliary variable z: (3) F(z)=∑n=0∞P(n)zn F(z) = \sum\limits_{n = 0}^\infty P(n){z^n} So, by formulas (2) and (3), we can wait until the default probability function of a loan portfolio within 1 year is: (4) F(z)=eμ(z−1) F(z) = {e^{\mu (z - 1)}}

Step 3: Calculation of the loss distribution of N loan portfolio. The expected default loss for each frequency band is: (5) φj=εj×μj(1≤j≤m) {\varphi _j} = {\varepsilon _j} \times {\mu _j}\left( {1 \le j \le m} \right) where ε_j represents the average value of risk exposure in the V_j band. μ_j represents the number of defaults in the V_j band. μ represents the number of defaults expected to occur within a year, namely: (6) μ=μ1+μ2+⋯+μj+⋯+μm \mu = {\mu _1} + {\mu _2} + \cdots + {\mu _j} + \cdots + {\mu _m} From formulas (5) and (6), we get: (7) μ=∑j=0mμj=∑j=1mφjεj \mu = \sum\limits_{j = 0}^m {\mu _j} = \sum\limits_{j = 1}^m {{{\varphi _j}} \over {{\varepsilon _j}}} Then turn the number of default events into a loss distribution. This requires the introduction of the probability function G(z) for the occurrence of loss. We use this function to derive the default loss distribution function. The loss is equal to the multiple of the risk exposure, that is, the loss = L_n = L × n. Then we can set the default loss distribution function as: (8) G(z)=∑n=0∞P(Ln)zn G(z) = \sum\limits_{n = 0}^\infty P({L_n}){z^n} Because the various risk exposures in the portfolio are independent of each other, the formula for calculating the probability of the loan portfolio distribution is: (9) G(Z)=∏j=1mexp{−uj+ujZεj}=exp(∑j=1muj+∑j=1mujZεj) G(Z) = \prod\limits_{j = 1}^m \exp \left\{ { - {u_j} + {u_j}{Z^{{\varepsilon _j}}}} \right\} = \exp \left( {\sum\limits_{j = 1}^m {u_j} + \sum\limits_{j = 1}^m {u_j}{Z^{{\varepsilon _j}}}} \right) We define the expression P(z) as follows: (10) P(z)=∑j=1mujZεjμ=∑j=1m(φjεj)Zεj∑j=1m(φjεj) P(z) = {{\sum\limits_{j = 1}^m {u_j}{Z^{{\varepsilon _j}}}} \over \mu } = {{\sum\limits_{j = 1}^m \left( {{{{\varphi _j}} \over {{\varepsilon _j}}}} \right){Z^{{\varepsilon _j}}}} \over {\sum\limits_{j = 1}^m \left( {{{{\varphi _j}} \over {{\varepsilon _j}}}} \right)}} From formula (7) and formula (10), we can simplify formula (9) as (11) G(Z)=eμ(p(z)−1)=F(P(z)) G(Z) = {e^{\mu (p(z) - 1)}} = F(P(z)) The formula for calculating the probability of portfolio loss is: (12) P(Li)=1n!×dnG(z)dzn|z=0=An P({L_i}) = {1 \over {n!}} \times {{{d^n}G(z)} \over {d{z^n}}}{|_{z = 0}} = {A_n} Which represents the probability that the loss is (n × L). We expand formula (12) with Taylor's formula to get: (13) An=∑j:εj≤nφjnAn−εj, n=1,2,3⋯ {A_n} = \sum\limits_{j:{\varepsilon _{j \le n}}} {{{\varphi _j}} \over n}{A_{n - {\varepsilon _j}}},\quad n = 1,2,3 \cdots (14) A0=e−μ=e−∑j=1mφjεj {A_0} = {e^{ - \mu }} = {e^{ - \sum\limits_{j = 1}^m {{{\varphi _j}} \over {{\varepsilon _j}}}}} The loss distribution of the loan portfolio can be calculated by formulas (10), (13) and (14).

Step 4: Calculate the economic capital, expected loss and unexpected loss of the loan portfolio. The VaR of the loan portfolio can be calculated given a confidence level in the calculations of the second and third steps. That is the unexpected loss of the loan portfolio. For example, when the confidence level = 95%, we get the maximum loss under the confidence level. VaR (unexpected loss) = maximum loss-expected loss.

Among the 1,117 loan samples selected by Sichuan KK Rural Commercial Bank from 2018 to 2020, there are 328 agricultural loans. Sichuan KK Rural Commercial Bank's agricultural loans accounted for 29.7% of total loans from 2018 to 2020. The ratio of agricultural loans to total loans in the sample data is 29.3%, basically in line with the ratio of the overall data.

4.2.2

Distribution of agricultural loan losses

According to the principle of the linear fractional differential equation model, the loss distribution of agricultural loans can be calculated. The loss distribution is shown in Figure 3.

Fig. 3