# Fuzzy Fractional Brownian Motion: Review and Extension
## 3 Proofs and Discussion
### Main Results--Fuzzy Stochastic Differential Equation When H \(>\) %
Using this inequality, we will split the integral over the time interval [s,t] into two parts, one from s to s + \(\varepsilon\) and the second from s + \(\varepsilon\) to t.  
\(\int_{s}^{t}\Big{(}\Big{(}r-s+\varepsilon)^{a-1}-\Big{(}r-s)^{a-1}\Big{)}dr\)\[\leq\int_{s}^{s+\varepsilon}|((r-s+\varepsilon)^{a-1}-\left(r-s)^{a-1} \right)|dr+\int_{s+\varepsilon}^{t}\left|(r-s+\varepsilon)^{a-1}-\left(r-s)^{a- 1}\right)dr\right.\]  
Notice that \(\left(r-s+\varepsilon)^{a-1}-\left(r-s\right)^{a-1}\right)\) can be written as \([(r-s)+\varepsilon]^{a-1}-\left(r-s\right)^{a-1})\)  
\(==\)\([\left(r-s\right)^{a-1}+(a-1)(r-s)^{a-2}\varepsilon+\cdots+\varepsilon^{a-1} \right]\)\(-\)\(\left(r-s\right)^{a-1}\leq\)\(\left|(a-1)\left(r-s\right)^{a-2}\varepsilon+\)  
Just to clarify, \(\left(\mathbf{r}-s\right)\leq\varepsilon\) comes from the fact that we are integrating over the interval [s,t] and splitting it into two parts: [s, s + \(\varepsilon\)] and [s + \(\varepsilon\), t]. Since r is in the interval [s, s + \(\varepsilon\)], \(r-s+\varepsilon\ \leq\ s+\varepsilon-s=\varepsilon\), so \(r-s\ \leq\ \varepsilon\), and since r is in the interval [s + \(\varepsilon\), t], we have \(r-s+\varepsilon\ \leq\ t-s\). Since we are interested in bounding the integrand over the entire interval [s,t], we can take the maximum of the upper bounds for \(\left(\mathbf{r}-s\right)\) over both parts, which gives \(\left(r-s\right)\ \leq\ max\ \left\{\varepsilon,\ t-s\right\}\). We can now use these bounds to re-write the integrals:  
\[\int_{s}^{s+\varepsilon}|((r-s+\varepsilon)^{a-1}-\left(r-s\right)^{a-1})|dr+ \int_{s+\varepsilon}^{t}|(r-s+\varepsilon)^{a-1}-\left(r-s\right)^{a-1})dr\]  
\[\leq\int_{s}^{s+\varepsilon}|(2(r-s))^{a-1}-(r-s)^{a-1}|dr+|a-1|\varepsilon\int _{s+\varepsilon}^{t}|r-s|^{a-2}dr\ \leq[\frac{(r-s)^{a}}{a}]_{s}^{s+\varepsilon}\]  
\[+(a-1)\varepsilon[\frac{(r-s)^{a-1}}{a-1}]_{s+\varepsilon}^{t}\leq\frac{ \varepsilon}{a}+\varepsilon^{a}\]  
We will now use this result to prove the convergence of the solution and show how to apply it for the purposes of this paper. We want to prove that \(X^{\varepsilon}(t)=X_{0}+\int_{0}^{t}f(s,\ X^{\varepsilon}(s))ds+\left\langle \int_{0}^{t}g(s,\ X^{\varepsilon}(s))dB^{\varepsilon}_{H}(s)\right\rangle)\) converges to \(X(t)=X_{0}+\int_{0}^{t}f(s,\ X(s))ds+\left\langle\int_{0}^{t}g(s,\ X(s))dB_{H }(s)\right\rangle)\). We will utilise Equation (12) and the Euler-Maruyama numerical method for that. This is a reminder of the Equation (12):  
\[\sup_{u\in[0,t]}d_{\infty}^{2}(\langle\int_{0}^{u}X(s)dW(s)\rangle,\langle\int _{0}^{u}Y(s)dW(s)\rangle)\leq 4E\int_{0}^{t}d_{\infty}^{2}(\langle X(s)\rangle,\ \langle Y(s)\rangle)ds\]  
The Euler-Maruyama method is a modification of the standard Euler method for solving ordinary differential equations (ODEs) to solve stochastic differential equations (SDEs). The solution of an SDE is itself a random process, and therefore cannot be computed exactly in closed form. The Euler-Maruyama method approximates a solution of an SDE as a piecewise linear function. At each step, the approximation is updated by taking a small step in time and adding a random perturbation. Specifically, it takes the form dX(t) = f(t,X(t))dt + g(t,X(t))dBH(t). Choose a time step size At and define the time points t\({}_{\text{x}}\) = kAt for k = 0,1,2,..,n. Intialise the approximation by setting X\({}_{0}\) = X(t\({}_{0}\)). For k = 1,2,..,n, compute the approximation at time t\({}_{\text{x}}\) as X(t\({}_{\text{x}}\)) = X(t\({}_{\text{x-1}}\)) + f(t\({}_{\text{x-1}}\)X(t\({}_{\text{x-1}}\)))At + g(t\({}_{\text{x-1}}\)X(t\({}_{\text{x-1}}\)))(B\({}^{\text{H}}\)(t\({}_{\text{x}}\)) - B\({}^{\text{H}}\)(t\({}_{\text{x-1}}\))), where B\({}^{\text{H}}\)(t\({}_{\text{x}}\)) - B\({}^{\text{H}}\)(t\({}_{\text{x-1}}\)) is the increment of the Brownian motion over the time interval [t\({}_{\text{x-1}}\), t\({}_{\text{x}}\)].