# Fuzzy Fractional Brownian Motion: Review and Extension
## 3 Proofs and Discussion
### Main Results--Fuzzy Stochastic Differential Equation When H \(>\) %
Going back to our proof of uniqueness, consider that  
\(J(t)=E[\sup_{u\in[0,t]}d_{\infty}^{c}\ (X^{\varepsilon}(u),\ Y^{\varepsilon}(u))]\), where we already have an estimation or upper bound, which is  
\[j(t)\leq 3\big{[}(T+4^{2a} +4(T+\varepsilon)^{2a}\big{]}\ \ L^{2}\ E\int_{0}^{t}d_{\infty}^{2}\ [(X^{ \varepsilon}(s),\ Y^{\varepsilon}(s))ds\big{]}\] \[\leq\ 3\Big{[}(T+4^{2a}+4(T+\varepsilon)^{2a}\Big{]}\ \ L^{2}\ \int_{0}^{t}j_{n}(s)ds\]  
In the context of the j(t) function, we apply [PERSON]'s inequality with \(A=0\) and \(B=3\Big{[}(T+4^{2a}+4\Big{(}T+\varepsilon)^{2a}\Big{]}\ \ L^{2}\). Using \(f(t)\leq A\int_{a}^{b}e^{g(s)ds}\), we can state that  
\(j(t)\leq 0*e^{3[(T+4^{2a}+4(T+\varepsilon)^{2a}]\ L^{2}dt}=0\). Therefore, j(t) = 0 for all t within the given interval. This bound shows that the function j(t) grows exponentially in time, with the rate of growth determined by the constant \(3\Big{[}(T+4^{2a}+4\Big{(}T+\varepsilon)^{2a}\Big{]}\ L^{2}\). However, the growth of j(t) is limited by its initial value j(0). If j(t) = 0 for all t in I, this means that the distance between the two solutions of the stochastic differential equation is zero for all times in the interval I. But since j(t) is non-negative, this implies that j(t) = 0 for all t \(\geq\) t\({}_{0}\), where t\({}_{0}\) is the endpoint of the interval I. Therefore, the two solutions of the equation coincide for all times t \(\geq\) t\({}_{0}\), which means that they become arbitrarily close to each other as time goes to infinity. The proof of uniqueness is complete:  
\(J(t)=E[\sup_{u\in[0,t]}d_{\infty}^{2}\ (X^{\varepsilon}(u),\ Y^{\varepsilon} (u))\ ]=0\ a.e.\),  
We will next prove that X\({}^{\varepsilon}\)(t) converges to the solution X(t) in L\({}^{2}\) space as \(\varepsilon\)>0 and t \(\subset\) [0, T], which is fundamental if we want to apply [PERSON]'s lemma later on. However, first, we need to prove that, for every \(\varepsilon>0\) and \(0<\) a < %,  
\(\int_{s}^{t}\ ((r-s+\varepsilon)^{a-1}-\Big{(}r-s)^{a-1}\Big{)}dr\leq\frac{a+1} {a}\varepsilon^{a}\)  
We will need this inequality for use in the former proof. We will employ the finite-increments formula in the proof. Let f(x) be a differentiable function defined on an interval [a,b] and let \(\delta\) be a small increment. The formula states that \(f(x+\delta)-f(x)=f^{\prime}(x)\delta+\theta\delta\), where f'(x) represents the first derivative of \(f\) evaluated between some point x and x + \(\delta\), and \(\theta\) is a value between 0 and 1. In other words, the change in the function's value over the interval [x, x + \(\delta\)] is approximately equal to the derivative of the function times the increment \(\delta\), plus an additional term \(\theta\delta\) accounting for the difference between the true change and the linear approximation. In the context of this proof, we will use the finite-increments formula to analyse the function f(x) = x\({}^{a-1}\) to obtain \((\mbox{x}+\varepsilon)^{a-1}-\mbox{x}^{a-1}\)=[(a \(-\) 1)(x + \(\theta\varepsilon)^{(a-2)}\varepsilon]\). By applying the formula to this function, we can derive relationships between the increments in the function's values and the derivative of the function evaluated at intermediate points. These relationships are then used to establish the inequality for the integral in the proof. In summary, the finite-increments formula allows us to approximate the change in a function over small increments and provides a useful tool for bounding the behaviour of the functions. Here are the steps of the proof. (i) Apply the finite-increments formula to obtain an upper bound for the absolute value of the following: \(|\left((r-s+\varepsilon)^{a-1}-\big{(}r-s)^{a-1}\right)|\leq|a-1||r-s|^{a-2}\varepsilon\).