# Fuzzy Fractional Brownian Motion: Review and Extension
## 2 Methods
Now, substitute the given expression for d\(\hat{S}\):  
\[\hat{d}\hat{S}/\hat{S}=(\hat{u}dt+\hat{\sigma}dB_{t}+\hat{\sigma}dB_{t}^{H}+d \left(\overset{\hat{N}}{\underset{1}{\sum}}\left(\overset{\sim}{V_{i}}-1 \right)\right)).\]  
Expanding this expression, we have  
\[\hat{d}\hat{S}/\hat{S}=\hat{u}dt+\hat{\sigma}dB_{t}+\hat{\sigma}dB_{t}^{H}+d \left(\overset{\hat{N}}{\underset{1}{\sum}}\left(\overset{\sim}{V_{i}}-1 \right)\right)\]Integrating both sides of the expression, we obtain  
\[\ln\left(\mathcal{S}_{t}\right)=\ln\left(\mathcal{S}(0)+\int_{0}^{t} \hat{u}dt+\int_{0}^{t}\hat{\sigma}dB_{t}+\int_{0}^{t}\hat{\sigma}dB_{t}^{H}+\int _{0}^{t}\frac{1}{2}\left\{-\frac{1}{\mathcal{S}^{2}}\right\}\partial^{2}\hat{S }^{2}\text{dt}+\right.\] \[\left.\int_{0}^{t}\frac{1}{2}\left\{-\frac{1}{\mathcal{S}^{2}} \right\}\partial^{2}\hat{S}^{2}\text{dt}^{H}+\int_{0}^{t}d(\sum_{1}^{\mathcal{ N}}\left(\overset{\sim}{V}_{i}-1\right))\text{dV}\right).\]  
In the crisp case, we used Ito isometry to show that \((dX)^{2}=(\mu dt)^{2}+2\mu orddW+\sigma(dW)^{2}=\sigma dt\). Now, recall that \(f\left(\hat{S}\right)=ln\left(\hat{S}\right)\), so \(\hat{S}=exp\left(f\left(\hat{S}\right)\right)\). Replacing \(\hat{S}\) with exp(f(\(\hat{S}\))) in the above expression,  
\[\ln\left(\mathcal{S}_{t}\right)=\ln\left(\mathcal{S}(0)+\hat{u}t+\sigma B_{t }+\hat{\sigma}dB_{t}^{H}-\frac{1}{2}\partial^{2}\text{t}+\frac{1}{2}\partial^ {2}\hat{\text{t}}^{2H}+\int_{0}^{t}d\left(\frac{\hat{N}}{1}\left(\overset{\sim }{V}_{i}-1\right)\right)\text{dV}\right).\]  
Recall that \(\overset{\sim}{V}_{i}\) are lognormally distributed such that \(\text{E}(\overset{\sim}{V}_{i})=E\left(e^{\overset{\sim}{V}_{i}}\right)=e^{ \mu+1/2\sigma^{2}}\); then, exponentiate both sides and simplify:  
\[\hat{S}_{t}=\mathcal{S}(0)e^{\hat{u}t+\sigma B_{t}+\hat{\sigma}dB_{t}^{H}- \frac{1}{2}\partial^{2}\text{t}+\frac{1}{2}\partial^{2}\text{t}^{2H}}\prod_{ i=1}^{\mathcal{N}}\overset{\sim}{V}_{i}.\]  
If we now take an expectation of \(\hat{S}_{t}\), we have  
\[\text{E}[\hat{S}_{t}]=\text{E}[\mathcal{S}(0)e^{\hat{u}t+\sigma B_{t}+\hat{ \sigma}dB_{t}^{H}-\frac{1}{2}\partial^{2}\text{t}+\frac{1}{2}\partial^{2}\hat {\text{t}}^{2H}}\prod_{i=1}^{\mathcal{N}}\overset{\sim}{V}_{i}].\]  
In stochastic calculus, \(\text{B}_{t}\) represents a standard Brownian motion and the key property of Brownian motion is that its variance increases linearly with time. The variance of \(\text{B}_{t}\) at time t is given by \(t\), which is a result of its definition and properties. The variance of a random variable X is defined as--\(Var(X)=E[(X-E[X])2]\). For a standard Brownian motion \(\text{B}_{t}\), its expected value \(\text{E}[\text{B}_{t}]\) is \(0\) at any time t. Therefore, we have \(Var(B_{t})=E\left[B_{t}^{2}\right]=\int_{-\infty}^{\infty}x^{2}P_{B(t)}(x)dx\). Since \(\text{B}_{t}\) follows a Gaussian distribution with mean \(0\) and variance \(t\), its probability density function is the Gaussian density function \(P_{B(t)}=\frac{1}{\sqrt{2\pi t}}\text{exp}\left(-\frac{x^{2}}{2t}\right)\). Plug this into the integral for the variance \(Var(B_{t})=E\left[B_{t}^{2}\right]=\int_{-\infty}^{\infty}x^{2}\frac{1}{\sqrt{ 2\pi t}}\text{exp}\left(-\frac{x^{2}}{2t}\right)dx\). The integrand \(\frac{1}{\sqrt{2\pi t}}\text{exp}\left(-\frac{x^{2}}{2t}\right)\) is the Gaussian p.d.f. with mean \(0\) and variance t. To integrate the Gaussian p.d.f over the entire real line, \(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi t}}\text{exp}\left(-\frac{x^{2}}{2t }\right)dx\), we will use the Gamma function. The integrand is an even function so \(\int_{-\infty}^{\infty}e^{-x^{2}}dx=2\int_{0}^{\infty}e^{-x^{2}}dx\). Thus, after a change in variable x= \(\sqrt{t}\), we turn this into gamma function  
\[2\int_{0}^{\infty}\frac{1}{2}e^{-t}t^{-1/2}dt=\sqrt{\pi} \tag{1}\]  
Similarly, \(\int_{-\infty}^{\infty}\frac{e^{-x^{2}}}{2}dx=\sqrt{2\pi}\).