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A Levy process, \(\{L(t),t\in \mathbb{R}\}\) is a process with the following \(4\) properties:
\(L(0)=0\)
\(L(t)-L(s)\) has the same distribution as \(L(t-s)\) for \(\forall\ s,t\) such that \(s\leq t\)
If \((s,t)\) and \((u,v)\) are disjoint intervals then \(L(t)-L(s)\) and \(L(v)-L(u)\) are independent
\(\{L(t)\}\) is continuous in probability, that is, for \(\forall\ \epsilon>0\) and for \(\forall\ t\in \mathbb{R}\) \[ \underset{s\rightarrow t}{lim}P(\left|L(t)-L(s)\right|>\epsilon)=0 \]
A well-known example of the levy process is the Brownian Motion \[ \begin{gather} L(t)\sim N(\mu t,\sigma^2t)\\ t\geq0,\ \mu\in \mathbb{R}\ and\ \sigma>0 \end{gather} \] To show that it satisfies property (ii) of Levy process, we can calculate the corresponding expectation and variance \[ \begin{gather} E[L(t)-L(s)]\\ =E[L(t)]-E[L(s)]\\ =\mu(t-s)\\ =E[L(t-s)] \end{gather} \]
\[ \begin{gather} Var[L(t)-L(s)]\\ =E[(L(t)-L(s)-E[L(t)-L(s)])^2]\\ =E[[L(t)-L(s)]^2-2[L(t)-L(s)]\mu(t-s)+\mu^2(t-s)^2]\\ =E[L^2(t)+L^2(s)-2L(t)L(s)]-\mu^2(t-s)^2\\ =\sigma^2t+\mu^2t^2+\sigma^2s+\mu^2s^2-\mu^2(t-s)^2-2E[L(t)L(s)]\\ =\sigma^2(t+s)+\mu^2(t^2+s^2-t^2-s^2+2ts)-2E[L(t)L(s)]\\ =\sigma^2(t+s)+2\mu^2ts-2E[L(t)L(s)]\\ =\sigma^2(t+s)+2\mu^2ts-2E[[L(s)+L(t)-L(s)]L(s)] \end{gather} \]
Since \(L(t)-L(s)\text{ is independent to }L(s)\), we have \[ \begin{gather} Var[L(t)-L(s)]\\ =\sigma^2(t+s)+2\mu^2ts-2[E[L^2(s)]+E[L(t)-L(s)]E[L(s)]]\\ =\sigma^2(t+s)+2\mu^2ts-2(\mu^2s^2+\sigma^2s+\mu(t-s)\mu s)\\ =\sigma^2(t+s)-2\sigma^2s\\ =\sigma^2(t-s)\\ =Var[L(t-s)]\ \end{gather} \] Hence \(L(t)-L(s)\) has the same distribution of \(L(t-s)\)