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How to calculate $E \big[(\int ^{t} _{-\infty} e^{\lambda u} d\tilde{L_\alpha}(u))^A (\int ^{t+h} _{-\infty} e^{\lambda u} d\tilde{L_\alpha}(u))^B\big]$, where \begin{align} \tilde{L}_\alpha (t) = \left\{ \begin{array}{l} L_\alpha(t),\quad t\geq 0 \\ L_\alpha(-t),\quad \text{otherwise} \end{array} \right. \end{align} and a $\{L(t),t\geq0\}$ is $\alpha$-stable Levy motion, thus $L(0)=0$ a.s., has stationary increments and $L(t)-L(s)\sim S_{\alpha}((t-s)^{\frac{1}{\alpha}},\beta,0)$ for any $0\leq s\leq t\leq \infty$ and for some $0<\alpha\leq 2$, $-1\leq\beta\leq1$.

I have to calculate fractional lower order covariance, thus I have \begin{align} FLOC(X(t),X(t+h),A,B)=E \big[\big(\int ^{t} _{-\infty} e^{-\lambda (t-u)} d\tilde{L_\alpha}(u)\big)^A \big(\int ^{t+h} _{-\infty} e^{-\lambda (t+h-u)} d\tilde{L_\alpha}(u)\big)^B\big]=e^{-\lambda t A}e^{-\lambda (t+h) B} E \big[\big(\int ^{t} _{-\infty} e^{\lambda u} d\tilde{L_\alpha}(u)\big)^A \big(\int ^{t+h} _{-\infty} e^{\lambda u} d\tilde{L_\alpha}(u)\big)^B\big], \end{align} with the parameters $A, B\geq 0$ satifying $A+B <\alpha$.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$ – dantopa Mar 4 at 1:29
  • $\begingroup$ I'd like to be able to help, but I'm not able to pick out details that I'd need to calculate the expected value of such integrals. You mention the dependence of integrals on a Lévy process (motion), but the capitalized variables $A,B$ deserve more of an explanation. It is customary to use capitalized variables for random variables, but I don't think that is the intention here in that you've pulled factors depending on $A,B$ outside the expected value operator. $\endgroup$ – hardmath Mar 4 at 2:48
  • $\begingroup$ A and B are not random variables. They are just parameters, such that $A,B\geq 0$ and $A+B< \alpha$. $\endgroup$ – Ksawka Mar 4 at 8:20

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