The background of this question is a paper written by Morten O.Ravn and Harald Uhlig, titled "On Adjusting The HODRICK-PRESCOTT Filter For The Frequency of Observations"
I will very much appreciate your help :D
Consider the decomposition of some given time series $y_t$ into a trend $\tau_t$ and a cycle $c_t$: $$y_t=\tau_t+c_t.$$ Now let $y_t$ be the "flow" $dz_t$ of some stochastic process $z_t$ with $$dz_t=\tau_t\ dt+\sigma_c\ dW_t^1$$ where $$d\tau_t=\mu_t\ dt,\\ d\mu_t=\sigma_\tau\ dW_t^2$$ and $dW_t^1$ and $dW_t^2$ are two independent Brownian motions.
We consider time aggregation; that is, for some length $\alpha>0$, consider observing $$y_{t;\alpha}=\int_{s=0}^\alpha dz_{t-s}=\tau_{t;\alpha}+c_{t;\alpha}$$ where $$\tau_{t;\alpha}=\int_{s=0}^\alpha \mu_{t-s}\ ds\\ c_{t;\alpha}=\int_{s=0}^\alpha \sigma_c\ dW_t^1.$$
Now, for any stochastic process $x_t$, define $\alpha$- differencing operator $$\Delta_\alpha x_t=x_t-x_{t-\alpha}.$$
Calculating second $\alpha$ difference, we will get $$\Delta_{\alpha}^2\tau_{t;\alpha}=\sigma_\tau\int_{s=0}^{3\alpha} A(s;\alpha)\ dW_{t-s}^2$$ where $$A(s;\alpha)=\int_{s_1=0}^\alpha \int_{s_2=0}^\alpha 1_{[0,\alpha]}(s-s_1-s_2)\ ds_2\ ds_1$$ and where $s=s_1+s_2+s_3.$ It is then said that the variance therefore given by $$\sigma^2(\Delta_{\alpha}^2\tau_{t;\alpha})=\sigma_\tau\int_{s=0}^{3\alpha} A(s;\alpha)\ ds.$$
I have several questions:
- Why the time aggregate $\tau_{t;\alpha}$ is written in the sense that the integrand walk backward, from $\mu_t$ to $\mu_{t-\alpha}$? The case is also for $c_{t;\alpha}$.
- What I understand is that the variance of the second difference of $\tau_{t;\alpha}$ is derived from Ito Isometry. However, according to Bernt Oksendal's book "Stochastic Differential Equations", it is said that the integrand of the Ito Integral have to be a bounded and elementary function. I think the function $A(s;\alpha)$ is surely bounded, but I doubt that it is also elementary.$\\$ NB: A function $\phi$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega).\chi_{[t_j,t_{j+1}]}(t) $$ where $\chi$ denotes the characteristic (indicator) function
- On the variance of the second difference of $\tau_{t;\alpha}$, am I right that the $\sigma_\tau$ is treated as constant and could be directly 'out' from the variance calculation? But, if so, it should be $\sigma_\tau^2$, isn't it?
Thank you so much :D