I have a persistent problem, which I'm almost certain can be answered using elementary probabilistic arguments, but for some reason I've been stuck for some time. Here is the problem.
Let $(B_s, s \in [0,1])$ be a fractional Brownian motion (here is the Wikipedia link for this process: http://en.wikipedia.org/wiki/Fractional_Brownian_motion). Consider the random variable:
$$Z = \int_0^1 B_s^4 \, ds.$$
Does $Z$ have a density? In other words, is the measure induced on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ by $Z$ absolutley continuous?
Many thanks in advance for some thoughts or advice! :)
John
Yes $Z$ has a density. For each $s$ the random variable $B_s^4$ is a linear combination of Hermite polynomials of degree at most $4$. Therefore the random variable $$ I_n = \frac{1}{2^n} \sum_{i=1}^{2^n} B_{i/2^n}^4 - B_{(i-1)/2^n}^4 $$ is also is a finite sum of Wiener chaos. Now $I_{n} \to Z$ in $L^2$ therefore $Z$ is also in a finite sum of chaos. You can then you can use a result of Shigekawa to conclude.
But I would not be surprise if a simpler argument exists.