Let $H \in (0, 1)$, $D \in \mathbb{R}$ and assume that the following function $$ r ( t, s ) = \frac{1}{2} \, \Big[ t^{2H} + s^{2H} - | t - s |^{2H} \Big] + D \, t^H s^H, \quad t, \, s \geq 0 $$ is positive definite. Then, clearly $D \geq -1$, since otherwise $r(t, t)$ is negative. It is also clear that if $D \geq 0$, then this is a covariance of $B_H (t) + \sqrt{D} \, t^H \, \xi$, where $\xi$ is independent of $B_H$ standard normal rv. Hence, $r$ is again positive definite.
Question: can $D \in (-1,0)$?
May be related to A simple decomposition for fractional Brownian motion with parameter $H<1/2$, but I didn't find the connection.
Update. If $r$ is to be positive definite, it has to satisfy in particular $$ g(t)= \det \begin{pmatrix} r ( t, t ) & r ( t, t + 1 ) \\ r ( t, t + 1 ) & r ( t+1, t+1) \end{pmatrix} \geq 0. $$ Expanding this function near $t = 0$, we obtain $$ g ( t ) \sim ( 1 + 2 D ) \, t^{2H} - D \, t^{3H} -2D \, t^{H+1}. $$ Hence, $D$ has to satisfy $D \geq -1/2$.
Same approach applied to $3 \times 3$ matrix $r (t_i, t_j)$ with $t_i \in \{ t, t+1, t+2 \}$ leads to the following lower bound: $$ D \geq -\frac{2 + 2^H}{6+2^H}, $$ which is even better than $D \geq -1/2$.
Taking four points $t, t+1, t+2, t+s$, constructing the lower bound for a fixed $s$ and letting $s\to\infty$ leads to the following bound: $$ D \geq -\frac{2+2^H}{8+2^{H+1}}. $$ This bound is much better than the one above. It is close to $1/3$.
