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## EXISTENCE OF EQUIVALENT MARTINGALE MEASURE FOR FRACTIONAL HESTON

The existence of an equivalent martingale measure for the fractional Heston model

$dX(t) = \mu X(t) + \sqrt{V(t)} dw^1(t)$

$D_t^\alpha V(t) = -\kappa(V(t)-\theta) + \nu \sqrt{V(t)} dw^2(t)$

$\langle dw^1(t), dw^2(t)\rangle = \rho dt$

in analogy with equivalent condition for the Heston model (see MeasureChange-HestonModel and Rydberg-EMM-1997 for discussions on this issue) is the necessary condition

$\mu-r = \sqrt{V(t)}(\rho \gamma_1(t) + \sqrt{1-\rho^2} \gamma_2(t))$

The equivalent martingale measure is

$dQ/dP = \exp( -\int_0^t \gamma_1(u) dw^1(u) -\int_0^t \gamma_2(u) dw^2(u) - \frac{1}{2}(\int_0^t \gamma_1^2(u) du + \int_0^t \gamma_2^2(u) du)$

So by Novikov’s condition, we want to check that $E(\exp[\int_0^t \frac{\mu-r}{V(s)} ds]) < \infty$.  Now Mainardi (mainardi-mittag-leffler-properties-2014)gives us asymptotic approximations of the Mittag-Leffler function $e_{\alpha}(t) =E_{\alpha}(-t^\alpha)$:

$e_{\alpha}(t) \sim \exp(-t^\alpha/\Gamma(1+\alpha))$ as $t\downarrow 0$
$e_{\alpha}(t) \sim \frac{t^{-\alpha}}{\Gamma(1-\alpha)}$ as $t \rightarrow \infty$

Now consider the solution of the stochastic fractional differential equation

$V(t) = T_{\alpha}(t) v_0 + \kappa\theta \int_0^t S_{\alpha}(s) ds + \nu \int_0^t S_{\alpha}(s) \sqrt{V(s)} dw^2(s)$

with $v_0>0$.  First consider the simpler case when $\nu=0$.  We have to show

$\int \frac{ds}{T_{\alpha}(t) v_0 + \kappa\theta\int_0^t S_{\alpha}(s)} <\infty$

When $\kappa=1$,

$\int_0^t S_{\alpha}(s)ds \sim \int_0^t s^{\alpha-1} e^{-s^\alpha}ds = 1-e^{-t^\alpha}$

and

$\int_0^t \frac{1}{s^{-\alpha} v_0 + \theta(1-e^{s^{-\alpha}})}ds \le t e^{t^\alpha}/v_0 <\infty$

With the stochastic part turned on with nonzero $\nu$ we should be more careful as even for the standard Heston model positivity of $V(t)$ is guaranteed only under restriction

$2\kappa\theta \ge \nu^2$