is from Lemma \[lemma list\](iv) and the integration by parts, the second equality is from Lemma \[L:2.3\] and the third equality is from Lemma \[lemma list\](v). Therefore $u(t,x)$ is equal to $$\begin{aligned}
&&D_t^a v(t,x)+ \int^t_0\frac{\partial}{\partial t}\int_{{\mathbb{R}}^d}
p(t-s,x-y) D_t^a v (s,y) dyds\\
&=&\frac{\partial}{\partial t}\int^t_0\int_{{\mathbb{R}}^d} p(t-s,x-y) D_t^a v
(s,y) dyds\\
&=&\frac{1}{\Gamma(1-a)}\sum_{k=1}^\infty\frac{\partial}{\partial t}\int^{t}_0\int_{{\mathbb{R}}^d} p(t-s,x-y)
\int^s_0(s-r)^{-a}g^k(r,y)dW^k_r dyds\\
&=&\frac{1}{\Gamma(1-a)}\sum_{k=1}^\infty\frac{\partial}{\partial t}\int^{t}_0\int_{{\mathbb{R}}^d} p(s,x-y)
\int^{t-s}_0(t-s-r)^{-a}g^k(r,y)dW^k_r dyds.\end{aligned}$$

Hence it is enough to prove $$\begin{aligned}
          \nonumber
&&\frac{1}{\Gamma(1-a)}\sum_{k=1}^\infty \int^{t}_0\int_{{\mathbb{R}}^d} p(s,x-y) \int^{t-s}_0(t-s-r)^{-a}g^k(r,y)dW^k_r
dyds\\
& =&\sum_{k=1}^\infty\int^t_0 \int^s_0 \int_{{\mathbb{R}}^d} D_t^ap (s-r,x-y)g^k(r,y)dy dW^k_r ds.
\label{imp}\end{aligned}$$ The latter equals $$\begin{aligned}
&&\sum_{k=1}^\infty\int^t_0 \int_{{\mathbb{R}}^d}\int^{t-r}_0  D_t^ap (s,x-y) ds g^k(r,y)dy dW^k_r \nonumber\\
&=&\sum_{k=1}^\infty\int^t_0 \int_{{\mathbb{R}}^d}I^{1-a}p (t-r,x-y) g^k(r,y)dy dW^k_r \label{new 10}\\
&=&\frac{1}{\Gamma(1-a)}\sum_{k=1}^\infty\int^t_0\int_{{\mathbb{R}}^d}\int^{t-r}_0(t-r-s)^{-a}p(s,x-y)ds g^k(r,y) dy dW^k_r.
\nonumber\end{aligned}$$ For (\[new 10\]) above we used the fact that $\int^t_0 D_t^a p ds=I^{1-a}p$. We thus get using the stochastic Fubini’s theorem (see [@Kr11]).

We now consider the case $\beta\geq \gamma$. Put $a=\beta-\gamma$ and define $$v(t,x)=\int^t_0 g^k(s,x)dW^k_s, \quad w(t,x)=I^{a}v(t,x)=\frac{1}{a\Gamma(a)}\int^{t}_0
(t-s)^a g^k(s,x)dW^k_s.$$ From this point on it is enough to repeat the case $\beta\leq \gamma$. Indeed, following the previous steps, we get $$u(t,x)=\frac{1}{a\Gamma(a)}\sum_{k=1}^\infty\frac{\partial}{\partial t}\int^{t}_0\int_{{\mathbb{R}}^d} p(s,x-y)
\int^{t-s}_0(t-s-r)^{a}g^k(r,y)dW^k_r dyds.$$ Note that $$\begin{aligned}
&&\sum_{k=1}^\infty\int^t_0 \int^s_0 \int_{{\mathbb{R}}^d} I^ap (s-r,x-y)g^k(r,y)dy dW^k_r ds\\
&=&\sum_{k=1}^\infty\int^t_0 \int^{t-r}_0 \int_{{\mathbb{R}}^d} I^ap (s,x-y)g^k(r,y)dy dsdW^k_r\\
&=&\frac{1}{a\Gamma(a)}\sum_{k=1}^\infty\int^t_0 \int^{t-r}_0 \int_{{\mathbb{R}}^d} (t-s-r)^ap (r,x-y)g^k(r,y)dy
dsdW^k_r.\end{aligned}$$ This clearly proves the case $\gamma\geq \beta$.

On the 