$f\in {\mathbb{L}}_2(T)$, $g\in {\mathbb{H}}^{\infty}_0(T,\ell_2)$ and $u\in {\mathbb{H}}^2_2(T)$. Then $u$ satisfies (\[time-space\]) with initial data $u_0\in U^1_2$ in the sense distributions (see Definition \[def 3.23\]) if and only if $$\begin{aligned}
u(t,x)&=&\int_{{\mathbb{R}}^d}p(t,x-y)u_0(y)dy+\int^t_0\int_{{\mathbb{R}}^d} q(t-s,x-y)f(s,y)dyds    \nonumber\\
&+&\sum_{k=1}^\infty \int^t_0 \int_{{\mathbb{R}}^d}P_{\beta,\gamma}(t-s,x-y)g^k(s,y)dydW^k_s.     \label{right term}\end{aligned}$$

Suppose $u$ satisfies (\[time-space\]). Recall that for the solution of the (deterministic) equation $$\partial^{\beta}_t\bar{u}=\Delta \bar{u}+f, \quad u(0)=u_0$$ is given by the formula $$\label{eqn deterministic}
\bar{u}(t,x):=\int_{{\mathbb{R}}^d}p(t,x-y)u_0(y)dy+ \int^t_0\int_{{\mathbb{R}}^d} q(t-s,x-y)f(s,y)dyds.$$ In fact, in [@EIK Section 5.2] the representation (\[eqn deterministic\]) is proved for sufficiently smooth $f$. This and the estimate of the solution obtained in [@Za Theorem 3.1] allow us to use an approximation (see the proof of Theorem \[thm laplace\]) and get (\[eqn deterministic\]) for general $f\in {\mathbb{L}}_2(T)$.

Thus by considering $u-\bar{u}$, where $\bar{u}$ is defined above, we may assume without loss of generality that $u_0=0$ and $f=0$. Suppose first $\beta \leq \gamma$. Set $a=\gamma-\beta$, $$v(t,x):=\sum_{k=1}^\infty\int^t_0 g^k(s,y)dW^k_s,$$ and $$w(t,x):=D^{a}_tv(t,x)=\frac{1}{\Gamma(1-a)}\sum_{k=1}^\infty\int^t_0(t-s)^{-a}g^k(s,x)dW^k_s.$$ Then $u-w$ satisfies the following fractional diffusion equation $$\partial^{\beta}_t(u-w)=\Delta u=\Delta (u-w)+ \Delta w, \quad \quad (u-v)(0)=0.$$ Thus by (\[eqn deterministic\]) with $\Delta w$ in place of $f$, we have $$u(t,x)=
w(t, x)
+\int^t_0\int_{{\mathbb{R}}^d}q(t-s,x-y)\Delta w(s,y)dyds.$$ Nota that for any $s<t$ $$\begin{aligned}
\int_{{\mathbb{R}}^d}q(t-s,x-y)\Delta w(s,y)dy
&=&\int_{{\mathbb{R}}^d} \Delta q(t-s,x-y) D_t^a v
(s,y) dy\\&=&\int_{{\mathbb{R}}^d}\frac{\partial}{\partial t}p(t-s,x-y)D^a_tv(s,y)dy\\
&=&\frac{\partial}{\partial t}\int_{{\mathbb{R}}^d}p(t-s,x-y)D^a_tv(s,y)dy.\end{aligned}$$ Indeed, the first equality 