\Phi_{P}(s,s)=I,
\end{array}
\right.$$ then we can write $$\begin{array}{l}
    r(t)=\displaystyle\Phi_{P}^{*}(T,t)(S_{T}^{*}\bar{Q}_{T}S_{T}\,{\bf 1}-S_{T}^{*}\bar{Q}_{T}-\bar{Q}_{T}S_{T})y(T)\\
    \qquad\qquad+\displaystyle\int_{t}^{T}\Phi_{P}^{*}(s,t)(S^{*}\bar{Q}S-\bar{Q}S-S^{*}\bar{Q}+P(s)\bar{A}+\bar{A}\,^{*}P(s))y(s)ds-\beta\int_{t}^{T}\Phi_{P}^{*}(s,t)\Sigma(s)db(s).
\end{array}$$ This relation implies $$\begin{split}
    &r(t)=\Phi_{P}^{*}(T,t)(S_{T}^{*}\bar{Q}_{T}S_{T}\,{\bf 1}-S_{T}^{*}\bar{Q}_{T}-\bar{Q}_{T}S_{T}) \mathbb{E} ^{\mathcal{B}^{t}}y(T)\label{eq:5.29}\\
    &\qquad\qquad+\int_{t}^{T}\Phi_{P}^{*}(s,t)(S^{*}\bar{Q}S-\bar{Q}S-S^{*}\bar{Q}+P(s)\bar{A}+\bar{A}\,^{*}P(s)) \mathbb{E} ^{\mathcal{B}^{t}}y(s)ds.
\end{split}$$ In the system (\[eq:5.28\]), (\[eq:5.29\]) the external function $\Sigma(s)$ does not appear anymore, but $r(t)$ is the solution of an integral equation, instead of an backward SDE. Finally, from (\[add\_1\]), by taking differentiation and then integrating from $t$ to $T$, we have $$\begin{aligned}
s(t) & =\frac{1}{2}y(T)^{*}S_{T}^{*}\bar{Q}_{T}S_{T}y(T)+\int_{t}^{T}(\frac{1}{2}\text{tr}\, aP(s)+\frac{\beta^{2}}{2}\text{tr}P(s)+\beta^{2}\text{tr }\Sigma(s))ds \nonumber \\
&\qquad+ \int_{t}^{T}(\frac{1}{2}y(s)^{*}S^{*}QSy(s)-\frac{1}{2}r(s)^{*}BR^{-1}B^{*}r(s)+r(s)^{*}\bar{A}y(s))ds-\beta\int_{t}^{T}y(s)^{*}\Gamma(s)db(s). \label{eq:5.30-1}\end{aligned}$$

Since $s(t)$ is adapted to $\mathcal{B}^{t}$, we can write $$\begin{aligned}
s(t) & =\int_{t}^{T}(\frac{1}{2}\text{tr}\, aP(s)+\frac{\beta^{2}}{2}\text{tr}P(s)+\beta^{2}\text{tr }\Sigma(s))ds+\frac{1}{2} \mathbb{E} ^{\mathcal{B}^{t}}y(T)^{*}S_{T}^{*}\bar{Q}_{T}S_{T}y(T)\label{eq:5.31-1}\\
&\qquad+  \mathbb{E} ^{\mathcal{B}^{t}}\int_{t}^{T}(\frac{1}{2}y(s)^{*}S^{*}QSy(s)-\frac{1}{2}r(s)^{*}BR^{-1}B^{*}r(s)+r(s)^{*}\bar{A}y(s))ds.\nonumber \end{aligned}$$

The results obtained contains that in Bensoussan et al. [@BSYY] as a special case when $\beta=0$.

MEAN FIELD GAMES MASTER EQUATION 
---------------------------------

The Master equation (\[eq:4.7\]) under the LQ setting is $$\label{eq:5.31}
\left\{
\begin{array}{l}
    -\displaystyle\frac{\partial U}{\partial t}-\frac{1}{2}\text{tr}\, aD^{2}U-\frac{1}{2}\beta^{2}\Delta U\\
    \qquad-\displaystyle\int_{\mathbb{R}^{n}}\Big(\frac{1}{2}\text{tr}\, aD_{\xi}^{2}\frac{\partial}{\partial m}U(x,m,t)(\xi)+\frac{1}{2}\beta^{2}\Delta_{\xi}\frac{\partial}{\partial m}U(x,m,t)(\xi)\Big)m(\xi)d\xi\\
    \qquad -\displaystyle\int_{\mathbb{R}^{n}}D_{\xi}\frac{\partial}{\partial m}U(x,m,t)(\xi).G(\xi,m,DU(\xi,m,t))m(\xi)d\xi\\
    \qquad-\displaystyle\frac{1}{2}\beta^{2}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\sum_{i=1}^{n}\frac{\partial}{\partial\xi_{i}}\frac{\partial}{\partial\eta_{i}}(\frac{\partial^{2}U(x,m,t)}{\partial m^{2}}(\xi,\eta))m(\xi)m(\eta)d\xi d\eta-\beta^{2}\displaystyle\sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}\left(\int_{\mathbb{R}^{n}}\frac{\partial}{\partial\xi_{i}}\frac{\partial U(x,m,t)}{\partial m}(\xi)m(\xi)d\xi\right)\\
    =\displaystyle\frac{1}{2}x^{*}(Q+\bar{Q})x-x^{*}\bar{Q}Sy+\frac{1}{2}y^{*}S^{*}\bar{Q}Sy\\
    \qquad -\displaystyle\frac{1}{2}(DU(x,m,t))^{*}BR^{-1}B^{*}DU(x,m,t)+(DU(x,m,t))^{*}(Ax+\bar{A}y),\\
    U(x,m,T)=\displaystyle\frac{1}{2}x^{*}(Q_{T}+\bar{Q}_{T})x-x^{*}\bar{Q}_{T}S_{T}y+\frac{1}{2}y^{*}S_{T}^{*}\bar{Q}_{T}S_{T}y.
\end{array}
\right.$$

We look for a solution of the form $$U(x,m,t)=\frac{1}{2}x^{*}P(t)x+x^{*}\Sigma(t,m_{1})y+\frac{1}{2}y^{*}\Gamma(t,m_{1})y+\mu(t,m_{1}).\label{eq:5.33}$$

Using the terminal condition, we cannot have a symmetric $\Sigma(t,m_{1})$ this time. We can also compute the derivatives $$\begin{split}
    DU(x,m,t)&=P(t)x+\Sigma(t,m_{1})y,\\
    D^2U(x,m,t)&=P(t),\\
    \frac{\partial}{\partial m}U(x,m,t)(\xi)&=x^{*}\Sigma(t,m_{1})\xi+y^{*}\Gamma(t,m_{1})\xi+x^{*}\frac{\partial}{\partial m_{1}}\Sigma(t,m_{1})y\\
    &\qquad+\frac{1}{2}y^{*}\frac{\partial}{\partial m_{1}}\Gamma(t,m_{1})y+\frac{\partial}{\partial m_{1}}\mu(t,m_{1}),\\
    D_{\xi}\frac{\partial}{\partial m}U(x,m,t)(\xi)&=\Sigma^{*}(t,m_{1})x+\Gamma(t,m_{1})y,\\
    D_{\xi}^{2}\frac{\partial}{\partial m}U(x,m,t)(\xi)&=0,\\
    \frac{\partial^{2}}{\partial m^{2}}U(x,m,t)(\xi,\eta)&=(x^{*}\frac{\partial}{\partial m_{1}}\Sigma(t,m_{1})+y^{*}\frac{\partial}{\partial m_{1}}\Gamma(t,m_{1}))(\xi+\eta)\\
    