7. Given the ellipse $\frac{x^{2}}{4}+y^{2}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, and $M$ as a point on the ellipse, and $I$ as the incenter of $\triangle M F_{1} F_{2}$. Then the value of $\overrightarrow{O I} \cdot \overrightarrow{O M}$ is $\qquad$ .
7. $2 \sqrt{2}-2$.

Given that $I$ is the incenter of $\triangle M F_{1} F_{2}$, we have
$$
\begin{array}{l}
\frac{\overrightarrow{M I} \cdot \overrightarrow{M F_{2}}}{\left|\overrightarrow{M F_{2}}\right|}=\frac{|\overrightarrow{M I}|\left|\overrightarrow{M F_{2}}\right| \cos \frac{1}{2} \angle F_{1} M F_{2}}{\left|\overrightarrow{M F_{2}}\right|} \\
=|\overrightarrow{M I}| \cos \frac{1}{2} \angle F_{1} M F_{2} .
\end{array}
$$

Let the incircle of $\triangle M F_{1} F_{2}$ touch the sides $M F_{1}$, $M F_{2}$, and $F_{1} F_{2}$ at points $D$, $E$, and $H$ respectively. Then
$$
\begin{aligned}
& F_{1} D=F_{1} H, H F_{2}=F_{2} E, \\
& |M D|=|M E| \\
& =\frac{1}{2}\left(\left|M F_{1}\right|+\left|M F_{2}\right|-\left|F_{1} F_{2}\right|\right) . \\
& \text { Also, }|M D|=|M I| \cos \frac{\angle F_{1} M F_{2}}{2}, \\
& \left|M F_{1}\right|+\left|M F_{2}\right|-\left|F_{1} F_{2}\right| \\
& =2|M I| \cos \frac{\angle F_{1} M F_{2}}{2} \\
\Rightarrow & |M I| \cos \frac{\angle F_{1} M F_{2}}{2} \\
= & \frac{1}{2}\left(\left|M F_{1}\right|+\left|M F_{2}\right|-\left|F_{1} F_{2}\right|\right) \\
= & \frac{1}{2}(2 \sqrt{2}-2)=\sqrt{2}-1 \\
\Rightarrow & \overrightarrow{M O} \cdot \overrightarrow{M I}=\frac{1}{2} \overrightarrow{M I} \cdot\left(\overrightarrow{M F_{1}}+\overrightarrow{M F_{2}}\right) \\
= & \frac{1}{2}\left(\overrightarrow{M I} \cdot \overrightarrow{M F_{1}}+\overrightarrow{M I} \cdot \overrightarrow{M F_{2}}\right) \\
= & \frac{1}{2}\left(\left|\overrightarrow{M F_{1}}\right|(\sqrt{2}-1)+\left|\overrightarrow{M F_{2}}\right|(\sqrt{2}-1)\right) \\
= & 2 \sqrt{2}-2 .
\end{aligned}
$$