4. Given a regular tetrahedron $P-ABC$ with a base edge length of 1 and a height of $\sqrt{2}$. Then its inscribed sphere radius is $\qquad$ .
4. $\frac{\sqrt{2}}{6}$.

As shown in Figure 2, let the projections of the center of the sphere $O$ onto the planes $ABC$ and $ABP$ be $H$ and $K$ respectively, and let the midpoint of side $AB$ be $M$. The inradius of the sphere is $r$. Then, $P, K, M$ and $P, O, H$ are collinear, $\angle PHM = \angle PKO = \frac{\pi}{2}$, and
$OH = OK = r$,
$PO = PH - OH$
$= \sqrt{2} - r$,
$MH = \frac{\sqrt{3}}{6} AB = \frac{\sqrt{3}}{6}$,
$PM = \sqrt{MH^2 + PH^2} = \sqrt{\frac{1}{12} + 2} = \frac{5 \sqrt{3}}{6}$.
Thus, $\frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \frac{MH}{PM} = \frac{1}{5}$.
Solving for $r$ gives $r = \frac{\sqrt{2}}{6}$.