4. Let the volume of a tetrahedron be $V_{1}$, and the midpoints of its edges form a convex polyhedron with volume $V_{2}$. Then $\frac{V_{2}}{V_{1}}$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) a constant, but not equal to $\frac{1}{2}$ and $\frac{2}{3}$
(D) uncertain, its value depends on the specific shape of the tetrahedron
4. (A).

As shown in Figure 2, $E, F, G, H, P$, and $Q$ are the midpoints of the respective edges, forming a convex polyhedron $P-EFGH-Q$. It is easy to see that the volumes of $A-EPH$, $B-EFQ$, $C-FGP$, and $D-GHQ$ are all $\left(\frac{1}{2}\right)^{3}$ of $V_{1}$, so,
$$
\begin{aligned}
V_{2} & =V_{1}-4\left(\frac{1}{2}\right)^{3} V_{1} \\
& =\frac{1}{2} V_{1}, \\
\text { i.e., } \frac{V_{2}}{V_{1}} & =\frac{1}{2} .
\end{aligned}
$$