Two babies are born on a moving train; one is in the compartment...

Let the train frame be $S'$ moving with velocity $v$ relative to the ground frame $S$ in the positive $x$-direction.

Let the position of the compartment adjacent to the guard (rear of the train) be $x'_{1} = 0$ and the compartment adjacent to the engine (front of the train) be $x'_{2} = L$, where $L$ is the proper length of the train.

According to the train frame, the babies are born at the exact same instant, so $t'_{1} = t'_{2} = 0$.

Using the Lorentz transformation for time, the time of an event in the ground frame is given by:
t = \gamma \left(t' +\dfrac{v x'}{c^2}\right)

For the baby born adjacent to the guard:
$t_{1} = \gamma \left(0 + 0\right) = 0$

For the baby born adjacent to the engine:
t_{2} = \gamma \left(0 +\dfrac{v L}{c^2}\right) = \dfrac{\gamma v L}{c^2}

Since $t_{2} > t_{1}$, the birth of the baby adjacent to the engine occurs later in the ground frame.

Because the baby adjacent to the guard is born earlier in the ground frame, this baby is elder according to the observers on the ground.