Expected area of a triangle in the unit square

Solution. Write $P_i = (X_i, Y_i)$. The signed area of $\triangle P_1 P_2 P_3$ is $\tfrac12 D$, where $$D = (X_2 - X_1)(Y_3 - Y_1) - (X_3 - X_1)(Y_2 - Y_1).$$ Expanding, $D = X_2 Y_3 - X_3 Y_2 + X_3 Y_1 - X_1 Y_3 + X_1 Y_2 - X_2 Y_1$.

Let $A = X_2 - X_1$, $B = X_3 - X_1$, $C = Y_2 - Y_1$, $E = Y_3 - Y_1$. Then $D = A E - B C$. The pairs $(A, B)$ and $(C, E)$ are independent (the $X$’s are independent of the $Y$’s), and both are distributed as the difference of two uniforms centred at $0$, giving a triangular density on $[-1, 1]$.

Step 1: $\mathbb{E}[D^2]$. Since the $X$’s are independent of the $Y$’s, the pair $(A, B)$ is independent of $(C, E)$, so $$\mathbb{E}[D^2] = \mathbb{E}[A^2 E^2] + \mathbb{E}[B^2 C^2] - 2 \mathbb{E}[A B C E] = \mathbb{E}[A^2]\mathbb{E}[E^2] + \mathbb{E}[B^2]\mathbb{E}[C^2] - 2\, \mathbb{E}[A B]\, \mathbb{E}[C E].$$ Here $A = X_2 - X_1$ and $B = X_3 - X_1$ are not independent: they share the term $X_1$. With the $X_i$ i.i.d. uniform on $[0,1]$, $$\mathbb{E}[A^2] = \operatorname{Var}(X_2 - X_1) = \tfrac16, \qquad \mathbb{E}[A B] = \mathbb{E}[(X_2 - X_1)(X_3 - X_1)] = \tfrac14 - \tfrac14 - \tfrac14 + \tfrac13 = \tfrac1{12},$$ and identically $\mathbb{E}[E^2] = \mathbb{E}[C^2] = \tfrac16$, $\mathbb{E}[C E] = \tfrac1{12}$. The mixed term does not vanish. Hence $$\mathbb{E}[D^2] = \tfrac16 \cdot \tfrac16 + \tfrac16 \cdot \tfrac16 - 2 \cdot \tfrac1{12} \cdot \tfrac1{12} = \tfrac1{18} - \tfrac1{72} = \tfrac1{24},$$ so the second moment of the signed area is $\mathbb{E}[(D/2)^2] = \tfrac1{96}$.

Step 2: $\mathbb{E}[|D|]$ directly. The expected value of $|D|$ is the genuinely hard step, and we give only the answer: classical computation (Woolhouse 1867, Alikoski 1939) yields $$\mathbb{E}[|D|] = \frac{11}{72}.$$ The calculation uses the six-fold expansion of $D$, Fubini, and a patient evaluation of the resulting polynomial integrals. The final answer is $$\boxed{\mathbb{E}[\operatorname{Area}(\triangle P_1 P_2 P_3)] = \tfrac12 \mathbb{E}[|D|] = \frac{11}{144}.} \qedhere$$

import numpy as np
u, v = np.random.rand(2, 3, 10**7)
area = 0.5 * np.abs((u[1]-u[0])*(v[2]-v[0]) - (u[2]-u[0])*(v[1]-v[0]))
print(f"sim: {area.mean():.5f}   exact: {11/144:.5f}")
# sim: 0.07636   exact: 0.07639