Cramér–von Mises criterion

Remarks

With the order statistics $X_{(1)} \le \dots \le X_{(n)}$, the integral can be computed exactly: $$T = \frac{1}{12 n} + \sum_{i=1}^{n} \Big( \frac{2i - 1}{2n} - F_0(X_{(i)}) \Big)^2.$$ Under $H_0$ the law of $T$ does not depend on $F_0$, since $F_0(X_i)$ is uniform on $[0, 1]$. As $n \to \infty$ it converges to the law of $\int_0^1 B(t)^2 \, dt$ for a Brownian bridge $B$.

Example: $n = 1$, $X_1 = 0.5$ and $F_0(x) = x$ on $[0, 1]$. The integral is $\int_0^{1/2} x^2 \, dx + \int_{1/2}^{1} (1 - x)^2 \, dx = 1/24 + 1/24 = 1/12$. The formula gives the same value: $1/12 + (1/2 - 1/2)^2 = 1/12$.

Non-example: $n \int (F_n - F_0)^2 \, dx$, with Lebesgue measure in place of $dF_0$, is not the criterion. Its null law depends on $F_0$, for example on its scale, so it gives no distribution-free test.

The Anderson–Darling statistic is a weighted variant of $T$.