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By continuity of the cumulative distribution function and nonnegativity, $F(0)=0$; strict increase and $F(x)\to 1$ allow inversion for every $0<t<1$. The probability integral transform follows fromThus $F(X)$ is uniformly distributed on $(0,1)$.

The order statistics arrange the sample increasingly: $X_{(1)}\le \cdots \le X_{(n)}$. Strict monotonicity means $F(X_{(i)})$ is the $i$th order statistic of $n$ independent uniform variables. Its density isobtained by choosing the $i-1$ observations below $t$ and the $n-i$ above it. This is the beta distribution with parameters $i,n+1-i$, soFor the given log-logistic distribution, rearrangement givesConsequently plot $y_i=\log (i/(n+1-i))$ vertically against $x_i^{\ast }=\log X_{(i)}$ horizontally. Under a good fit, replacing $F(X_{(i)})$ by its plotting position $i/(n+1)$ gives an approximate straight line withThis is a diagnostic probability plot, rather than an assertion that the expectation of the nonlinear transformed order statistic equals that transformed plotting position. Positive observations occur almost surely under the model.

A probability plot compares order statistics with plotting positions or model quantiles to diagnose whether a proposed probability distribution describes data. If the model can be transformed into a linear identity $T(F(x))=a+bS(x)$, plot $T(i/(n+1))$ against $S(X_{(i)})$. Approximate linearity supports that model's shape; systematic curvature suggests a mismatch.

The plotting position $i/(n+1)$ comes from the exact mean of the $i$th uniform order statistic, by the probability integral transform. It does not assert equality between the expectation of a nonlinear transformation and that transformation of the expectation. Fitted slope and intercept can estimate model parameters, but the visual plot alone is not a calibrated hypothesis test.