Past exam of the options II course of the University of Oxford / 2012 / O / O4 / c

For the shifted exponential model, $F(x)=1-e^{-\lambda (x-1)}$ for $x\ge 1$. It is increasing, so the probability integral transform preserves order: $F(X_{(r)})=Y_{(r)}$. The expected uniform plotting positions are $r/(n+1)$ by part (b). Replacing the random transformed order statistic by this representative position givesThus plotting $x_{(r)}$ against $t_r=-\log (1-r/(n+1))$ should give an approximate line with intercept one and slope $\boxed{1/\lambda }$.

The replacement is a quantile-plot approximation, not an assertion that expectation commutes exactly with the nonlinear inverse distribution function. The random points need not lie exactly on the line.

For the Pareto model, integration gives $F(x)=1-x^{-\lambda }$ on $x\ge 1$. The same plotting positions suggest $1-x_{(r)}^{-\lambda }\approx r/(n+1)$. Taking logarithms givesTherefore plot $\log x_{(r)}$ vertically against $t_r=-\log (1-r/(n+1))$ horizontally. A reasonable fit should be approximately a straight line through the origin, with slope $\boxed{1/\lambda }$. This logarithmic transformation distinguishes the Pareto tail model from the shifted exponential plot in part (i).