When the first (
n − 1) readings have been taken, the PLR slope is β
n−1. We are interested in the effect that
X n (and hence,
Y n ) has on β
n . For a pointwise linear least-squares regression, the slope is calculated by the equation
\[\left({{\sum}}\ t_{i}^{2}\ {-}\ \frac{({{\sum}}t_{i})^{2}}{n}\right)\ {\beta}_{n}{=}{{\sum}}\ t_{i}X_{i}\ {-}\ \frac{{{\sum}}t_{i}}{n}\ {{\sum}}\ X_{i}\]
Because
t i =
i in this case, and using the substitution
Y i =
X i −
c, we can simplify this to
\[\frac{1}{12}\ n(n{+}1)(n{-}1){\beta}_{n}{=}\ {{\sum}}\ iY_{i}\ {-}\ \frac{n{+}1}{2}\ {{\sum}}\ Y_{i}\]
Next, using our new Two-Omitting method, we can calculate the new PLR slope θ
n+1 from the equation
\[\frac{1}{12n}\ (n^{2}{-}1)(n^{2}{+}12){\theta}_{n{+}1}{=}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}(n{+}1)Y_{n{+}1}{-}\left(\frac{n{+}1}{2}{+}\frac{1}{n}\right)\ ({{\sum}_{i{=}1}^{n{-}1}}\ Y_{i}{+}Y_{n{+}1})\]
We are interested in the case in which
X n has been added to the series, and has made the PLR slope significant, and the clinician wants to perform a confirmation test. Therefore, we know that β
n > β
n−1 or, equivalently,
\[\frac{(n{-}2)(n{-}1)}{2}\ Y_{n}{>}3\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\]
From our definition of
Y i , given the first (
n − 1) readings, we know that the expected values
E(
Y n ) =
E(
nY n ) =
E(
Y n+1) =
E[(
n + 1)
Y n+1] = 0, and remembering that
\({{\sum}_{i{=}1}^{n{-}1}}\ Y_{i}{=}0\) , we see that when
\({{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\mathrm{{\geq}}\ 0\) \[E({\beta}_{n{+}1}\ \mathrm{given}\ Y_{1},{\ldots}Y_{n}){=}\ \frac{12}{n(n{+}1)(n{+}2)}\ \left({{\sum}_{i{=}1}^{n}}\ iY_{i}\ {-}\ \frac{n{+}2}{2}\ {{\sum}_{i{=}1}^{n}}\ Y_{i}\right)\]
\[{=}\frac{12}{n(n{+}1)(n{+}2)}\ \left({{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\ {-}\ \frac{n{+}2}{2}\ {{\sum}_{i{=}1}^{n{-}1}}\ Y_{i}{+}\left(n{-}\frac{n{+}2}{2}\right)\ Y_{n}\right)\]
\[{=}\frac{12}{n(n^{2}{-}1)}\ \left(\frac{n{-}1}{n{+}2}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}\ \frac{(n{-}2)(n{-}1)}{2(n{+}2)}\ Y_{n}\right)\]
\[{>}\frac{12}{n(n^{2}{-}1)}\ \left(\frac{n{-}1}{n{+}2}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}\ \frac{3}{n{+}2}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\right)\ {=}\frac{12n}{n^{2}(n^{2}{-}1)}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\]
\[{\geq}\frac{12n}{(n^{2}{+}12)(n^{2}{-}1)}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{=}E({\theta}_{n{+}1}{\vert}Y_{1},{\ldots}Y_{n})\]
Also, because β
n > 0 (because the PLR slope satisfies theprogression criteria), if
\({{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\) < 0 then [(
n − 1)/2]
Y n > −
\({{\sum}_{i{=}1}^{n{-}1}}\) iY i and in this case,
\[E({\beta}_{n{+}1}\ \mathrm{given}\ Y_{1},\ {\ldots}\ Y_{n}){=}\ \frac{12}{n(n{+}1)(n{+}2)}\ \left({{\sum}_{i{=}1}^{n}}\ iY_{i}{-}\ \frac{n{+}2}{2}\ {{\sum}_{i{=}1}^{n}}\ Y_{i}\right)\]
\[{=}\frac{12}{n(n{+}1)(n{+}2)}\ \left({{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{-}\ \frac{n{+}2}{2}\ {{\sum}_{i{=}1}^{n{-}1}}\ Y_{i}{+}\left(n\ {-}\ \frac{n{+}2}{2}\right)\ Y_{n}\right)\]
\[{=}\frac{12}{n(n^{2}{-}1)(n{+}2)}\ \left((n{-}1)\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}\ \frac{(n{-}2)(n{-}1)}{2}\ Y_{n}\right)\]
\[{>}\frac{12}{n(n^{2}{-}1)(n{+}2)}\ \left((n{-}1)\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{-}(n{-}2)\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\right)\]
\[{=}\frac{12n}{n^{2}(n^{2}{-}1)(n{+}2)}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{\geq}\ \frac{12n}{(n^{2}{+}12)(n^{2}{-}1)}\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\]
\[{=}E({\theta}_{n{+}1}{\vert}Y_{1},{\ldots}Y_{n})\]
Therefore, in either case, the Two-Omitting method for PLR would be expected to produce a slope closer to the actual slope for a stable eye than would the more standard Two-of-Two method.
The advantage to assuming the
Y i values to be normally distributed is that any linear combination of them is also normal. In particular, β
n+1 and θ
n+1 are, with variance
\[\mathrm{Var}\ ({\beta}_{n{+}1}){=}\ \frac{12{\varsigma}^{2}}{n(n{+}1)(n{+}2)}\ \mathrm{and\ Var}\ ({\theta}_{n{+}1}){=}\ \frac{12n{\varsigma}^{2}}{(n^{2}{+}12)(n^{2}{-}1)}\]
where ς
2 is the variance of each
Y i (assumed earlier to be constant). And so for testing the significance of the PLR slopes
\[E\left(\left|\ \frac{{\theta}_{n{+}1}}{se({\theta}_{n{+}1})}\right|\ {\vert}Y_{1},\ {\ldots}\ Y_{n}\right)\ {=}\sqrt{\left(\frac{12n}{(n^{2}{+}12)(n^{2}{-}1){\varsigma}^{2}}\right)}\left|{{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\right|\]
\[{<}\sqrt{\left(\frac{12(n{+}1)(n{+}2)}{n(n{-}1)^{2}(n{-}2)^{2}{\varsigma}^{2}}\right)}\left|{{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\right|\]
\[{=}\sqrt{\left(\frac{12}{n(n{+}1)(n{+}2){\varsigma}^{2}}\right)}\left|{{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}\ \frac{2}{(n{-}1)(n{-}2)}\ 3n\ {{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}\right|\]
\[{<}\sqrt{\left(\frac{12}{n(n{+}1)(n{+}2){\varsigma}^{2}}\right)}\left|{{\sum}_{i{=}1}^{n{-}1}}\ iY_{i}{+}nY_{n}\right|\]
\[{=}E\left(\left|\frac{{\beta}_{n{+}1}}{se({\beta}_{n{+}1})}\right|{\vert}Y_{1},{\ldots}Y_{n}\right)\]
Thus, the slope θ
n+1 would be expected to be less steep and also less significant than β
n+1, which shows that our new Two-Omitting method is more specific than the current Two-of-Two method.