[R] Plotting confidence bands around regression line

[Frank Harrell]

Frank E Harrell Jr   Professor and Chairman        School of Medicine
                      Department of Biostatistics   Vanderbilt University

On Wed, 11 Aug 2010, Michal Figurski wrote:

> Peter, Frank, David and others,
>
> Thank you all for your ideas. I understand your lack of trust in P&B
> method. Setting that aside (it's beyond me anyways), please see below
> what I have finally came up with to calculate the CI boundaries. Given
> the slope and intercept with their 05% & 95% CIs, and a range of x = 1:50 :
>
>    ints    = c(-1, 0, 1)      # c(int05%, int, int95%)
>    slos    = c(0.9, 1, 1.1)   # c(slo05%, slo, slo95%)
>    CIs     = data.frame(x=1:50, lo=NA, hi=NA)
>    for (i in 1:50) {
>        CIs$lo[i] = min(ints + slos * CIs$x[i])
>        CIs$hi[i] = max(ints + slos * CIs$x[i])
>        }
>
> It looks like it works to me. Does it make sense?

Doesn't look like it takes the correlation of slope and intercept into 
account but I may not understand.

>
> Now, what about a 4-parameter 'nls' model? Do you guys think I could use
> the same approach?

This problem seems to cry out for one of the many available robust 
regression methods in R.

Frank

>
> Best regards,
>
> --
> Michal J. Figurski, PhD
> HUP, Pathology & Laboratory Medicine
> Biomarker Research Laboratory
> 3400 Spruce St. 7 Maloney
> Philadelphia, PA 19104
> tel. (215) 662-3413
>
> On 2010-08-10 13:12, Peter Dalgaard wrote:
>> Michal Figurski wrote:
>>
>>> # And the result of the Passing-Bablok regression on this data frame:
>>>              Estimate      5%CI     95%CI
>>> Intercept -4.306197 -9.948438 -1.374663
>>> Slope      1.257584  1.052696  1.679290
>>>
>>> The original Passing&  Bablok article on this method has an easy
>>> prescription for CIs on coefficients, so I implemented that. Now I need
>>> a way to calculate CI boundaries for individual points - this may be a
>>> basic handbook stuff - I just don't know it (I'm not a statistician).
>>
>> The answer is that you can't. You can't even do it with ordinary linear
>> regression without knowing the correlation between slope and intercept.
>> However, if you can get a CI for the intercept then you could subtract
>> x0 from all the x and get a CI for the value at x0.
>>
>> (This brings echos from a distant past. My master's thesis was about
>> some similar median-type estimators. I can't remember whether I looked
>> at the Passing-Bablok paper at the time (1985!!) but my general
>> recollection is that this group of methods is littered with unstated
>> assumptions.)
>>
>