---
author: Stéphane Laurent
date: '2019-11-20'
highlighter: 'pandoc-solarized'
output:
  html_document:
    highlight: kate
    keep_md: no
  md_document:
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    variant: markdown
rbloggers: yes
tags: 'maths, statistics, R'
title: 'Five-parameters logistic regression'
---

The five-parameters logistic curve is commonly defined by $$
f(x) = A + \frac{D-A}{\Bigl(1+\exp\bigl(B(C-x)\bigr)\Bigr)^S}.
$$ Assuming $B>0$ and $S>0$,

-   $A$ is the value of the horizontal asymptote when $x \to -\infty$;

-   $D$ is the value of the horizontal asymptote when $x \to +\infty$;

-   $B$ describes how rapidly the curve makes its transition between the
    two asymptotes;

-   $C$ is a location parameter, which does not have a nice
    interpretation (except if $S=1$);

-   $S$ describes the asymmetry of the curve (the curve is symmetric
    when $S=1$).

In the case when $S=1$, the parameter $C$ is the value of $x$ for which
the corresponding value $f(x)$ is the midpoint between the two
asymptotes; moreover, the curve has an inflection point at $x = C$.

In the general case, the value of $x$ for which the corresponding value
$f(x)$ is the midpoint between the two asymptotes is $$
x_{\text{mid}} = C - \frac{\log\Bigl(2^{\frac{1}{S}}-1\Bigr)}{B}.
$$ It is obtained by solving
$\Bigl(1+\exp\bigl(B(C-x)\bigr)\Bigr)^S = 2$.

``` {.r .numberLines}
n <- 100
x <- seq(49, 60, length.out = n)
A <- 30; D <- 100; B <- 1; C <- 50; S <- 10
f <- function(x) A + (D-A) / (1 + exp(B*(C-x)))^S
y0 <- f(x) 
par(mar = c(4, 4, 0.5, 1))
plot(x, y0, type = "l", cex.axis = 0.5, ylab = "f(x)")
abline(v = C, col = "green", lty = "dashed")
( xmid <- C - log(2^(1/S) - 1)/B )
## [1] 52.63424
```

``` {.r .numberLines}
abline(v = xmid, col = "red", lwd = 2) 
abline(h = (A+D)/2, col = "red", lwd = 2)
```

<img src="./figures/5pl-plot_5pl-1.png" style="display: block; margin: auto;" />

Note that the inflection point of the curve is *not* the point
correspoding to $x_{\text{mid}}$:

``` {.r .numberLines}
library(numDeriv)
df <- grad(f, x)
par(mar = c(4, 4, 0.5, 1))
plot(x, df, type = "l", cex.axis = 0.5, ylab="f'(x)")
abline(v = xmid, col = "red", lwd = 2) 
```

<img src="./figures/5pl-inflection-1.png" style="display: block; margin: auto;" />

In practice, we are often interested in estimating $x_{\text{mid}}$. So
it is better to use this other parameterization of the five-parameters
logistic curve: $$
g(x) = 
A + \frac{D-A}{{\biggl(1+\exp\Bigl(\log\bigl(2^{\frac{1}{S}}-1\bigr) + B(x_{\text{mid}}-x)\Bigr)\biggr)}^S}
$$ because fitting this curve will directly give the estimate of
$x_{\text{mid}}$ and its standard error.

Another advantage of this parameterization is that there is a way to get
a good starting value of $x_{\text{mid}}$ when one wants to fit the
five-parameters logistic regression model:

``` {.r .numberLines}
getInitial1 <- function(x, y){
  s <- getInitial(y ~ SSfpl(x, A, D, xmid, inverseB),
             data = data.frame(x = x, y = y))
  c(A = s[["A"]], 
    B = 1/s[["inverseB"]], 
    xmid = s[["xmid"]], 
    D = s[["D"]], 
    S = 1)
}
```

I don't know how to get a good starting value for $S$, so I always take
$1$.

Sometimes, `SSfpl` can fail. Here is another function which returns some
starting values:

``` {.r .numberLines}
getInitial2 <- function(x, y){
  NAs <- union(which(is.na(x)), which(is.na(y)))
  if(length(NAs)){
    x <- x[-NAs]
    y <- y[-NAs]
  }
  low_init <- min(y)
  high_init <- max(y)
  minmax <- c(which(y == low_init), which(y == high_init))
  X <- cbind(1, x[-minmax])
  Y <- log((high_init-y[-minmax])/(y[-minmax]-low_init))
  fit <- lm.fit(x = X, y = Y)
  b_init <- fit$coefficients[[2]]
  xmid_init <- -fit$coefficients[[1]] / b_init
  if(b_init < 0){
    b_init <- -b_init
    A <- low_init
    D <- high_init
  }else{
    A <- high_init
    D <- low_init
  }
  c(A = A, B = b_init, xmid = xmid_init, D = D, S = 1)
}
```

Now we wrap these two functions into a single one:

``` {.r .numberLines}
getInitial5PL <- function(x, y){
  tryCatch({
    getInitial1(x, y)
  }, error = function(e){
    getInitial2(x, y)
  })
}
```

And finally we can write a function for the fitting:

``` {.r .numberLines}
library(minpack.lm)
fit5pl <- function(x, y){
  startingValues <- getInitial5PL(x, y)
  fit <- tryCatch({
    nlsLM(
      y ~ A + (D-A)/(1 + exp(log(2^(1/S)-1) + B*(xmid-x)))^S,
      data = data.frame(x = x, y = y),
      start = startingValues,
      lower = c(-Inf, 0, -Inf, -Inf, 0),
      control = nls.lm.control(maxiter = 1024, maxfev=10000))
  }, error = function(e){
    paste0("Failure of model fitting: ", e$message)
  })
  if(class(fit) == "nls" && fit[["convInfo"]][["isConv"]]){
    fit
  }else if(class(fit) == "nls" && !fit[["convInfo"]][["isConv"]]){
    "Convergence not achieved"
  }else{ # in this case, 'fit' is the error message
    fit
  }
}
```

Let's try it on a couple of simulated samples:

``` {.r .numberLines}
set.seed(666)
nsims <- 25
epsilon <- matrix(rnorm(nsims*n, 0, 5), nrow = nsims, ncol = n)
estimates <- matrix(NA_real_, nrow = nsims, ncol = 5)
colnames(estimates) <- c("A", "B", "xmid", "D", "S")
for(i in 1:nsims){
  fit <- fit5pl(x, y0 + epsilon[i,])
  if(class(fit) == "nls"){
    estimates[i, ] <- coef(fit)
  }else{
    estimates[i, ] <- c(NaN, NaN, NaN, NaN, NaN)
  }
}
summary(estimates)
```

    ##        A               B               xmid             D         
    ##  Min.   :24.19   Min.   :0.8918   Min.   :52.52   Min.   : 98.63  
    ##  1st Qu.:27.99   1st Qu.:0.9566   1st Qu.:52.58   1st Qu.: 99.71  
    ##  Median :29.54   Median :1.0121   Median :52.64   Median :100.31  
    ##  Mean   :29.22   Mean   :1.0367   Mean   :52.63   Mean   :100.22  
    ##  3rd Qu.:30.18   3rd Qu.:1.1207   3rd Qu.:52.67   3rd Qu.:100.64  
    ##  Max.   :32.23   Max.   :1.2599   Max.   :52.76   Max.   :101.80  
    ##        S           
    ##  Min.   :   1.001  
    ##  1st Qu.:   2.262  
    ##  Median :  36.444  
    ##  Mean   :1357.639  
    ##  3rd Qu.:2003.601  
    ##  Max.   :7261.694

The estimate of $x_{\text{mid}}$ is excellent. As you can see, the
estimate of $S$ is sometimes much larger than the true value. Let's have
a look at the worst case:

``` {.r .numberLines}
i0 <- match(max(estimates[, "S"]), estimates[, "S"])
estimates[i0, ]
##            A            B         xmid            D            S 
##   29.9159582    0.8917679   52.5992848  100.0519760 7261.6944532
```

``` {.r .numberLines}
# sample
par(mar = c(4, 4, 0.5, 1))
plot(x, y0 + epsilon[i0, ], col = "yellow", cex.axis = 0.6)
# true curve
curve(A + (D-A)/(1 + exp(log(2^(1/S)-1) + B*(xmid-x)))^S, 
      add = TRUE, col = "red", lwd = 2)
# fitted curve
with(as.list(estimates[i0, ]), 
     curve(A + (D-A)/(1 + exp(log(2^(1/S)-1) + B*(xmid-x)))^S, 
           add = TRUE, col = "blue", lwd = 2, lty = "dashed")
)
```

<img src="./figures/5pl-worstcase-1.png" style="display: block; margin: auto;" />

Thus, while the estimate of $S$ is very far from the true value of $S$,
the fitted curve correctly estimates the true curve. And in such cases,
the standard error of the estimate of $S$ is big:

``` {.r .numberLines}
fit <- fit5pl(x, y0 + epsilon[i0,])
summary(fit)
```

    ## 
    ## Formula: y ~ A + (D - A)/(1 + exp(log(2^(1/S) - 1) + B * (xmid - x)))^S
    ## 
    ## Parameters:
    ##       Estimate Std. Error t value Pr(>|t|)    
    ## A    2.992e+01  1.334e+00  22.424   <2e-16 ***
    ## B    8.918e-01  7.058e-02  12.635   <2e-16 ***
    ## xmid 5.260e+01  7.310e-02 719.542   <2e-16 ***
    ## D    1.001e+02  9.347e-01 107.038   <2e-16 ***
    ## S    7.262e+03  1.757e+06   0.004    0.997    
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
    ## 
    ## Residual standard error: 5.224 on 95 degrees of freedom
    ## 
    ## Number of iterations to convergence: 27 
    ## Achieved convergence tolerance: 1.49e-08

Note that `nlsLM` provides a test of the nullity of $S$. This is not
interesting, whereas the equality $S = 1$ is of interest. So it is
better to parametrize the logistic function with $L = \log(S)$ instead
of $S$: $$
h(x) = 
A + \frac{D-A}{{\biggl(1+\exp\Bigl(\log\bigl(2^{\exp(-L)}-1\bigr) + B(x_{\text{mid}}-x)\Bigr)\biggr)}^{\exp(L)}}.
$$ In this way we can get a test of $L = 0$, that is $S = 1$.