Bioequivalence and Bioavailability Forum

Hi Shuanghe,

diving into the formulas. Ugly as shit in the ICH M9 guideline$$\text{f2}=50\phantom{.}\tiny{\bullet}\normalsize\phantom{.}\log\{\text{[}1+(1/\text{n}){\Sigma_{\text{t=1}}}^\text{n}(\text{R}_\text{t}-\text{T}_\text{t})^2\text{]}^{-0.5}\phantom{.}\tiny{\bullet}\normalsize\phantom{.}100\}\tag{1}$$ which we can beautify to your $$f_2=50\cdot\log_{10}\frac{100}{\sqrt{1+\frac{1}{n}\sum_{t=1}^{t=n}(\text{R}_t-\text{T}_t)^2}}\tag{2}$$ Thanks a lot! I screwed up in my post above because I had overlooked that the exponent of the term in square brackets of \((1)\) is –\(0.5\) and not \(0.5\)… If we don’t trust algebra, we can compare the formulas by brute force.

f2 <- function(R, T, formula) {
  if (!length(R) == length(T))
    stop ("R and T vectors must have the same length.")
  if (missing(formula)) formula <- 2 # default
  if (!formula %in% 1:2)
    stop ("formula must be 1 or 2.")
  ms <- mean((R - T)^2) # instead of the clumsy formula
  if (formula == 1) {
    x <- 50 * log10((1 + ms)^-0.5 * 100)
  } else {
    x <- 50 * log10(100 / sqrt(1 + ms))
  }
  return(x)
}

n       <- 20
mu_R    <- mu_T    <- 100
sigma_R <- sigma_T <- 20
# is the package "truncnorm" installed?
inst    <- installed.packages()["truncnorm", "Package"]
if (length(inst) == 1) library(truncnorm)  # if yes, attach it
set.seed(123456)                           # for reproducibility
if (length(inst) == 1) { # truncated normal distribution
  R <- sort(rtruncnorm(n = n, a = 0, b = Inf,
                       mean = mu_R, sd = sigma_R))
  T <- sort(rtruncnorm(n = n, a = 0, b = Inf,
                       mean = mu_T, sd = sigma_T))
} else {                 # workaround to avoid negative values
  R <- sort(abs(rnorm(n = n, mean = mu_R, sd = sigma_R)))
  T <- sort(abs(rnorm(n = n, mean = mu_T, sd = sigma_T)))
}
f2_1 <- f2(R, T, 1)
f2_2 <- f2(R, T, 2)
cat(n, "sorted random values of R and T",
    "\nmean (R) =", sprintf("%8.3f", mean(R)),
    "range:", paste(signif(signif(range(R), 5)), collapse = " – "),
    "\nmean (T) =", sprintf("%8.3f", mean(T)),
    "range:", paste(signif(signif(range(T), 5)), collapse = " – "),
    "\n  f2 (1) =", sprintf("%.4g", f2_1),
    "\n  f2 (2) =", sprintf("%.4g", f2_2),
    "\nAre the results by the two formulas identical?",
    isTRUE(all.equal(f2_1, f2_2)), "\n")

20 sorted random values of R and T
mean (R) =  110.477 range: 77.721 – 150.05
mean (T) =   97.769 range: 45.023 – 131.2
  f2 (1) = 42.37
  f2 (2) = 42.37
Are the results by the two formulas identical? TRUE

Try very different values:

mu_R    <- 100
sigma_R <- 20
mu_T    <- 10
sigma_T <- 2
...
20 sorted random values of R and T
mean (R) =  110.477 range: 77.721 – 150.05
mean (T) =    9.777 range: 4.5023 – 13.12
  f2 (1) = -0.4946
  f2 (2) = -0.4946
Are the results by the two formulas identical? TRUE

Bingo, negative \(\small{f_2}\)!

❝ An example: if the average difference is 100%, then you'll have $$f_2=50\,\log_{10} \frac{100}{\sqrt{1+\frac{1}{n}\sum_{i=1}^{i=n}100^2}} = -0.001.$$

❝ With average difference of 110%, \(f_2=-2.07\).

IMHO, if you have the average difference, you should not square it. You need the average of squared difference. Then:

f2_ms <- function(ms) {
  if (ms < 0) stop ("ms must not be negative.")
  50 * log10(100 / sqrt(1 + ms))
}
ms <- c(0, 99, 100, 110, 99^2, 100^2, 110^2)
for (j in seq_along(ms)) {
  cat(sprintf("Mean sq. diff. = %5.0f, f2 = %+8.3f\n",
              ms[j], f2_ms(ms[j])))
}

Mean sq. diff. =     0, f2 = +100.000
Mean sq. diff. =    99, f2 =  +50.000
Mean sq. diff. =   100, f2 =  +49.892
Mean sq. diff. =   110, f2 =  +48.867
Mean sq. diff. =  9801, f2 =   +0.217
Mean sq. diff. = 10000, f2 =   -0.001
Mean sq. diff. = 12100, f2 =   -2.071

Of course, I can be wrong. Given all that, I think that \(\small{f_2}\) is not useful for comparing (plasma) concentration profiles.