How is this formula derived

Question

I have carried out an experiment with several repeats and my program has returned the values of EC50 and its CI(95%) confidence intervals, LogEC50 and the standard error of (LogEC50). The logs are from the Hill equation. The standard error in this case referring to curve fitting and not the SEM.

Following the Cochrane guidelines, I have calculated the SD's from the EC50 value and CI. https://handbook-5-1.cochrane.org/chapter_7/7_7_3_2_obtaining_standard_deviations_from_standard_errors_and.htm

I also have read on here (https://www.ncbi.nlm.nih.gov/books/NBK91994/) that a fitting error of an estimate (percentage wise) can be calculated from the the error of the LogEC50 by multiplying it by ln(10) * 100:

%FE(EC50)= FE(Log(EC50))*ln(10)*100

From which I assume you can infer the absolute fitting error by multiplying the EC50 by FE(Log(EC50))*ln(10).

The values I got seemed to closely allign with the SD's calculated earlier.

What I don't understand is what this formula actually does

I know Log10(EC50) can be rewritten as ln(EC50)/ln10 so I guess the ln(10)'s cancel out, but then what.

I understand that since the SE(Log(EC50)) is added and subtracted from the Log(EC50) it makes sense that we would be multiplying the absolute value by the error value (log(xy)=log(x)+log(y) and all that). I just can't seem to connect this intuition to whats actually happening in the formula provided

Thanks in advance

Answer 1

TL;DR:

• The factor $\ln(10)$ arises because changing the base of a logarithm involves multiplying by the natural logarithm of the new base (in this case, 10).
• The multiplication by (100) converts the relative error to a percentage.
• Therefore, the formula converts the error in the logarithmic domain to an error in the original domain (EC50) and expresses it as a percentage of EC50, giving a meaningful measure of variability in the original scale of measurement. This allows for a direct comparison with the standard deviation calculated earlier.

Key Concepts

EC50 and LogEC50}

• $\text{EC50}$ is the concentration of a drug that gives half-maximal response.
• $\text{LogEC50}$ is the logarithm (usually base 10) of EC50, used for fitting the Hill equation.

Standard Error and Standard Deviation

• $\text{Standard Error (SE)}$ of LogEC50 refers to the variability in the estimated LogEC50 due to curve fitting.
• $\text{Standard Deviation (SD)}$ is a measure of variability of the data points around the mean EC50.

Key Relationships

• Transforming LogEC50 to EC50: The relationship between EC50 and LogEC50 (base 10) is given by: $$ \text{LogEC50} = \log_{10}(\text{EC50}) $$ Using natural logarithms: $$ \text{LogEC50} = \frac{\ln(\text{EC50})}{\ln(10)} $$

Calculating Standard Deviation from Standard Error

The Cochrane guidelines provide a method to obtain SD from SE: $$ SD = SE \times \sqrt{n} $$ where $n$ is the number of observations or repeats.

Error Propagation from LogEC50 to EC50

To propagate the error from LogEC50 to EC50, we use the fact that: $$ \text{EC50} = 10^{\text{LogEC50}} $$ When we apply an error in LogEC50, we need to consider how this affects EC50. A small change in LogEC50 corresponds to a multiplicative change in EC50: $$ \Delta \text{EC50} \approx \text{EC50} \times \Delta (\log_{10}(\text{EC50})) \times \ln(10) $$ Here, $\Delta (\log_{10}(\text{EC50}))$ is the standard error of LogEC50.

Derivation of the Formula

To understand why $\%FE(\text{EC50}) = \text{FE}(\log(\text{EC50})) \times \ln(10) \times 100$:

1. Logarithmic Scale Conversion:

• When converting from LogEC50 to EC50, the variability (error) in LogEC50 gets scaled by a factor of $\ln(10)$ because of the derivative of the exponential function (since we are dealing with logarithms of base 10).

2. Percentage Error:

• The percentage fitting error is expressed as a percentage of the EC50 value, hence the factor of 100.

Putting it all together, the fitting error of EC50 in percentage terms is derived from the fitting error of LogEC50 scaled by $\ln(10) \times 100$: $$ \%FE(\text{EC50}) = \text{FE}(\log_{10}(\text{EC50})) \times \ln(10) \times 100 $$