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I have 2 populations with their survival at n years and their 95% confidence interval.

Under the assumption of exponential distribution, the hazard ratio can be expressed as:

$HR_{AB} = \frac{log(S_A)}{log(S_B)} $

However, how to infer the confidence interval of the HR (or $log(HR)$) from these data?

If there is no analytical solution, could a bootstrap procedure be an alternative?

PS: this would be a very exploratory analysis, hence assumption on the distribution is not highly significant.

EDIT - Let's consider these notations:

  • $p_A [95\%CI :p_{A_{inf}} ; p_{A_{sup}}]$ and $p_B [95\%CI :p_{B_{inf}} ; p_{B_{sup}}]$at time $t$
  • measured on respectively $n_A$ and $n_B$ individuals
  • with $\Delta_A = p_{A_{sup}}-p_{A_{inf}}$ and $\Delta_B = p_{B_{sup}}-p_{B_{inf}}$
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With an exponential distribution, $\log S_A(t)= -\lambda_A t$ and $\log S_B(t)=-\lambda_B t$. The hazard ratio ($\text{HR}$) you request is:

$$ \frac{\log S_A}{\log S_B} = \frac{\lambda_A}{\lambda_B},$$

and the $\log \text{HR}$ is:

$$ \log \text{HR} = \log \lambda_A - \log \lambda_B.$$

The confidence interval (CI) for the $\text{HR}$ is related to the variance of the its estimate. The formula for the variance of a difference gives:

$$\text{Var} (\log \text{HR }) = \text{Var}(\log \lambda_A) +\text{Var} (\log \lambda_B) - 2 \text{ Cov}(\log \lambda_A,\log \lambda_B). $$

If the estimates of $\lambda_A$ and $\lambda_B$ were obtained independently, then the last covariance term is zero so all you need are the variances of the individual $\lambda$ estimates.

What you have are survival probabilities $p_A$ and $p_B$ and their 95% CI at some time $t$. A survival probability $p$ is reached at $t= -\frac{\log p}{\lambda}$, $\lambda = - \frac{\log p}{t}$ or $$\log \lambda = \log(-\log p) - \log t.$$

Thus $\text{Var}(\log \lambda) = \text{Var}(\log(-\log p))$.

That result suggests converting the CI on the survival probabilities $p$ to CI on $\log(-\log p)$ to get estimates for the CI of the individual $\lambda$ estimates. If those latter CI estimates are sufficiently symmetrical about the point estimates, you can then use a normal approximation, dividing the half-widths of the 95% CI by 1.96 to estimate the standard deviations of the $\lambda$ estimates and then squaring to get their estimated variances. Adding those variances, taking the square root, and multiplying by 1.96 then give estimated CI half-widths for $\log \text{HR}$.

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  • $\begingroup$ great, perfect answer as usual! $\endgroup$ Commented May 29, 2021 at 18:39
  • $\begingroup$ Just a little question: when you say "dividing the half-widths of the 95% CI by 1.96", are your referring to $p_{A_{inf}}$ or to $log(-log(p_{A_{inf}}))$? $\endgroup$ Commented May 31, 2021 at 9:41
  • $\begingroup$ Also, shouldn't the formula take $n_A$ and $n_B$ into account? $\endgroup$ Commented May 31, 2021 at 15:49
  • $\begingroup$ @DanChaltiel $n_a$ and $n_b$ are implicitly taken into account in the original CI. More cases, narrower CI. The way I formulated the answer, I was referring to half-widths on the $\log (-\log p)$ scale, as the variance on that scale is what's most directly related to the variance of $\log \text{HR}$. $\endgroup$
    – EdM
    Commented May 31, 2021 at 16:00

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